question_answer

In this circuit, when certain current flows, the heat produced in \[5\Omega \], is \[4.05\text{ }J\]in a time t. The heat produced in \[2\Omega \]. coil in the same time interval is

A)  \[5.76\]   

B)  \[1.44\]

C)  \[2.88\]  

D)  \[2.02\]

Correct Answer: C

Solution :

: As per question, Heat produced in \[5\Omega \]. in time t is \[4.05=I_{1}^{2}5\] \[{{I}_{1}}=\sqrt{\frac{4.05}{5}}=0.9A\] Potential difference across B and C is \[=5{{I}_{1}}=5\times 0.9V=4.5V\] \[{{I}_{2}}=\frac{4.5V}{15\Omega }=0.3A\] \[\therefore \] \[I={{I}_{1}}+{{I}_{2}}=0.9A+0.3A=1.2A\] Heat produced in \[2\Omega \] in same time interval t is \[={{(1.2A)}^{2}}(2\Omega )=2.88J\]