Question

For the given circuit the current i through the battery when the key in closed and the steady state has been reached is __________.

• A. 6 A
• B. 25 A
• C. 10 A
• D. 0 A

Hint:

At steady state an ideal inductor offers no resistance (acts like a plain wire), so the three \(3\,\Omega\) branches are simply in parallel.

Answer 1

The Correct Option is C

In the steady state \(\frac{dI}{dt} = 0\), so the induced emf \(L\frac{dI}{dt} = 0\) and each inductor (0.5 mH and 0.2 H) behaves like a plain conducting wire.
The three parallel branches then each have \(3\,\Omega\): \(\frac{1}{R_p} = \frac{1}{3}+\frac{1}{3}+\frac{1}{3} = 1 \Rightarrow R_p = 1\,\Omega\).
Total resistance with the \(2\,\Omega\) resistor in series: \(R = 2 + 1 = 3\,\Omega\).
\(i = \frac{30}{3} = 10\) A.
Answer: (3) 10 A