Question

1. Your are given the circuit below which contains four resistors R1 = 1.0 . , R2 = 4.0 . , R3 = 4.0 . , R4 = 5.0 . , and one capacitor C1 = 2.0 µF. (a) What is the current through the resistor R4? (b) What is the potential difference across the capacitor? (c) What is the charge on the capacitor? (d) What is the power dissipated by resistor R4?

Answer 1

You have to apply Kirchhoff's rules to find the current and potential difference. Let's take first loop:
R4
C1
V
R1
R3
R2
I1
I3
I2
I1
I1
fully charged capacitor-no
current going through it

we have that:
V=I1R1 + I3R3 + I1R4
second loop:
I2R2=I3R3
and we know that
I1=I2+I3
If you solve that for I1 you will have the current going through resistor R4
to find the potential difference across the capacitor and its charge we also apply K's rule
We have then:
q/C=I1R4
solve for q , and potential difference is simply V=q/C
Power dissipated by resistor 4:
P=(I1)^2*R