Maharashtra Board Class 9 Science Solution Chapter 3 – Current Electricity
Balbharati Maharashtra Board Class 9 Science Solution Chapter 3: Current Electricity. Marathi or English Medium Students of Class 9 get here Current Electricity full Exercise Solution.
|
Std |
Maharashtra Class 9 |
| Subject |
Science Solution |
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Chapter |
Current Electricity |
1.) The accompanying figure shows some electrical appliances connected in a circuit in a house. Answer the following questions.
A) By which method appliances are connected?
B) What must be the potential difference across each individual appliance?
C) Will the current passing through each appliance be the same? Justify your answer.
D) Why are domestic appliances are connected in this way?
E) If the TV stops working, will the other appliances also stop working? Explain your answer.
Ans:
A) In our houses, all the domestic appliances are connected in parallel combination.
B) As all the domestic appliances are connected in parallel combination, we know that in parallel combination the voltage across each appliance will be same and it is nearly 220 V.
C) We know that, in parallel combination the potential difference across each appliance will be same but resistance of each appliance will be different.
And we know that, according to Ohm’s law
V= IR
Hence, if voltage may remains constant but the resistance of each appliance will be different and hence current through each appliance will be different.
D) The domestic appliances are connected in parallel combination because in parallel combination the voltage across each appliance will be same and hence the each appliance will works according to their requirements. And if any one of the appliance may be damaged then that appliance only stops working but all other circuit is working till. Due to this advantage the domestic appliances are connected in parallel combination.
E) If the TV stops working, there will be no current through that part of the circuit but other appliances are working till now. And this is possible only due to the parallel combination of appliances.
2.) The following figure shows symbols used for components used in the accompanying circuit. Place them at proper places and complete the circuit. Which law can you prove with the help of above circuit.
Ans:
Ohm’s law can be proved by using the above electrical circuit drawn.
3.) Umesh has two bulbs having resistance of 15ohm and 30 ohm. He wants to connect them in circuit, but he connects them one at a time the filament gets burnt. Answer the following sentences.
A) Which method he should use to connect the bulbs?
B) What are the characteristics of this way of connecting the bulbs depending on the answer of question A above.
C) What will be the effective resistance in the above circuit.
Ans:
As we know that, the heat produced in the circuit is directly proportional to the current flowing through the circuit.
Here, Umesh has connected bulb one at a time due to which current flowing through them is more and hence large heat is produced which causes to burn the filament.
A) Umesh should have to connect the bulbs in series combination because effective resistance in series combination is more than in parallel combination. Hence current flowing through the bulbs will be controlled which does not burnt the bulbs.
B) The characteristics of series combination are as follows:
In series combination, the current through each resistor will be same but potential difference across each resistor will be different.
The effective resistance in series combination will be more than the value of resistance which is high in the circuit.
And the effective resistance in series combination is obtained by directly taking sum of the resistances.
C) The effective resistance in series combination is obtained by directly taking sum of the resistances.
Here, 15+30= 45 ohm
4.) The following table shows current in amperes and potential difference in volts.
a) Find the average resistance.
b) What will be the nature of the graph between the current and potential difference? Do not draw a graph.
c) Which law will the graph prove. Explain the law.
Voltage given is 4V, 5V, 6V and corresponding current in amperes will be 9, 11.25, 13.5 respectively.
Ans:
a)
According to Ohm’s law,
V= IR
Hence, here
R= V/I= 4/9=0.44 ohm
R= 5/11.25=0.44 ohm
R= 6/13.5= 0.44 ohm
Thus, average resistance is given by,
(0.44+0.44+0.44)/3= 0.44 ohm
b) We know that, when we plot the VI graph then the slope of that graph gives us the value of resistance.
Here the value of resistance i.e. slope is constant. Which indicates the graph will be straight line with constant slope.
c) The graph shows the Ohm’s law.
According to Ohm’s law, the current through the resistor of resistance R is directly proportional to the potential difference across it.
Hence, V a I
V= IR
This is well known formula of Ohm’s law.
5.) Match the pairs.
Ans:
1) Free electrons: Weakly attached
2) Current : V/R
3) Resistivity : VA/LI
4) Resistance in series: Increases the resistance in the circuit.
6.) The resistance of conductor of length X is r. It’s area of cross section is a, what is its resistivity? What is its SI unit?
Ans:
At a given temperature the resistance of the material r depends on its length X, area of cross section a and the material from which it is made.
And resistivity is nothing but the specific resistance of the specific material.
If the resistance of the conductor of length X is r and area of cross section is a then it’s resistivity is given by,
q= (r*a)/X
And the SI unit of resistivity is Ohm meter.
