Question

2. The capacitors have values C₁ = 2.0 μF and C₂ = 4.0 μF, C₂ = 5.0 μF C4 = 7.0 μF and the potential difference across the battery is 9.0 V. Assume that the capacitors are connected in series.

a) Find the equivalent capacitance of the circuit.

b) Solve for the potential difference across each capacitors.

3. Given the same problem in #2. Assume that the capacitors are connected in parallel.

a) Find the equivalent capacitance of the circuit.

b) Solve for the charge across each capacitors.

Answer 1

(a)

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}$

$\frac{1}{C}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}$

$\frac{1}{C}=\frac{70+35+28+20}{140}$

$C=\frac{153}{140}=1.09\mu F$

(b)Charge

$Q=CV$

$Q=1.09\times10^{-6}\times9=9.84\mu C$

$V_1=\frac{9.84\times10^{-6}}{2\times10^{-6}}=4.92V$

$V_2=\frac{9.84\times10^{-6}}{4\times10^{-6}}=2.46V$

$V_3=\frac{9.84\times10^{-6}}{5\times10^{-6}}=1.968V$

$V_4=\frac{9.84\times10^{-6}}{7\times10^{-6}}=1.40V$

3(a)

$C_{eq}=C_1+C_2+C_3+C_4$

$$C_{eq}=(2+4+5+7)\times10^{-6}=18\times10^{-6}C$$

(b)

Charge

$$Q_1=C_1V\\Q_1=2\times9\times10^{-6}C=18\mu C$$

$Q_2=C_2V\\Q_2=4\times9\times10^{-6}C=36\mu C$

$Q_3=C_3V\\Q_3=5\times9\times10^{-6}C=45\mu C$

$Q_4=C_4V\\Q_4=7\times9\times10^{-6}C=63\mu C$