Selina Concise Class 10 Physics Solution Chapter No. 8 – ‘Current Electricity’ For ICSE Board Students.
Selina Concise Class 10 Physics Chapter 8 Current Electricity Exercise All Questions and Answers by Physics Teacher here in this post.
Exercise 8 (A)
Question: 1
Define the term current and state its S.I. unit.
Solution: The rate of flow charge is current. Ampere is the S.I. unit of current.
Question: 2
Define the term electric potential. State it’s S.I. unit.
Solution: The potential at a point is defined as the amount of work done per unit charge in bringing s positive test charge from infinity to the point. Volt is the S.I. unit of electric potential.
Question: 3
How is the electric potential difference between the two points defined? State its S.I. unit.
Solution: The potential difference between two point is equal to the work done per unit charge in moving a positive test charge from one point to the other.Joule per coulomb is the S.I unit potential difference.
Question: 4
Explain the statement ‘the potential difference between two points is 1 volt’.
Solution: The potential difference between two points is said to be 1 volt if the work done in moving 1 coulomb charge from 1 point to the other is 1 joule.
Question: 5
(a.) State whether the current is a scalar or vector? What does the direction of current convey?
(b.) State whether the potential is a scalar or vector? What does the positive and negative sign of potential convey?
Solution:
a.) Current is the scalar quantity. The direction of flow of current is opposite to the direction of electrons.
b.) Potential is the scalar quantity. The work done due to attractive force is negative and work done due to repulsive force is positive.
Question: 6
Define the term resistance. State its S.I. unit.
Solution: The obstruction offered to the flow of current by the conductor is called its resistance. Ohm is the S.I. unit of resistance.
Question: 7
(a) Name the particles which are responsible for the flow of current in a metallic wire.
(b) Explain the flow of current in a metallic wire on the basis of movement of the particles named by you above in part (a).
(c) What is the cause of resistance offered by the metallic wire in the flow of current through it?
Solution:
a.) Free electrons in the metallic wire are responsible for the flow of current.
b.) The free electrons available in the metallic wire will carry current. N number of electrons pass through time t.
Charge = N x charge on electrons.
c.) A metal is electrically neutral. Hence it contains both positive and negative charge. The flow of free electrons is repelled by the positive charge in the metal. Hence, the resistance is offered by the metallic wire in the flow of current through.
Question: 8
State Ohm’s law and draw a neat labelled circuit diagram containing a battery, a key, a voltmeter, an ammeter, a rheostat and an unknown resistance to verify it.
Solutions: The current flowing in a conductor is directly proportional to the potential difference applied across its ends provided that the physical conditions and the temperature of the conductor remain constant, this is ohm’s law.
Question: 9
(a.) Name and state the law which relates the potential difference and current in a conductor.
(b.) What is the necessary condition for a conductor to obey the law named above in part (a)?
Solution:
a.) The current flowing in a conductor is directly proportional to the potential difference applied across its ends provided that the physical conditions and the temperature of the conductor remain constant, this is ohm’s law.
b.) The condition for the ohm’s law is that the temperature of the conductor shouldremain constant.
Question: 10
(a.) Draw a V-I graph for a conductor obeying Ohm’s law.
(b.) What does the slope of V-I graph for a conductor represent?
Solution: (a) voltage to current graph for conductor obeying ohm’s law.
(b.) Slope represents the resistance.
Question: 11
Draw a I-V graph for a linear resistor. What does its slope represent?
Solution: current voltage graph for a linear resistor.
Question: 12
What is an ohmic resistor? Give one example of an ohmic resistor. Draw a graph to show its current – voltage relationship. How is the resistance of the resistor determined from this graph?
Solution: the conductor behaving according ohm’s law is known as ohmic resistors.
The resistance can be determined by the slope of above graph.
Question: 13
What are non-ohmic resistors? Give one example and draw a graph to show its current-voltage relationship.
Solution: The conductor which do not obey ohm’s law is known as non-ohmic resistors
Question: 14
Give two differences between an ohmic and non-ohmic resistor
Solution:
Ohmic resistor | Non-ohmic resistor |
It obeys ohm’s law. | It does not obey ohm’s law. |
The graph for the potential difference V versus current I is a straight line passing through the origin. | The graph for the potential difference V versus current I is not a straight line, but a curve which may pass through the origin. |
Question: 15
Fig. below shows the I-V curves for two resistors. Identify the ohmic and non-ohmic resistors. Give a reason for your answer.
Solution: Graph (a) shows the non-ohmic resistor and graph (b) is ohmic resistor. The graph (a) shows curve which is characteristic of non-ohmic resistor and graph (b) shows the straight line which is characteristic of ohmic resistor.
Question: 16
Draw a V – I graph for a conductor at two different temperatures. What conclusion do you draw from your graph for the variation of resistance of conductor with temperature?
Solution:
Based on the graph the temperature of the wire A is more than the temperature of wire B as the straight line of A is steeper than line of B. hence, the wire A offers more resistance than wire B hence wire A is thinner as the thinner wire more resistance to raise the temperature of the wire.
Question: 17
(a.) How does the resistance of a wire depend on its radius? Explain your answer.
(b.) Two copper wires are of same length, but one is thicker than the other. Which will have more resistance?
Solution:
a.) The resistance of wire is inversely proportional to the radius of the wire. The thin wire offers more resistance than thin wire.
b.) The copper wire with thin cross section will offer more resistance. As they have same length.
Question: 18
How does the resistance of a wire depend on its length? Give a reason of your answer.
Solution: The resistance of a wire is directly proportional length of wire. The resistance offered by the wire is depends upon the repulsion between differently charged particles. Hence, the more the length more the resistance offered by the wire as the electrons have to travel more distance under influence of repulsion.
Question: 19
How does the resistance of a metallic wire depend on its temperature? Explain with reason.
Solution: The resistance of wire increases as the temperature increases. The increase in the temperature causes more random motion of electrons which ultimately increases the resistance of the wire.
Question: 20
Two wires, one of copper and other of iron, are of the same length and same radius. Which will have more resistance? Give reason.
Solution: Resistivity of the material is directly proportional to the resistance offered by that material. The resistivity of iron is more iron is more than that of copper. hence, the resistance of iron is more than wire of copper as the same length and radius is same.
Question: 21
Name three factors on which the resistance of a given wire depends and state how it is affected by the factors stated by you?
Solution:
a.) The length of wire. The resistance of wire is directly proportional to the length of wire. More the length of wire more the resistance.
b.) Cross section of wire. The resistance of wire is inversely proportional to the cross section of wire. Thinner the wire more the resistance offered by the wire.
