Question

2Ω Four resistors and a capacitor are connected to an 18 V battery with negligible internal resistance, as shown on the diagram. Initially the capacitor is disconnected from the battery- the switch is open Calculate the net resistance of the circuit. a) b) Calculate the current in the 2-Ω resistor. Calculate the current in the 3-Ω resistor. c) The switch is then closed and the current reaches a constant value. d) Calculate the charge on the capacitor. e) Calculate the energy stored in the capacitor.

Answer 1

a)

9 and 3 ohms are in series

R₁ = 9+3 =12 ohms

6 and 12 ohms are in parallel

1/R₂ = 1/6+1/12

R₂ = 4 ohms

2 and R2 are in series ,therefore net resistance of the circuit is18/

R_net=2+4 =6 ohms

b)

current flowing through 2 ohms is

I=18/6 =3 A

c)

Current through 3 ohms

I₃ = 3*(6/(6+12)) =1 A

d)

Voltage across capacitor

V_C = 1*9 =9 V

Charge on the capacitor

Q=CV=2*9 =18 uC or 1.8*10^-5 C

e)

Energy stored in capacitor

U=(1/2)CV_C² =(1/2)*(2*10^-6)*9²

U=8.1*10^-5 J or 81 uJ