Question

For the system of capacitors shown in the figure below, find the following. (Let C1 = 4.00 µF and C2 = 5.00 µF.)

(a) the equivalent capacitance of the system

(b) the charge on each capacitor

on C1 = ?

on C2 = ?

on the 6.00 µF capacitor = ?

on the 2.00 µF capacitor = ?

(c) the potential difference across each capacitor

across C1 = ?

across C2 = ?

across the 6.00 µF capacitor = ?

across the 2.00 µF capacitor = ?

Answer 1

PART (a):

The equivalent capacitance in upper branch (consisting of \(\mathrm{C}_{1}\) and 6 uF capacitor) is given by:

$$ C^{\prime}=\frac{4 \times 6}{4+6} \mu F=2.4 \mu F $$

The equivalent capacitance in lower branch (consisting of \(\mathrm{C}_{2}\) and 2 uF capacitor) is given by:

$$ C^{\prime \prime}=\frac{5 \times 2}{5+2} \mu F=10 / 7 \mu F $$

Therefore, the equivalent capacitance of the system is given by:

$$ \begin{array}{c} C_{e q}=C^{\prime}+C^{\prime \prime}=2.4 \mu F+10 / 7 \mu F \\ \mathbf{C}_{\mathrm{eq}}=3.829 \mu \mathrm{F} \rightarrow(\text { answer }) \end{array} $$

PART (b):

Charge on \(\mathrm{C}_{1}\) :

$$ \begin{aligned} Q_{C_{1}} &=C_{1} \cdot V_{C_{1}}=(4 \mu F) \times(54 V) \\ \therefore Q_{C_{1}} &=216 \mu C=2.16 \times 10^{-4} \mathrm{C} \end{aligned} $$

Charge on 6 uF capacitor:

$$ Q_{6}=(6 \mu F) \times(36 \mathrm{~V}) $$

$$ \therefore Q_{6}=216 \mu C=2.16 \times 10^{-4} \mathrm{C} $$

Charge on \(\mathrm{C}_{2}\) :

$$ \begin{aligned} Q_{C_{2}} &=C_{2} \cdot V_{C_{2}}=(5 \mu F) \times(25.7 \mathrm{~V}) \\ \therefore \mathrm{Q}_{\mathrm{C}_{2}} &=128.5 \mu \mathrm{C}=1.285 \times 10^{-4} \mathrm{C} \end{aligned} $$

Charge on 2 uF capacitor:

$$ Q_{2}=(2 \mu F) \times(64.3 \mathrm{~V}) $$

$$ \therefore Q_{2}=128.5 \mu C=1.285 \times 10^{-4} \mathrm{C} $$

PART (c):

Voltage across \(\mathrm{C}_{1}\) :

$$ \begin{array}{c} V_{C_{1}}=(90 \mathrm{~V}) \times\left(\frac{C^{\prime}}{C_{1}}\right) \\ \therefore V_{C_{1}}=(90 \mathrm{~V}) \times\left(\frac{2.4}{4}\right) \\ \therefore \mathbf{V}_{\mathrm{C}_{1}}=54 \mathrm{~V} \end{array} $$

Voltage across 6 uF capacitor:

$$ \begin{array}{c} V_{6}=(90 \mathrm{~V}) \times\left(\frac{C^{\prime}}{6 \mu F}\right) \\ \therefore V_{6}=(90 \mathrm{~V}) \times\left(\frac{2.4}{6}\right) \\ \therefore \mathrm{V}_{6}=36 \mathrm{~V} \end{array} $$

Voltage across \(\mathrm{C}_{2}\) :

$$ V_{C_{2}}=(90 \mathrm{~V}) \times\left(\frac{C^{\prime \prime}}{C_{2}}\right) $$

$$ \begin{array}{c} \therefore V_{C_{2}}=(90 \mathrm{~V}) \times\left(\frac{10 / 7}{5}\right) \\ \therefore \mathbf{V}_{\mathrm{C}_{2}}=25.7 \mathrm{~V} \end{array} $$

Voltage across 2 uF capacitor:

$$ \begin{aligned} V_{2}=&(90 \mathrm{~V})-(25.7 \mathrm{~V}) \\ \therefore \mathbf{V}_{2}=\mathbf{6 4 . 3} \mathbf{V} \end{aligned} $$