Question

5,6,8 plz will rate :)

Use the RC circuit below to answer the following questions. class-0 In the RC circuit, the emf 8 = 12.0 V, resistance R=1.40 MS2, and capacitance C=1.80 uF. 5. What is the current in the circuit the instant the switch is closed? 6. What is the initial charge on the capacitor? 7. What is the time constant for the RC circuit? 8. How much charge is on the capacitor after on time constant?

Answer 1

5) The voltage across the capacitor is 0 V when t=0 sec.

The current in the circuit will be = (V/R) = 12/(1.4×10^3) = 8.57×10^(-3) Amp (approx)

6) Initially the capacitor doesn't have any charge. As time goes on the charge is accumulated in the capacitor.

So, initial charge on the capacitor is 0 C

8)We know,

Q = C.V.[1-e^(-t)]

here, Q = charge on the capacitor.

C =capacitance of the capacitor= 1.8 F

V = supplied voltage= 12 V

t = number of time constant= 1.

Now, Q= 1.8×12 [1- e^(-1)]

=> Q = 13.654 C

The charge on the capacitor after one time constant is 13.654 C.

Please comment if you have any doubt and like if it helps.

Happy learning.