Question

A series circuit is comprised of a battery with a voltage $v$, an open switch, solenoid with and inductance $L$ and a resistor with the resistance $R$.
At time $t=0$, the switch is closed, what is the current through the resistor immediately after the switch is closed?

• A. $0$
• B. $\frac{V}{R}$
• C. $\frac{V}{R}\left(1e^{\left(\right(R/L)t}\right)$
• D. $\frac{V}{R}\left(1e^{\left(\right(RLt)}\right)$

Answer 1

Answer: (A)

$0$

Just at the moment , when the switch is closed , current increases in inductor and due to Lenz's law an opposite current is induced in inductor which cancels the current given by battery , therefore total current (current in resistor) is zero just after the switch is closed , after some time applied voltage overcome the induced voltage and current starts flowing in circuit .