Question: 22
Define the term specific resistance and state its S.I. unit.
Solution:The resistance of a wire of a material of unit length and unit area of cross section. Ohm x meter is the S.I. unit of specific resistance.
Question: 23
Write an expression connecting the resistance of a wire and specific resistance of its material. State the meaning of symbols used.
Solution:
Specific resistance (ρ) = R a / l
Where,
R = resistance of wire
a = area of cross section
l = length of wire.
Question: 24
State the order of specific resistance of (i) a metal, (ii) a semiconductor and (iii) an insulator.
Solution:
i.) The specific resistance of metal is low as it has high conductivity.
ii.) The specific resistance of semi-conductor is more than metal
iii.) The specific resistance of insulator is highest, as it doesn’t show any conductivity.
Question: 25
(a.) Name two factors on which the specific resistance of a wire depends?
(b.) Two wires A and B are made of copper. The wire A is long and thin while the wire B is Short and thick. Which will have more specific resistance?
Solution:
a.) Material and temperature of the substance are the two factors specific resistance of a wire depends upon.
b.) The specific resistance is same as the specific resistance is depends upon the material of the wire.
Question: 26
Name a substance of which the specific resistance remains almost unchanged by the increase in temperature.
Solution: Manganin is the substance of which the specific resistance almost remains unchanged by the increase in the temperature.
Question: 27
How does specific resistance of a semi-conductor change with the increase in temperature?
Solution: The specific resistance of semi-conductors decreases with increase in the temperature.
Question: 28
How does (a) resistance, and (b) specific resistance of a wire depend on its (i) length, and (ii) radius?
Solution:
a.) Resistance is directly proportional to the length of wire and inversely proportional to the cross section.
b.) Specific resistance is independent of length and cross section of the wire.
Question: 29
(a) Name the material used for making the connection wires. Give reason for your answer.
(b) Why should a connection wire be thick?
Solution:
a.) Copper or aluminium are used in making of wire as the they have low specific resistance. Hence, there wire offers very low resistance.
b.) As the resistance of a wire is inversely proportional to the cross section of the wire. Connections are made thick as the resistance decreases with it.
Question: 30
Name the material used for making a fuse wire. Give a reason.
Solution: Alloy of tin and lead is used to make fuse wire as it has high resistivity and low melting point. Which is important for working of the fuse.
MULTIPLE CHOICE TYPE
Question: 1
Which of the following is an ohmic resistance?
(a) LED
(b) Junction diode
(c) Filament of a bulb
(d) Nichrome wire
Solution: (d)Nichrome wire
Question: 2
For which of the following substances, resistance decreases with increase in temperature?
(a) Copper
(b) Mercury
(c) Carbon
(d) Platinum
Solution: (c) Carbon
NUMERICALS
Question: 1
In a conductor, 6.25 × 1016 electrons flow from its end A to B in 2 s. Find the current flowing through the conductor. (e = 1.6 × 10-19 C)
Solution: Given
Number of electrons, n = 6.25 x 1016
Charge, e = 1.6 x 10-16
Time, t = 2s
Current = ne / t
I = 6.25 x 1016 x 1.6 x 10-16 / 2
I = 5 x 10-3 A = 5 mA.
Question: 2
A current of 1.6 mA flows through a conductor. If charge of an electron is -1.6 x 10-19 coulomb, find the number of electrons that will pass each second through the cross section of that conductor.
Solution: Given
Current, I = 1.6 mA = 1.6 x 10-3 A
Charge, e = 1.6 x 10-16C
Time = 1s
number of electrons = I x t / e
n= 1.6 x 10-3 x 1 / 1.6 x 10-16
n = 1016
Question: 3
Find the potential difference required to flow a current of 200 mA in a wire of resistance 20 ohm.
Solution: Given
Current, I = 200 mA = 0.2 A
Resistance, R = 20 ohm
Potential difference, V = I x R
V = 0.2 x 20
V = 4 V
Question: 4
An electric bulb draws 1.2 A current at 6.0 V. Find the resistance of filament of bulb while glowing.
Solution: Given
Current, I = 1.2 A
Potential difference, V = 6 V
Resistance, R = V / I
R = 6 / 1.2
R = 5 ohm.
Question: 5
A car bulb connected to a 12 volt battery draws 2 A current when glowing. What is the resistance of the filament of the bulb? Will the resistance be more, same or less when the bulb is not glowing.
Solution: Given
Potential difference, V = 12 V
Current, I = 2 A
Resistance of filament, R = V / I
R = 12 / 2
R = 6 ohm
When the bulb is not glowing, the resistance will be less.
Question: 6
Calculate the current flowing through a wire of resistance 5 Ohm connected to a battery of potential difference 3 V.
Solution: Given
Resistance, R = 5 ohm
Potential difference, V = 3V
Current, I = V / R
I = 3 / 5
I = 0.6 A
Question: 7
In an experiment of verification of Ohm’s law, following observations are obtained.
Potential difference V (in volt) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 |
Current I (in amp) | 0.2 | 0.4 | 0.6 | 0.8 | 1.0 |
Draw a characteristic V-I graph and use this graph to find:
(a.) potential difference V when the current I is 0.5 A.
(b.) current I when the potential difference V is 0.75 V.
(c.) resistance in circuit
Solution: (a)
Potential different is 1.25V when current is 0.5 A.
(b) when potential difference is 0.75 V current is 0.3 A.
(c) resistance = voltage / current
R = 1.25 / 0.5
R = 2.5 ohm
Question: 8
Two wires of the same material and same length have radii 1 mm and 2 mm respectively. Compare (i) their resistances (ii) their specific resistance.
Solution:
i.) The resistance
R1 = ρ (I / πr12)
R2 = ρ (I / πr22)
R1 / R2 = r22 / r12
R1 / R2 = 4/1
Ratio of resistance = 4:1
ii.) The ratio of specific resistance will be 1:1 as the materials is same.
Question: 9
A given wire of resistance 1 Ohm is stretched to double its length. What will be its new resistance?
Solution:
Resistance, R1 = ρ (I / πr2)
Now, length of wire
Resistance, R2 =ρ (2I /πr12/2)
R2 = 4 ρ (I / πr2)
Therefore,
R2 = 4 x R1
R2 = 4 ohm.
Question: 10
A wire of resistance 3 Ohm and length 10 cm is stretched to length 30 cm. Assuming that it has a uniform cross-section, what will be its new resistance?
