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. (ii) A 4 V battery in parallel with 12 Ω and 2 Ω resistors.

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

(ii) A 4 V battery in parallel with 12 Ω and 2 Ω resistors

Answer:

(i) Potential difference V₁ = 6V ; Resistance R₁= 1Ω and R₂= 2Ω

As per ohm's law

V=IR

P=I²R

Equivalent Resistance= R₁+R₂ (connected in series)

R = 1 + 2

R= 3Ω

As per ohm's law

V =IR

I = V₁ /R

I = 6/3

I = 2A

Since there is no division in the series circuit, the current will pass through every component. 2Ω resistor has a current of 2A will flow through it as a result.

Since the current remains constant

P₁ = I²R

P₁ = 2² X 2

P₁ = 8 W

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(ii) Potential difference V₂ = 4V ; Resistance R₃= 1Ω and R₄= 2Ω

Potential difference V₂ = 4V

Since it is a parallel circuit, the voltage will remain constant. The voltage across 2Ω resistor will be 4V.

P₂ = (V₂)²/R

P₂ = (4)²/2

P₂ = 8 W

Power used by both circuits is 8 W.

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