Question

A battery of voltage V_0 is hooked up to a network of 4 capacitors, connected together as shown below. The values are as follows, C_1 = 9.0 nF, C_2 = 4.0 nF, C_3 = 2.0 nF, C_4 = 3.0 nF, and V_0 = 9.0 V. Determine the voltage across C_4

Answer 1

Given data

C1 C2 C3 C4

$C_{1}=9nF$

$C_{2}=4nF$

$C_{3}=2nF$

$C_{4}=3nF$

$V_{0}=9volt$

Equivalent capacitance between point C and D can be given by

$C_{CD}=C_{3}+C_{4}=(2+3)nF=5nF$

Equivalent capacitance between A and B can be given by

$C_{AB}=C_{CD}+C_{2}=(5+4)nF=9nF$

Equivalent capacitance of the total circuit can be given by

$C=\frac{C_{AB}\times C_{1}}{C_{AB}+C_{1}}=\frac{9\times 9}{9+9}nF=4.5nF$

Total charge can be given by

$Q=CV_{0}=(4.5\times 10^{-9}\times 9)C=40.5\times 10^{-9}C$

Voltage accross A and B can be given by

$V_{AB}=\frac{Q}{C_{AB}}=\frac{40.5\times 10^{-9}}{9\times 10^{-9}}volt=4.5volt$

Voltage accross $C_{4}$ can be given by

$V_{C_{4}}=4.5volt$