Solution: Given
Resistance, R = 3 ohm
Length, l1 = 10cm
Length, l2 = 30 cm = 3l
Area will be πr2/3
New resistance = ρ (I / πr2)
R2 = ρ (3I / πr2/3)
R2 = 9 ρ (I / πr2)
R2 = 9 R
R2 = 9 x 3
R2 = 27 ohm.
Question: 11
A wire of resistance 9 Ohm having length 30 cm is tripled on itself. What is its new resistance?
Solution: Given
Resistance = 9 ohm
Length = 30 cm
New length = 1/3L
Area will be 3πr2
New resistance = ρ (I / πr2)
R2 = ρ (I/3 / 3πr2)
R2 = 1/9 ρ (I / πr2)
R2 = 1/9 R
R2 = 9 /9
R2 = 1 ohm.
Question: 12
What length of copper wire of specific resistance 1.7 x 10-8 ohmand radius 1 mm is required so that its resistance is 1 ohm.
Solution: Given
specific resistance = 1.7 x 10-8 ohm
radius = 1mm = 10-3 m
resistance = ρ (I / πr2)
l = R πr2/ ρ
l = 1 x 3.14 x 10-6/ 1.7 x 10-8
l = 184.7 m
Question: 13
The filament of a bulb takes a current 100 mA when potential difference across it is 0.2 V. When the potential difference across it becomes 1.0 V, the current becomes 400 mA. Calculate the resistance of filament in each case and account for the difference.
Solution: Given
Current, I = 100 mA = 0.1 A
Potential difference, V = 0.2 V
Resistance = V / I
= 0.2 / 0.1
= 2 ohm
Current, I = 400 mA = 0.4 A
Potential difference, V = 1 V
Resistance = V / I
= 1 / 0.4
= 2.5 ohm
Exercise 8 (B)
Question: 1
Explain the meaning of the terms e.m.f., terminal voltage and internal resistance of cell.
Solution: The e.m.f. of a cell defined as the energy per unit charge in to positive test charge around the complete of the cell.
The work done per unit charge in carrying a positive test charge around the circuit connected across the terminal of the cell is called as terminal voltage.
The resistance offered by the electrolyte inside the cell, to the flow of current is called as the internal resistance of the cell.
Question: 2
State two differences between the e.m.f. and terminal voltage of a cell.
Solution:
e.m.f. of the cell. | Terminal voltage of the cell. |
The energy per unit charge in to positive test charge around the complete of the cell. | The work done per unit charge in carrying a positive test charge around the circuit connected across the terminal of the cell |
It is characteristic of an cell. | It depends upon the amount of current drawn from the cell. |
Question: 3
Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.
Solution:
The surface area of the electrodes will affect the internal resistance as the surface area of the electrode increases the internal resistance of the cell decreases. The internal resistance is directly proportional to the distance between the electrodes.
Question: 4
A cell of e.m.f. ε and internal resistance r is used to send current to an external resistance R. Write expressions for (a) the total resistance of circuit, (b) the current drawn from the cell, (c) the p.d. across the cell, and (d) voltage drop inside the cell.
Solution:
a.) Total resistance = R + r
b.) Current drawn =
e.m.f.ε = V + v
ε = IR + Ir
ε = I (R + r)
I = (R + r)
C.) Potential difference across the cell = (ε/ R+ r) / R
d.) Voltage drop inside the cell = (ε/ R+ r) / r
Question: 5
A cell is used to send current to an external circuit. (a) How does the voltage across its terminal compare with its emf? (b) Under what condition is the emf of the cell equal to its terminal voltage?
Solution:
a.) The terminal voltage V is less than e.m.f. of the cell as the current is drawn from a cell.
b.) The emf of the cell becomes equal to its terminal voltage when there is no current drawn.
Question: 6
Explain why the p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
Solution: When the circuit is closed current flows through the circuit. Which causes the drop in the potential difference across the internal resistance of the cell. Hence, potential difference across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
Question: 7
Write the expressions for the equivalent resistance R of three resistors R1, R2 and R3 joined in (a) parallel, (b) series.
Solution:
a.) Total resistance in parallel
1/R = 1/R1 + 1/ R2 + 1/R3
b.) Total resistance in the series
R = R1 + R2 + R3
Question: 8
How would you connect two resistors in series? Draw a diagram. Calculate the total equivalent resistance.
Solution:
We can apply ohms law on the resistors separately
V1 = IR1
V2 = IR2
V = V1 + V1
IR = IR1 + IR2
R = R1 + R2
Question: 9
Show by a diagram how two resistors R1 and R2 are joined in parallel. Obtain an expression for the total resistance of the combination.
Solution:
We can apply ohms law on the resistors separately
V= I1R1
I1 = V / R1
V = I2R2
I2 = V / R2
I = I1 + I1
V / R = V / R1 + V / R2
1 / R = 1 / R1 + 1 / R2
Question: 10
State how are the two resistors joined with a battery in each of the following cases when:
(a.) same current flows in each resistor
(b.) potential difference is same across each resistor.
(c.) equivalent resistance is less than either of the two resistances.
(d.) equivalent resistance is more than either of the two resistances.
Solution:
(a.) When two resistors are connected in series.
(b.) When two resistors are connected in parallel.
(c.) When two resistors are connected in series.
(d.) When two resistors are connected in parallel.
Question: 11
The V-I graph for a series combination and for a parallel combination of two resistors is shown in fig. Which of the two, A or B, represents the parallel combination? Give a reason for your answer.
Solution:
The line B represent more change in voltage which is a result of more resistance. On the other hand line A represents lower change in voltage which accounts to lower resistance. When resistors are connected in parallel they show less resistance than when connected in series. Hence, line A represents parallel combination.
MULTIPLE CHOICE TYPE
Question: 1
In series combination of resistances:
(a.) P.d. is same across each resistance
(b.) Total resistance is reduced
(c.) Current is same in each resistance
(d.) All of the above are true
Solution: (c) Current is same in each resistance
Question: 2
In parallel combination of resistances:
(a.) P.D. is same across each resistance
(b.) Total resistance is increased
(c.) Current is same in each resistance
(d.) All of the above are true
Solution: (a) P.D. is same across each resistance
Question: 3
Which of the following combinations have the same equivalent resistance between X and Y?
Solution:
a.) Resistances are connected in parallel
1/R = ½ + ½ =1 ohm
b.) Resistances are connected in parallel
1/R = 1/1 + 1/1 +1/2 = 2.5 ohm
c.) Resistances are connected in parallel
1/R = 1/1 + 1/1 = 0.5 ohm
d.) Pair of resistors are connected in parallel
1/R = ½ + ½ = 1 ohm
Therefore, the (a) and (d) combination has same resistance across between X and Y.
