An RC circuit has an R of 10 kΩ and C = 0.05 μF. The applied voltage for charging is 36 V. How long will it take C to charge to 24 V?

Concept:

The expression for capacitor voltage is given by;

V_C = V(1-e^-t/τ)

Where;

V_C → Capacitor voltage

V→ Voltage at steady state(Applied voltage)

τ → Time constant (RC)

Calculation:

Given;

R = 10 KΩ 

C= 0.05 μF

V_C = 24 V

V = 36 V;

Then;

     VC = V(1-e^-t/τ)

⇒ t = -τln(1 - V_C/V) = -(10× 10³×0.05 × 10^-6)ln(1 - 24/36) = 0.549 msec