# Chhattisgarh State Class 10 Science Chapter 6 Solution

Chhattisgarh State Board Class 10 Science Chapter 6 Electric Current and the Circuit Exercise Multiple Choice, Fill in the Blanks, Questions and Answers here.

## Chhattisgarh State Class 10 Science Chapter 6 Solution

1) Choose the correct option:

(i) Which one of the options given below is a good conductor of electricity–

(a) Wood piece (b) Sheet of paper (c) Copper wire (d) glass rod.

Ans: – (c) Copper wire.

(ii) The instrument used to measure current in a circuit is–

a) voltmeter (b) ammeter (c) odometer (d) none of these.

Ans: – (b) ammeter.

(iii) The relation between V and I for a conducting wire is–

(a) a variable ratio (b) a fixed ratio

(c) sometimes variable sometime fixed (d) none of these.

Ans: – (a) a variable ratio.

(iv) The resistance of an electrical instrument is 2.2 ohm and potential difference is 220volts. Find out the value of current on connecting the instrument to a electric source–

a) 5 A (b) 20 A (c) 25 A (d) 100 A.

Ans: –  (d) 100A.

(v) The equivalent resistance of the following diagram would be–

(a) 25 ohm (b) 20 ohm (c) 10 ohm (d) 35 ohm.

Ans: – (b) 20 ohms.

2) Fill in the blanks–

(i) The potential difference of resistors connected in a series would be … different ………………….

(ii) In resistors connected in a parallel combination, the current flowing across each resistor would be ………different……………..

(iii) We show the consumption of electricity in Kwh or in ……….no of units………………….

(iv) The filament used in electric bulbs is made of …..tungsten…………….. metal.

(v) The ……….. resistance…………….. of a conductor increases with temperature.

3) Find out the equivalent resistance of the circuit given along side– 4) What is a fuse and of what material is it made?

Ans: – The fuse wire has low melting point which melted when the current in the circuit suddenly rises and by this the remaining electrical equipment saved from shot circuit. Metal alloy like silver, aluminium, lead is used for making fuse wire.

5) Explain the Ohm’s law in your own words and establish the relation between V & I by drawing a graph.

Ans: – The current in a wire is directly proportional to the potential difference between two points of the wire if temperature and other condition are constant. 6) If an electric bulb consumes 2400 J energy in a minute, calculate the power of the bulb?

Ans: – The power of the bulb is p and we know the power is the ratio of consumed electricity to the time.

So, power (p) = 2400/60 = 40 Watt.

7) An electric heater has 3 kw, 220 v, stamped on it. Find values of the following–

(a) Electric current (b) Resistance of the heater

(c) Cost of electricity if unit cost is Rs. 1.50/kwh and heater is used for 10 hours.

Ans: – (a) As we know that the power = voltage × current.

So, 3kw = 220v × I.

Or, I = 3000/ 220 = 13.63A.

b) From the formula voltage = current × resistance;

So, resistance (r) = 220/13.63;

Or, r = 16.13 ohm.

c) Cost of electricity =power × time × rupees

Or, cost of electricity = 3× 10 × 60 × 1.50 = 2700.

Then the cost of the electricity is 2700 Rs.

(i) Why are domestic circuits not arranged in a series combination?

(ii) Why is only tungsten metal used to make filaments of bulbs?

Ans: – (i) The domestic circuit are not arranged in series combination because if any one equipment is not gotten current then the other also doesn’t get. And by chance if short circuit is happening then all the equipment will affect so they’re connected in parallel.

(ii) The tungsten metal has some special quality that differs it from the other. The tungsten has high resistivity by which it can passes high amount of electricity. And it has high melting point also.

9) If a current of 0.5 A flows for 20 minutes in a filament of a bulb, then what would be the quantity of electrical charge flowing in the circuit?

Ans: – We know that, i = q/ t ;

Here q = electrical charge = ? ; i = current = 0.5A ; and t = time = 20 min = 20×60 = 1200 sec.

So, q = it = 0.5 × 1200 = 600 Columb.

So, the electrical charge in the circuit will be 600C.

10) In a house, 4 electric bulbs of 40 watt burn for 5 hours, 2 bulbs of 60 watts for 6 hours and 3fans of 80 watts run for 6 hours, daily. What would be the cost of electricity, for one month of that home, if the rate is 50 paisa per unit?

Ans: – 11) Discuss the following questions –

(i) What would be the effect on the circuit if one bulb out of three bulbs arranged in a series combination becomes fused?

Ans: – If any one bulb in a series arrangements become fused then the other bulb will not glow. As the current will not passes to the other bulb so we have to replace the fuse bulb to a new one for getting the glow.

(ii) What would be the effect if one bulb out of three arranged in parallel combination malfunctions?

Ans: – In a parallel connection the current is passes to every circuit is same so if any one gets disconnected the remaining will be glow. In this case if one bulb out of the three then the remaining two will glowing until they disconnected from the connection.

12) Find out the equivalent resistance between A & B in the circuit diagram given below?

Ans: – 13) If the potential difference generated by a battery across a circuit of two bulbs, combined in series, is 6 volts, and the potential difference at the first bulb is 2 volts, find out the potential difference at the second bulb?

Ans: – Let the potential difference of the second bulb is x.

As we know the total potential difference across the two bulb is 6.

So, total PD = potential difference at first bulb + potential difference in second bulb;

Or, 6 = 2 + x;

X = 4 volt.

14) If two conducting wires of the same length and cross section are combined first in a series and then in a parallel combination, what would be the ratio between the quantity of heat generated in the two series and parallel circuits?

Ans: – 15) A 9-volt battery is arranged in series combination with resistors of 2ohm, 3 ohm, 4 ohm, 5 ohm and 12 ohm, what would be the current flowing across the 2 ohm resistor?

Ans: – The equivalent resistance of the circuit is R.

So, R = 2 + 3 + 4 + 5 + 12 = 26;

As we know from ohms law, V = IR;

Here v = 9V; I =? ; R = 26 ohm.

So,I = 9/ 26 = 0.34 A.

As all are connected is series so the current passing through each resistor be same so the current passing through 2 ohm is 0.34A.

16) List the various appliances working on the thermal effect produced by the electric current, which are used in our homes, and discuss their working?

Ans: – The various appliances working on the thermal effect produced by the electric current which are used in our homes is heater, mixer grinder, Washing machine, pressure cooker etc. The heater is used for warming the water in colder place, in which a metal is tempted by using electricity. In a mixture grinder a single-phase moto is connected which blade which cuts the vegetables.

17) On a bulb, 200 V – 100 W is printed. What would be the resistance of the bulb? If 5 of these bulbs burn for 4 hours, what would be the amount of electricity consumed and the cost of electricity, if the rate is 50 paisa per unit?

Ans: – As we all know that the resistance R and voltage V and power P is related by the formula P= V^2 / R;

Here P = 100W;V = 200 v; R =?; R = 200^2 / 100 =  4 ohm.

The electricity consumes is E  .

So, E = Power × no of bulb × time = 100 × 5 × 4 = 2000 wh.

Then the electricity consume by the bulb is 2 Kwh.

The cost of the electricity = 2000 × 50/100  = 1000 Rs.

18) How would you combine 3 resistances of 2 ohm, 3 ohm, and 5 ohm, so that the equivalent resistance is 2.5 ohm. Draw a diagram to show the circuit.

Ans: – Updated: December 25, 2021 — 6:43 pm