Question

Three capacitors $C_1,C_2$ and $C_3$ whose values are $10\mu F,5\mu F$ and 2$\mu F$ respectively, have breakdown voltages of 10 V, 5 V and 2 V respectively. For the interconnection shown below, the maximum safe voltage in volts that can be applied across the combination, and the corresponding total charge in $\mu C$ stored in the effective capacitance across the terminals are, respectively

• A. 2.8 and 36
• B. 7 and 119
• C. 2.8 and 32
• D. 7 and 80

Answer 1

The correct option is C 2.8 and 32

Q = CV

$Q_1=C_1{V}_1=10\times {10}^{-6}\times 10$

= 100 $\mu C$

$Q_2=C_2{V}_2=5\times {10}^{-6}\times 5$

= 25 $\mu C$

$Q_3=C_3{V}_3=2\times {10}^{-6}\times 2$

= 4 $\mu C$

Capacitors $C_2$ and $C_3$ are in series.

In series charge is same.

So, the maximum charge on $C_2$ and $C_3$ will be minimum of $(Q_2,Q_3)$ = $min(25\mu C,4\mu C)$

$=4\mu C=Q_{23}$

in series the equivalent capacitance of $C_2$ and $C_3$ is

$C_{23}=\frac{C_2{C}_3}{C_2+C_3}=\frac{5\times 2}{5+2}=\frac{10}{7}\mu F$

So, the equivalent voltage

$V_{23}=\frac{Q_{23}}{C_{23}}=\frac{4\times {10}^{-6}}{\frac{10}{7}\times {10}^{-6}}=\frac{28}{10}=2.8V$

In parallel, the voltage is same.

$V_1=V_{23}=2.8V$

Charge in capacitor $C_1$

$Q_1=C_1{V}_1$

$=10\times {10}^{-6}\times 2.8=28\mu C$

In parallel, the total charge

$Q=Q_1+Q_{23}$

$Q=4+28$

$Q=32\mu C$