Question

Two capacitors are connected to a battery. The battery voltage is V = 90 V, and the capacitances are C1 = 2.00 μF and C2 = 4.00 μF. Determine the total energy stored by the two capacitors when they are wired (a) in parallel and (b) in series.

Answer 1

Answer:$E=24.3\times 10^{-3} J$

$E=5.4\times 10^{-3} J$

Explanation:$E=24.3\times 10^{-3} J$

$E=5.4\times 10^{-3} J$

Given

$C_1=2 \mu F$

$C_2=4\mu F$

Voltage$\left ( V\right )=90 V$

$\left ( a\right ) parallel$

$C_{eq}=C_1+C_2=2+4=6\mu F$

Energy stored in capacitor$\left ( E\right )=\frac{1}{2}C_{eq}V^2$

$E=\frac{1}{2}\times 6\times 10^{-6}\times 90^2$

$E=24.3\times 10^{-3} J$

$\left ( b\right )Series$

$C_{eq}=\frac{C_1C_2}{C_1+C_2}$

$C_{eq}=\frac{4}{3} \mu F$

Energy stored=$\frac{1}{2}C_{eq}V^2$

E=$\frac{1}{2}\times \frac{4}{3}\times \left ( 90\right )^2$

E=$5.4\times 10^{-3} J$