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A current of `3 A` flows through the `2 Omega` resistor as shown in the circuit. The power dissipated in the `5 Omega` resistor is A. `4 W`B. `2 W`C. `1 W`D. `5 W` |
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Answer» Correct Answer - D (d) Voltage across `2 Omega` is the same as voltage across arm containing `1 Omega` and `5 Omega` resistance. Voltage across `2 Omega` resistance, `V_(1) = 6 V` Current across `5 Omega, I = (6)/(1 + 5) = 1 A` Thus, power across `5 Omega`, `P = I^(2) R = (1)^(2) xx 5 = 5 W` |
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