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  4. In the figure shown the battery is ideal. The values are ϵ =10V,R=5Ω,L=2H . The current through the battery at t=2s after closing the switch is: <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-o1re09gbp4z1zp4h.jpg />

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Q. In the figure shown the battery is ideal. The values are $\epsilon =10V,R=5\Omega,L=2H$ . The current through the battery at $t=2s$ after closing the switch is:
Question

NTA AbhyasNTA Abhyas 2022

Solution:

From inductor, potential difference $V_{L}=L\frac{d i_{L}}{d t}\Rightarrow 10=2\frac{d i_{L}}{d t}$
Current through inductor, $i_{L}=5t$
At $t=2s$ , $i_{L}=10A$
Current through resistance, $i_{R}=2A$
Current through battery, $i=i_{L}+i_{R}=12A$