Question

You connect a battery, resistor, and capacitor as in (Figure 1) , where

E = 56.0 V , C = 5.00 μF , and R = 130 Ω . The switch S is closed at t = 0.

(A) When the voltage across the capacitor is 8.00 V, what is the magnitude of the current in the circuit?

(B) At what time t after the switch is closed is the voltage across the capacitor 8.00 V?

(C) When the voltage across the capacitor is 8.00 V, at what rate is energy being stored in the capacitor? (Solve for P = ?)

Answer 1

Apply kirchhoff's voltage law to the circuit, we get dt C The current in the circuit is defined as dt The voltage 1s deline as Hence, the above equaiton becomes as Rearrange the equati on for I E -V Hence, The magnitude of the current in the circuitis 56.0 V-8.0 V 130 S2 0.37 A The charge at time t is expressed as follows: 2-Ro(1-c) Rewrite the equation for t. -t/Rc Plug the values and solve it for t 80 V t=-(130 Ω)(5.00x10-6 F)In| 1-an 56.0 V = 100.2×10-6 s 0.1002 ms The rate of stored energy is given by -(8 V) (0.37 A) -2.96 Watts