NUMERICALS
Question: 1
The diagram in figure shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm connected to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammetere and voltmeter when (i) the key K is open, and (ii) the key K is closed
Solution: Given
Emf, ε = 2 volt
Internal resistance, r = 1 ohm
External resistance, R = 4 ohm
i.) Voltage = ε – Ir
As the key is open I = 0
Voltage = 2 – 0 x 1
Voltage = 2 volt
ii.) Ammeter reading
I = ε / (R + r)
I = 2 / (4 + 1)
I = 2/5
I = 0.4 A
Voltage = ε – Ir
Voltage = 2 – 0.4 x 1
Voltage = 1.6 V
Question: 2
A battery of e.m.f. 3.0 V supplies current through a circuit in which resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 1.5 A, the voltmeter reads 2.7 V. Find the internal resistance of the battery.
Solution: Given
Emf, ε = 3 V
Current, I = 1.5 A
Voltage = 2.7 V
V = ε – Ir
2.7 = 3 – 1.5r
0.3 / 1.5 = r
r = 0.2 ohm.
Question: 3
A cell of emf 1.8 V and internal resistance 2 ohm is connected in series with an ammeter of resistance 0.7 ohm and resistance of 4.5 ohm as shown in figure.
(a) What would be the reading of the ammeter?
(b) What is the potential difference across the terminals of the cell?
Solution:
Given
Emf, ε = 1.8V
Internal resistance, r = 2 ohm
Ammeter = 0.7 ohm
External resistance, R = 4.5 ohm
Total resistance = 2 + 0.7 + 4.5
Total resistance = 7.2 ohm
a.) Reading of ammeter
Current = ε / R
I = 1.8 / 7.2
I = 0.25 A
b.) Total resistance excluding internal resistance = 4.5 + 0.7 = 5.2 ohm
Voltage = IR
V = 0.25 x 5.2
V = 1.3 V
Question: 5
A cell of e.m.f. ε and internal resistance r sends current 1.0 A when it is connected to an external resistance 1.9 ohm. But its sends current 0.5 A when it is connected to an external resistance of 3.9 ohm. Calculate the values of e and r.
Solution: Given
Emf, ε = 15 V
External resistance, R1 = 1.9 ohm
Current, I1 = 1 A
Eternal resistance, R2 = 3.9 ohm
Current, I2 = 0.3 A
For first case, ε = I (R + r)
= 1 (1.9 + r)
For second case, ε = I (R + r)
= 0.5 (3.9 +r)
Emf of cell is same. Therefore,
1 (1.9 + r) = 0.5 (3.9 + r)
r = 0.05 / 0.5
r = 0.1 ohm
emf, ε = 1.9 + r
ε = 1.9 + 0.1
ε = 2 V
Question: 6
Two resistors having resistance 4 ohm and 6 ohm are connected in parallel. Find their equivalent resistance.
Solution: Given
R1 = 4 ohm
R2 = 6 ohm
As they are connected in parallel
1/R = 1/ R1 + 1/ R2
1/R = 1/ 4 + 1/6
1/R = 10/24
R = 2.4 ohm
Question: 7
Four resistors each of resistance 2 ohm are connected in parallel. What is the effective resistance?
Solution: Given
R1 = 2 ohm
R2 = 2 ohm
R3 = 2 ohm
R4 = 2 ohm
As they are connected in parallel
1/R = 1/ R1 + 1/ R2 + 1/R3 + 1/R4
1/R = ½ + ½ + ½ + ½
1/R = 2
R = 0.5 ohm.
Question: 8
You have three resistors of values 2 Ω, 3 Ω and 5 Ω. How will you join them so that the total resistance is less than 1 Ω? Draw diagram and find the total resistance.
Solution: We have to connect the resistors in parallel.
Total resistance, R
1 / R = 1 / 2 + 1 / 3 + 1 / 5
1 / R = 31 / 30
R = 0.97 Ohm
Question: 9
Three resistors each of 2 Ω are connected together so that their total resistance is 3 Ω. Draw a diagram to show this arrangement and check it by calculation.
Solution: to get total resistance of 3 ohm we have to connect two resistors in parallel and one in series as shown below.
For resistors in parallel
1 / r = 1/2 + ½
r = 1ohm
now this is connected to 2 ohm resistor in series
total resistance = 1 + 2
R = 3 ohm.
Question: 10
Calculate the equivalent resistance between the points A and B in figure if each resistance is 2.0 Ω
Solution: Given
Resistor value = 2 ohm
The total resistance in parallel, 1/R = ½ + ½ = 1ohm
Therefore, total resistance of the circuit = 2 + 2 + 1
= 5 ohm
Question: 11
A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.
Solution: Given
Resistor value = 2 ohm
Four sets of three of these resistors are connected in series
Resistance of each one = 2 + 2 + 2 = 6 ohm
The resistance of all four set is 6 ohm.
Now these four sets are connected in parallel.
1/R = 1/6 + 1/6 + 1/6+ 1/6
1/R = 4/6
R = 6/4
R = 1.5 ohm.
Question: 12
In the circuit shown below in figure, calculate the value of x if the equivalent resistance between the points A and B is 4 ohm.
Solution: Given
R = 4 ohm
R1 = 4 ohm
R2 = 8 ohm
R3 = x ohm
R4 = 5 ohm
Now R1 and R2 are connected in series and R3 and R4 are connected in series
R1 + R2 = 4+ 8 = 12 ohm
R3 + R4 = x + 5 ohm
Now these are connected in parallel,
1/R = 1/R1+ R2 + 1/ R3+R4
¼ = 1/12 + 1/ x + 5
¼ – 1/12 = 1/ x + 5
2/12 = 1/ x + 5
1/6 = 1 / x + 5
x = 6 – 5
x = 1 ohm.
Question: 13
Calculate the effective resistance between the points A and B in the circuit shown in figure
Solution: The resistance in series in upper area = 1 +1 + 1 = 3 ohm
The resistance in series in lower area = 2 + 2 + 2 = 6 ohm
The total resistance in the parallel
1/R = 1/3 + ½ + 1/6
1/R = 2 + 3 + 1/ 6
1/R = 1 ohm
Total resistance across A to B = 1 + 1 + 1 = 3 ohm.
