Answer:
The rate of current change in the circuit is [tex]\frac{dI}{dt} = 1500 A/s[/tex]
Explanation:
From the question we are told that
The capacitor has a value [tex]C = 80 \mu F[/tex]
The inductor has a value [tex]L = 0.020 H[/tex]
The capacitors voltage is [tex]V_c = 50V[/tex] at t = 0s
The new capacitor voltage is [tex]V_c__{n}} = 30 V[/tex]
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
[tex]\frac{dI}{dt} = \frac{V_l}{L}[/tex]
Where [tex]V_l[/tex] is the voltage across the inductor. [tex]V_l = V_c__{n}}[/tex]
Substituting value
[tex]\frac{dI}{dt} = \frac{30}{0.020}[/tex]
[tex]\frac{dI}{dt} = 1500 A/s[/tex]
