question_answer

In the following circuit, 5 W resistor develops 45 J/s due to current flowing through it. The power developed per second across 12 W resistor is                                        [AMU (Engg.) 1999]

A)            16 W

B)            192 W

C)            36 W

D)            64 W

Correct Answer: B

Solution :

                   \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{15}{5}=\frac{3}{1}\]                                    ? (i) Also \[\frac{H}{t}={{i}^{2}}R\Rightarrow 45={{({{i}_{1}})}^{2}}\times 5\] Þ \[{{i}_{1}}=3\,A\] and from equation (i) \[{{i}_{2}}=1\,A\] So \[i={{i}_{1}}+{{i}_{2}}=4\,A\] Hence power developed in \[12\,\Omega \] resistance \[P={{i}^{2}}R={{(4)}^{2}}\times 12=192\,W\]