Question: 14
A uniform wire with a resistance of 27 ohm is divided into three equal pieces and then they are joined in parallel. Find the equivalent resistance of the parallel combination.
Solution: As the resistance was split in three pieces. Therefore, the resulting resistance
27/3 = 9 ohm
As they are connected in parallel
1/R = 1/9 + 1/9 + 1/9
1/R = 3/9
1/R = 1/3
R = 3 ohm
Question: 15
A circuit consists of a 1 ohm resistor in series with a parallel arrangement of 6 ohm and 3 ohm resistors. Calculate the total resistance of the circuit. Draw a diagram of the arrangement.
Solution: As the resistance is connected in parallel
1/R = 1/6 + 1/3
1/R = 3/6
R = 2 ohm
Total resistance = 2 + 1
= 3 ohm
Question: 16
Calculate the effective resistance between the points A and B in the network shown below in figure.
Solution: The parallel resistance
1/R = 1/12 + 1/6 + ¼
1/R = 6 /12
R = 2 ohm
Total resistance across A to B = 2 + 2 + 5
= 9 ohm
Question: 17
Calculate the equivalent resistance between the points A and B in figure
Solution: The resistance connected in series = 3 +2 = 5 ohm
The resistance connected in series in series = 6 + 4 = 10 ohm
Therefore, total resistance across A to B
1/R = 1/5 + 1/ 30 + 1/10
1/R = 6+ 1 + 3 /30
1/R = 1/3
R = 3 ohm
Question: 18
In the network shown in adjacent figure, calculate the equivalent resistance between the points (a) A and B (b) A and C
Solution:
a.) Resistance between A and B.
The resistance connected series = 2 + 2 +2 = 6 ohm.
The resistance connected in parallel
1/R = 1/6 + ½
1/R = 4/6
R = 1.5 ohm
b.) The resistance between A and C
Resistance in series = 2 + 2 = 4 ohm
Two of the 4 ohm resistors are connected in parallel.
1/R = ¼ + ¼
1/R = ½
R = 2 ohm
Question: 19
Five resistors, each of 3 ohm, are connected as shown in figure. Calculate the resistance (a) between the points P and Q, and (b) between the points X and Y.
Solution:
a.) The resistance between P and Q.
The resistance in series = 3 + 3 = 6 ohm
The total resistance between P and Q
1/R = 1/6 + 1/3
1/R = 3/6
R = 2 ohm
b.) Total resistance across X and Y
All the resistors connected in series
Total resistance = 3 + 2 + 3
Total resistance = 8 ohm.
Question: 20
Two resistors of 2 ohm and 3 ohm are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.
Solution:
a.) In series
Total resistance of the circuit, R
R = 2 + 3 = 5 ohm
V = 6 v
Current = V / R
I = 6 / 5 = 1.2 A
b.) In parallel
Total resistance of the circuit, R
1/R = ½ + 1/3
1/R = 5 / 6
R = 1.2 ohm
V = 6 V
Current, I = V / R
I = 6 / 1.2
I = 5 A
Question: 21
A resistor of 6 ohm is connected in series with another resistor of 4 ohm. A potential difference of 20 V is applied across the combination. (a) Calculate the current in the circuit and (b) potential difference across the 6 ohm resistor.
Solution:
a.) As the resistors connected in series
R = 6 + 4 = 10 ohm
Current I = V / R
I = 20 / 10
I = 2 A
b.) To calculate potential difference across 6 ohm resistor
R = 6 ohm
I = 2 A
V = IR
V = 6 x 2
V = 12 V
Question: 22
Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.
(a) Draw a labelled diagram of the arrangement
(b) Calculate current in each resistor.
Solution:
a.)
b.) Total resistance of the circuit
1 / R = ¼ + 1/6
1/R = 10 /24
R = 2.4 ohm
I = 0.5 A
Voltage, V = IR
V = 0.5 x 2.4
V = 1.2
Therefore,
Current through resistor of 4 ohm
I = 1.2 / 4
I = 0.3 A
Current through resistor of 6 ohm
I = 1.2 / 6
I = 0.2 A
Question: 23
Calculate the current flowing through each of the resistors A and B in the circuit shown in figure?
Solution: Given
For resistor A = 1 ohm
Voltage = 2 V
Current across A,
I = V / R
I = 2 / 1
I = 2 A
For resistor B
Resistor = 2 ohm
I = V / R
I = 2 / 2
I = 1
Question: 24
In figure, calculate:
(a.) the total resistance of the circuit.
(b.) the value of R, and
(c.) the current flowing in R
Solution:
a.) The total resistance of the circuit
Given
Current, I = 0.4 A
Potential difference, V = 4 V
R1 = V / I
R1 = 4 / 0.4
R1 = 10 ohm.
b.) To calculate value of R
The total resistance is 10 ohm
As the two resistors are connected in parallel
1/10 = 1/R + 1/ 20
1/R = 1/20
R = 20 ohm
c.) To calculate current flowing through R
R = 20 ohm
V = 4 V
I = V / R
I = 4 / 20
I = 0.2 A
Question: 25
A particular resistance wire has a resistance of 3.0 ohm per meter. Find:
(a.) The total resistance of three lengths of this wire each 1.5 m long, in parallel.
(b.) The potential difference of the battery which gives a current of 2 A in each of the 1.5 m length when connected in the parallel to the battery (assume that resistance of the battery is negligible).
(c.) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.
Solution:
a.) Resistance of wire per metre is 3 ohm. Hence for 1.5m wire resistance is 4.5 ohm
Total resistance in parallel
1/R = 1/4.5 + 1/4.5 + 1/4.5
1/R = 3/4.5
R = 1.5 ohm
b.) Current, I = 2 A
Potential difference, V = IR
V = 2 x 4.5
V = 9 V
c.) Resistance of 1m wire is 3 ohm.
Therefore, 5 m length of wire resistance = 3 x 5 = 15 ohm
But as the cross section is doubled resistance is halved as they are inversely proportional.
Resistance = 15 /2 = 7.5 ohm.
Question: 26
A cell supplies a current of 1.2 A through two resistors each of 2 ohm connected in parallel. When resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell.
Solution: Given
Current, I = 1.2 A
Resistance as in parallel
1/R = ½ + ½ = 1
R = 1 ohm
ε = I (R + r)
ε = 1.2 (1 + r)
resistance as in series,
R = 2 + 2 = 4 ohm
ε = 0.4 (4 + r)
as the emf is same,
1.2 (1 + r) = 0.4 (4 + r)
1.2r – 0.4r = 1.6 – 1.2
0.8r = 0.4
r = 0.5 ohm
emf of cell, ε = I (R + r)
ε = 1.2 (1 + 0.5)
= 1.2 + 0.6
= 1.8 V
Question: 27
A battery of emf 15 V and internal resistance 3 ohm is connected to two resistors 3 ohm and 6 ohm connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the current in 3 ohm resistor (d) the current in 6 ohm resistor.
