In the following circuit, $ 5 \Omega $ resistor develops $ 45 \mathrm{~J} \mathrm{~s}^{-1} $ due to current flowing through it. The power developed per second across $ 12 \Omega $ resistor is:

Question

In the following circuit, $ 5 \Omega $ resistor develops $ 45 \mathrm{~J} \mathrm{~s}^{-1} $ due to current flowing through it. The power developed per second across $ 12 \Omega $ resistor is:, $ 15 \mathrm{~W} $, $ 192 \mathrm{~W} $, $ 36 \mathrm{~W} $, $ 64 \mathrm{~W} $

MARKS App · Accepted Answer

From the formula of power in 5 Ω , P = i 1 2 R ⇒ i 1 = 45 5 = 3 A In parallel combination, potential difference across resistors remain same. Hence, I 1 R 1 = I 2 R 2 ⇒ 3 × 5 = i 2 × ( 9 + 6 ) ⇒ i 2 = 1 A From KCL, i = i 1 + i 2 ⇒ i = 3 + 1 = 4 A Hence, power developed in 12 Ω resistance is, P = I 2 R = ( 4 ) 2 × 12 = 192 W

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