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:

Question

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" /> (A) 12A (B) 7A (C) 3A (D) None of these

Tardigrade Admin · Accepted Answer

From inductor, potential difference VL=L(d iL/d t)⇒ 10=2(d iL/d t) Current through inductor, iL=5t At t=2s , iL=10A Current through resistance, iR=2A Current through battery, i=iL+iR=12A

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

$12 A$

$7 A$

$3 A$

None of these

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, $i_{L} = 5 t$
At $t = 2 s$ , $i_{L} = 10 A$
Current through resistance, $i_{R} = 2 A$
Current through battery, $i = i_{L} + i_{R} = 12 A$

Q. 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:

12A

7A

3A