Solution: Given
Emf = 15 V
Internal resistance, r = 3 ohm
External resistance, 1/R = 1/3 + 1/6
1/R = 3/6
1/R = ½
R = 2 ohm
a.) Current through the battery
ε = I (R + r)
15 v = I (2 + 3)
I = 15 / 5
I = 3 A
b.) Potential difference
V = I R
V = 3 x 2
V = 6 V
c.) Current across 3 ohm resistor
I = V / R
I = 6 / 3
I = 2 A
d.) Current across 6 ohm resistor
I = V /R
I = 6 / 6
I = 1 A
Question: 28
The circuit diagram in figure shows three resistors 2 ohm, 4 ohm and R ohm connected to a battery of e.m.f. 2 V and internal resistance 3 ohm. If main current of 0.25 A flows through the circuit, find:
(a.) the p.d. across the 4 ohm resistor
(b.) the p.d. across the internal resistance of the cell,
(c.) the p.d. across the R ohm or 2 ohm resistor, and
(d.) the value of R.
Solution:
a.) Potential difference across 4 ohm resistor.
Resistance, R = 4 ohm
Current, I = 0.25 A
Potential difference, V= IR
V = 0.25 x 4
V = 1 V
b.) Potential difference across cell.
Internal Resistance, r= 3 ohm
Current, I = 0.25 A
Potential difference, V= Ir
V = 0.25 x 3
V = 0.75 V
c.) We have first find out the value of R
Total resistance of the circuit = (1/2 +1/R) + 4
Emf = 2V
r = 3 ohm
I = 0.25A
Emf = I (R total + r)
2 = 0.25 (Rtotal + 3)
R total = 5 ohm
Therefore,
(1/2 + 1/R) + 4 = 5
1/R = 1- ½
R = 2 ohm
Now for potential difference across 2 ohm resistor
Resultant resistance of 2 ohm resistor as two 2 ohm resistors are connected in parallel
= 1 ohm
V = IR
V = 0.25 x 1
V = 0.25 V
Question: 29
Three resistors of 6.0 ohm, 2.0 ohm and 4.0 ohm are joined to an ammeter A and a cell of emf 6.0 V as shown in figure. Calculate:
(a.) the efective resistance of the circuit.
(b.) the reading of ammeter
Solution: Given
R1 = 6 ohm
R2 = 2 ohm
R3 = 4 ohm
a.) As R2 and R3 are connected in series
R’ = 2+4 = 6 ohm
And R’ and R1 are connected in parallel
Total resistance of the circuit, 1/R = 1/6 +1/6
1/R = 2/6
R = 3 ohm
b.) Potential difference, V = 6 V
V = IR
I = V / R
I = 6 /3
I = 2 A
Question: 30
The diagram below in Fig., shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8 V. Calculate:
a.) The total resistance of the circuit
b.) The reading of ammeter A.
Solution:
The resistance of the resistor that are parallel in the upper part of the circuit
1/R’ = 1/10 + 1/40
1/R’ = 5 /40
R’ = 8 ohm
The resistance of the resistor that are parallel in the lower part of the circuit
1/R’’ = 1/20 + 1/30 + 1/60
1/R’’ = 6 /60
R’ = 10 ohm
Total resistance of the circuit = 10 + 8 = 18 ohm
For reading of ammeter
Current, I = V/R
I = 1.8 /18
I 0.1 A
Question: 31
A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in fig.
Find:
(a) The reading of the ammeter,
(b) The potential difference across the terminals of the cells, and
(c) The potential difference across the 4.5 ohm resistor.
Solution: Given
Emf = 2 V
Internal resistance, r = 1.2 ohm
Resistance of the ammeter = 0.8 ohm
The resultant resistance of the resistors connected in parallel
1/R1 = 1/4.5 + 1/9 =3/9
R1 = 3 ohm
Total resistance of the circuit, R= 3 + 0.8 + 1.2 = 5 ohm
a.) For reading of ammeter
Current, I = V / R
I = 2 / 5
I = 0.4 A
b.) Potential difference across the cell
V1 = emf – Ir
V1 =2 – 0.4 x 1.2
V1 = 1.52 V
c.) The potential difference across 4.5 ohm resistor
V = V1 – IR
V = 1.52 – 0.4 x 0.8
V = 1.2 V
Exercise 8 (C)
Question: 1
Write an expression for the electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time.
Solution:
Electrical energy
W = I2R
Where,
W = electrical power
I = current and
R = resistance
Question: 2
Write an expression for the electrical power spent in flow of current through a conductor in terms of (a) resistance and potential difference, (b) current and resistance.
Solution:
a.) Electrical power, P = V2/ R
Where, V = potential difference and R = resistance.
b.) Electrical power, P = I2R
Where, I = current and R = resistance.
Question: 3
Electrical power P is given by the expression P = (Q × V) ÷ time
(a.) What do the symbols Q and V represent?
(b.) Express the power P in terms of current and resistance explaining the meanings of symbols used their in.
Solution:
a.) Q represents charge and V represents potential difference.
b.) Electrical power, P = I2R
Where, I = current and R = resistance.
Question: 4
Name the S.I. unit of electrical energy. How is it related to Wh?
Solution: S.I. unit of electrical energy is joule.
1 W = 3600 J
Question: 5
Explain the meaning of the statement ‘the power of an appliance is 100 W’.
Solution: power = work done / time
Hence, 100 W energy was consumed in 1 second.
Question: 6
State the S.I. unit of electrical power.
Solution: S.I. unit of electrical power is watt.
Question: 7
(i.) State and define the household unit of electricity.
(ii.) What is the voltage of the electricity that is generally supplied to a house?
(iii.) What is consumed while using different electrical appliances, for which electricity bills are paid?
Solution:
i.) Unit of household electricity is kilowatt/hr. which means 1 kilowatt energy consumed by household appliances in an hour is called as 1 kilowatt/hr energy.
ii.) 220 -240 volts is generally supplied to a house.
iii.) Electrical energy.
Question: 8
Name the physical quantity which is measured in (i) kW, (ii) kWh. (iii) Wh
Solution:
i.) Electrical power is measured in kW.
ii.) Electrical energy is measured in kW/h
iii.) Electrical energy is measured in Wh.
