Question

A total of Q charge flows from resistance R in time interval T such that current vs time graph is obtained as shown in figure. The heat generated in resistance is:

• A. $I_0^{2}RT$
• B. $\frac{I_0^{2}RT}{3}$
• C. $\frac{I_0^{2}RT}{2}$
• D. $\frac{I_0^{2}RT}{4}$

Answer 1

The correct option is B $\frac{I_0^{2}RT}{3}$

Area under $I-t$ graph is charge.

So, $\frac{1}{2}\times T\times I_0=Q\Rightarrow I_0=\frac{2Q}{T}$
$H={\int }_0^T{I}^{2}Rdt=2{\int }_0^{\frac{T}{2}}(\frac{2I_{0}t}{T}{)}^{2}Rdt$
$H=\frac{I_0^{2}RT}{3}$