Chapter 3: Electricity

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Q
Electricity

Question:

Compare the power used in the 2 Ω resistor in each of the following circuits:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer:

(i) The circuit diagram is shown below.

Total resistance, R = 1 Omega + 2 Omega = 3 Omega

Potential difference, V = 6 V

Using Ohm’s law

V = IR

I = frac{V}{R}

I = frac{6}{3} = 2 A

I = 2 A

In series, the current does not divide, so current flowing through the 2 โ„ฆ resistance is 2 A.

Power used in 2 โ„ฆ resistor will be,

P = I^2R

P = (2)^2 	imes 2

P = 4 	imes 2 = 8 W

P = 8 W

 

(ii) The circuit diagram is shown below.

Potential difference, V = 4 V

In parallel, the voltage remains the same for each component connected. So, the voltage across 2 โ„ฆ resistance is 4 V.

Power used in 2 โ„ฆ resistor will be,

P = frac{V^2}{R}

P = frac{(4)^2}{2}

P = frac{16}{2} = 8 W

P = 8 W


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