Question: 9
Define the term kilowatt – hour and state its value in S.I. unit.
Solution: 1 kilowatt/ hour is 1 kilowatt energy consumed by appliances within an hour.
Vale of kilowatt/hour in S.I. unit = 3.6 x 106 J.
Question: 10
How do kilowatt and kilowatt-hour differ?
Solution: Kilowatt hour is the unit of electrical energy and kilowatt/ hour is the unit of electrical power.
Question: 11
Complete the following:
(a.) 1 kWh = (1 volt × 1 ampere × ……..) / 1000
(b.) 1 kWh= ________ J
Solution:
a.) 1 hour
b.) 3.6 x 106 J.
Question: 12
What do you mean by power rating of an electrical appliance? How do you use it to calculate (a.) the resistance of the appliance and (b.) the safe limit of the current in it, while in use?
Solution: The power rating of an electrical appliance is given buy its power consumption and voltage.
a.) Resistance of the appliance
R= V2/P
Voltage = V
Power = P
b.) Safe limit of current while in use
I = P /V
P = power
V = voltage.
Question: 13
An electric bulb is rated ‘100 W, 250 V’. What information does this convey?
Solution:
The power of the bulb is 100 W and it lights up when supplied with voltage of 250 V.
Question: 14
List the names of three electrical gadgets used in your house. Write their power, voltage rating and approximate time for which each one is used in a day. Hence find the electrical energy consumed by each in a month of 30 days.
Solution:
Appliance | Power
In watt |
Voltage, V | Time, hours | Electrical energy
E = P x t |
Led bulb | 18 | 220 | 12 | 0.22 kWh |
Television | 120 | 220 | 5 | 0.6 kWh |
Geyser | 1000 | 220 | 1 | 1 kWh |
Question: 15
Two lamps, one rated 220 V, 50 W and the other rated 220 V, 100 W, are connected in series with mains of voltage 220 V. Explain why does the 50 W lamp consume more power.
Solution: resistance of 50 W bulb
Resistance = V2/ P
R1 = 220 x 220 / 50
R1 = 968 ohm
Resistance of 100 W bulb
Resistance = V2/ P
R2 = 220 x 220 / 100
R2 = 484 ohm
As it stated above the resistance of the 50 W bulb is less than100 W bulb hence it consumed more power, as power is directly proportional to resistance.
Question: 16
Name the factors on which the heat produced in a wire depends when current is passed in it, and state how does it depend on the factors stated by you.
Solution:
a.) amount of heat produced in wire is directly proportional to the square of the amount of current flowing through it.
b.) The amount of heat produced by a wire is directly proportional to the resistance of the wire
c.) And lastly amount of heat is directly proportional to the time for which current flows through the wire.
MULTIPLE CHOICE TYPE
Question: 1
When a current I flows through a resistance R for time t, the electrical energy spent is:
(a.) IRt
(b.) I2Rt
(c.) IR2t
(d.) I2R / t
Solution: (b) I2Rt
Question: 2
An electrical appliance has a rating 100 W, 120 V. The resistance of element of appliance when in use is:
(a) 1.2 ohm
(b) 144 ohm
(c) 120 ohm
(d) 100 ohm
Solution: (b) 144 ohm
NUMERICALS
Question: 1
An electric bulb of resistance 500 ohm draws current 0.4 A from the source. Calculate: (a) the power of bulb and (b) the potential difference at its end.
Solution: Given
Resistance, R= 500 ohm
Current, I = 0.4 A
Potential difference, V = I x R
V = 0.4 x 500
V = 200 V
Power, P = V x I
P = 200 x 0.4
P = 80 W
Question: 2
A current of 2 A is passed through a coil of resistance 75 Ω for 2 minutes. (a) How much heat energy is produced? (b) How much charge is passed through the resistance?
Solution: Given
Resistance, R= 75 ohm
Current, I = 2 A
Time, t = 2 min = 120 s
Heat produced = I2Rt
= 2 x 2 x 75 x120
= 36000 J
Charge passed = I x t
Q = 2 x 120
Q = 240 C.
Question: 3
Calculate the current through a 60 W lamp rated for 250 V. If the line voltage falls to 200 V, how is power consumed by the lamp affected?
Solution: Given
Power, P = 60 W
Voltage, V = 250 V
As we know,
R = V2/ P
When the voltage drops 200 V
Power = V2/ R
= 200 x 200 / 250 x 250 / 60
Power reduced = 38.4 W
Question: 4
An electric bulb is rated ‘100 W, 250 V’. How much current will the bulb draw if connected to a 250 V supply?
Solution: Given
Power, P = 100 W
Voltage, V = 250 V
Current, I = P / V
I = 100 / 250
I = 0.4 A
Question: 5
An electric bulb is rated at 220 V, 100 W. (a) What is its resistance? (b) What safe current can be passed through it?
Solution: Given
Power, P = 100 W
Voltage, V = 220 V
Resistance, R = V2 / P
R = 220 x 220 / 100
R = 484 ohm
Safe limit of current
I = P / V
I = 100 / 220
I = 0.45 A
Question: 6
A bulb of power 40 W is used for 12.5 h each day for 30 days. Calculate the electrical energy consumed.
Solution: Given
Power, P = 40 W
Time, t= 12.5 hours
Energy consumed = P x t
E = 40 x 12.5
E = 500 Wh
Energy consumed in 30 days
E = 500 x 30
E = 15000 Wh = 15 kWh
Question: 7
An electric press is rated ‘750 W, 230 V’. Calculate the electrical energy consumed by the press in 16 hours
Solution: Given
Power, P = 750 W
Time, t = 16 hours
Energy consumed = 750 x 16
E = 12000 Wh = 12 kWh
Question: 8
An electrical appliance having a resistance of 200 ohm is operated at 200 V. Calculate the energy consumed by the appliance in 5 minutes (i) in joule, (ii) in kWh
Solution: Given
Resistance, R = 200 ohm
Voltage, V = 200 V
Time, t = 5 min = 300 s
Energy = V2t /R
E = 200 x 200 x 300 / 200
E = 60000 J
As we know,
3.6 x 106J = 1 kWh
60000 J = 60000 /3.6 x 106
= 0.0167 kWh
Question: 9
A bulb rated 12 V, 24 W operates on a 12 volt battery for 20 minutes. Calculate:
(i) the current flowing through it, and
(ii) the energy consumed.
