Question

LC Circuits: A series circuit contains an 80-μF capacitor, a 0.020-H inductor, and a switch. The resistance of the circuit is negligible. Initially, the switch is open, the capacitor voltage is 50 V, and the current in the inductor is zero. At time t = 0 s, the switch is closed. When the potential across the capacitor is 30 V, what is the rate of change of the current?

Answer 1

Answer:

The rate of current change in the circuit is $\frac{dI}{dt} = 1500 A/s$

Explanation:

From the question we are told that

     The capacitor has a value $C = 80 \mu F$

     The inductor has a value $L = 0.020 H$

     The capacitors voltage is  $V_c = 50V$ at t = 0s

     The new capacitor voltage is $V_c__{n}} = 30 V$

Generally when the switch is  closed the potential across the inductor is equal to the potential across the capacitor

So for the inductor

     rate of current change is given as

                     $\frac{dI}{dt} = \frac{V_l}{L}$

Where $V_l$ is the  voltage across the inductor.     $V_l = V_c__{n}}$

Substituting value

                    $\frac{dI}{dt} = \frac{30}{0.020}$

                     $\frac{dI}{dt} = 1500 A/s$