7.) Resistances R1, R2, R3 and R4 are connected as shown in figure. S1 and S2 are two keys. Discuss the current flowing in the circuit in the following cases.
a) both S1 and S2 are closed.
b) both S1 and S2 are open.
c) S1 is closed but S2 is open.
Ans:
a) when both S1 and S2 are closed.
The circuit diagram is as shown in figure.
In this resistances R1 and R2 are in parallel.
Hence, their effective resistance will be,
R= R1*R2/(R1 + R2)
And the resistance R4 is in parallel with the resistance of value 0.
Hence, their effective resistance should be zero.
Hence, the effective resistance of parallel combination is in series with the resistor R3.
Hence, R= R3 + R1*R2/(R1 + R2)
If V is the potential difference applied then in this case current flowing through circuit will be,
I=V/(R3 + R1*R2/R1+R2)
b) when both S1 and S2 are open then the circuit diagram is as shown in figure.
In this case, the resistances R1, R3 and R4 are directly connected in series combination.
Hence, their effective resistance will be,
R= R1 + R3 + R4
In this case,, the current flowing through the circuit is given by,
I= V/(R1 + R3 + R4)
c) when the S1 is closed but S2 is open then the circuit diagram is as follows.
In this circuit, resistances R1 and R2 are in parallel combination, hence their effective resistance will be
R’= R1*R2/(R1 + R2)
Now, this effective resistance R’ is in series with the resistances R3 and R4.
Hence, their effective resistance will be,
R= R’ + R3 + R4
Hence, in this case the current flowing through the circuit is given by,
I= V/(R’ + R3 + R4)
8.) Three resistances X1, X2 and X3 are connected in a circuit in different ways. X is the effective resistance. The properties observed for this different ways of connecting X1, X2 and X3 are given below. Write the way in which they are connected in each case. Where I is the current, V is the potential difference and X is the effective resistance.
a) Current I flows through X1, X2 and X3.
b) X is larger than X1, X2 and X3.
c) X is smaller than X1, X2 and X3.
d) The potential difference across the resistance X1, X2 and X3 is the same.
e) X = X1+ X2 + X3
f)1/X = 1/X1 + 1/X2 + 1/X3
Ans::
a) When current I flows through X1, X2 and X3, that means these resistances are connected in series.
Because in series combination only the current flowing through each resistor will be same.
b) When X is larger than X1, X2 and X3, it means the effective resistance X is larger than each individual resistance.
And this is possible only in case of series combination.
Because, in series combination the effective resistance is greater than each individual resistance connected in series.
c) when X is smaller than X1, X2 and X3, it means that the effective resistance is smaller than each individual resistance.
And this is possible only in parallel combination.
Because in parallel combination the effective resistance is less than each individual resistance.
d) when the potential difference across each resistance X1, X2 and X3 is same, that means these resistances are connected in parallel combination.
Because only in parallel combination the potential difference across each Resistance will be same.
e) X= X1 + X2 + X3
This is the effective resistance in series combination.
f) 1/X = 1/X1 + 1/X2 + 1/X3
This is the effective resistance in parallel combination.
9.) Solve the following problems.
a) The resistance of 1m long nichrome wire is 6ohm. If we reduce the length of nichrome wire to 70cm. What will it’s resistance be?
Ans:
We know that, the resistance of the material is directly proportional to the length of the wire.
Let R be the resistance of wire of length 1m= 100cm
And R’ be the resistance of wire of length 70cm.
Then we can write,
R a 100cm
And R’ a 70cm
Hence, R/R’ = 100/70
R/R’= 10/7
Given that, the resistance of wire of length 100cm is 6 ohm i.e. R= 6ohm
Hence, R’= (7/10)*6= 4.2ohm
Thus, the resistance of wire of length 70 cm is 4.2 ohm.
b) When two resistors are connected in series, their effective resistance is 80 ohm. When they are connected in parallel their effective resistance is 20 ohm. What are the values of two resistances.
Ans:
Let a and b be the two resistors which are connected in series then their effective resistance is 80ohm.
Hence, a + b = 80
Thus, a= 80-b
And the resistors a and b are connected in parallel then their effective resistance is 20ohm.
ab/(a+b)= 20
But, a= 80- b
(80-b)*b= 20*(80-b+b)
80b – b2 = 1600
b2 -80b + 1600= 0
This is the quadratic equation which has two roots, by solving it we get two values of b= 40, 40.
But, a= 80- b = 80-40= 40ohm
Thus, the two resistors used here are of 40 ohm and 40 ohm.
c) If a charge of 420C flows through a conducting wire in 5minutes what is the value of the current?
Ans:
Given that,
Charge = 420C
Time = 5*60= 300 seconds
Hence, the current flowing is given by,
I= 420/300
Thus, I= 1.4 A