Solution: Given
Power, P = 24 W
Voltage, V = 12 V
Time, t = 20 min = 120 s
Current, I = P / V
I = 24/12
I = 2 A
Energy = P x t
= 24 x 120
= 28,800 J
Question: 10
A current of 0.2 A flows through a wire whose ends are at a potential difference of 15 V. Calculate:
(i.)The resistance of the wire, and
(ii.)The heat energy produced in 1 minute.
Solution: Given
Current, I = 0.2 A
Potential difference, V = 15 V
Resistance, R = V / I
R = 15 / 0.2
R = 75 ohm
Heat produced in 1 min
Heat produced = I2Rt
= 0.2 x 0.2 x 75 x 60
= 180 J
Question: 11
What is the resistance, under normal working conditions, of an electric lamp rated at ‘240 v’, 60 W? If two such lamps are connected in series across a 240 V mains supply, explain why each one appears less bright.
Solution: Given
Voltage, V = 240 V
Power, P = 60 W
Resistance, R = V2 / P
R = 240 x 240 / 60
R = 960 ohm
Current, I = V / R
I = 240 / 960
I = 0.25 A
Now the bulbs are connected in series then resultant resistance becomes
960 + 960 = 1920 ohm
Therefore, current = 240 / 1920
Current = 0.125 A
As the current flowing through the bulbs is reduced they appear less bright.
Question: 12
Two bulbs are rated 60 W, 220 V and 60 W, 110 V, respectively. Calculate the ratio of their resistances.
Solution: Given
Power, P1 = 60 W
Voltage, V1 = 220V
Resistance, R1 = V2 / P
= 220 x 220 / 60
Power, P2 = 60 W
Voltage, V2 = 110 V
Resistance, R2 = V2 / P
R2 = 110 x 110 / 60
Therefore,
R1/R2 = 220 x 220 /60 / 110 x 110 /60
R1/R2 = 220 x 220 / 110 x 110
R1/R2 = 4/1
Question: 13
An electric bulb is rated 250 W, 230 V.
(i.) the energy consumed in one hour, and
(ii.) the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains?
Solution: Given
Power, P = 250 W
Voltage, V = 230 V
Time, t = 1 hours = 3600s
Energy consumed =250 x 3600
E = 9 x 105 J
Energy = 1 kWh
1000 Wh = power x time
Time = 1000 / 250
Time = 4 hours.
Question: 14
Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. Calculate:
(i.) The total current taken from the supply,
(ii.) The resistance of each heater, and
(iii.) The energy supplied in kWh to the three heaters in 5 hours.
Solution: Given
Voltage, V = 100 V
Power, P = 250 W
Current, I = P / V
I = 250 / 100
I = 2.5 A
Therefore, current taken by three heaters = 3 x 2.5 = 7.5 A
Resistance of each heater,
Resistance, R = V2 / P
R = 100 x 100 / 250
R = 400 ohm
Energy consumed = P x t
E = 250 x 5
E = 1250 Wh
E = 1.25 kWh
Therefore, energy of three heaters = 3 x 1.25
= 3.75 kWh.
Question: 15
A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 ohm. A steady current of 0.5 A flows through the circuit. Calculate:
(i.) The total energy supplied by the battery in 10 minutes,
(ii.) The resistance of the bulb, and
(iii.) The energy dissipated in the bulb in 10 minutes.
Solution: Given
Resistance of the battery, r = 2.5 ohm
Voltage, V = 4 V
Current, I = 0.5 A
Time, t = 10 min = 600 s
Resistance, R = V / I
R = 4 / 0.5
R = 8 ohm
Energy = V2t /R
E = 4 x 4 x 600 / 8
E = 1200 J
As total resistance is 8 ohm
Resistance of the bulb is 5.5 ohm.
Energy of bulb (heating) = I2Rt
= 0.5 x 0.5 x 5.5 x 600
= 825 J.
Question: 16
Two resistors A and B of 4 ohm and 6 ohm, respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate: (i) the power supplied by the battery, (ii) the power dissipated in each resistor.
Solution: Given
Resistance connected in parallel resultant resistance
1/ R = ¼ + 1/6 = 10 /24
R = 2.4 ohm
Voltage, V = 6 V
Power supplied by the battery = V2/ R
P = 6 x 6 / 2.4
P = 15 W
Power at resistor of 4 ohm = V2/ R
= 36 / 4
= 9 W
Power at resistor of 6 ohm = V2/ R
= 36 / 6
= 6 W
Question: 17
A battery of e.m.f. 15 V and internal resistance 2 ohm is connected to two resistors of resistances 4 ohm and 6 ohm joined in series. Find the electrical energy spent per minute in 6 ohm resistor.
Solution: Given
Voltage, V = 15 V
Internal resistance, r = 2 ohm
Time, t = 1 min = 60s
Resultant resistance of two resistance in series and battery = 4 + 6 + 2 = 12 ohm.
Current, I = V / R
I = 15 / 12
I = 1.25 A
Voltage across 6 ohm resistor
V = I x R = 1.25 x 6 = 7.5 V
Energy = V2t / R
E = 7.5 x 7.5 x 60 / 6
E = 562.5 J
Question: 18
Water in an electric kettle connected to a 220 V supply took 5 minutes to reach its boiling point. How long will it take if the supply had been of 200 V?
Solution: Given
Voltage, V1 = 220 V
Voltage, V2 = 200 V
Time, t1 = 5 min
Heat energy = (V2 / R)t
As heat gained to boil water is same
(V12/R)t1= (V22/ R)t2
t2 = (220 x 220 /200 x 200) x 5
t2 = 6.05 min.
Question: 19
An electric toaster draws current 8 A in a circuit with source of voltage 220 V. It is used for 2 h. Find the cost of operating the toaster if the cost of electrical energy is ₹ 4.50 per kWh
Solution: Given
Current, I = 8 A
Voltage, V = 220 V
Time, t = 2 h
Energy used = P x t
E = 220 x 8 x 2
E = 3520 Wh = 3.52 kWh
Cost of operating = 3.52 x 4.50
= 15.84 rupees
Question: 21
A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. Calculate:
(i)The current drawn,
(ii)The energy consumed in 50 hours, and
(iii)The cost of energy consumed at ₹ 4.20 per kWh.
Solution: Given
Power, P = 1500 W
Voltage = 250 V
Current, I = P / V
I = 1500 / 250
I = 6 A
Energy consumed in 50 hours
Energy = P x t
E = 1.5 x 50 = 75 kWh
Cost of use = 75 x 4.20
= 315 rupees.