Question

For the capacitor network shown in (Figure 1) , the potential difference across ab is 48 V .

Find the charge on the 150 nFcapacitor.

Find the charge on the 120 nFcapacitor.

Find the total energy stored in the network.

Find the energy stored in the 150nF capacitor.

Find the energy stored in the 120nF capacitor.

Find the potential difference across the 150 nF capacitor.

Find the potential difference across the 120 nF capacitor.

Answer 1

The equivalent capacitance when the capacitors are

connected in series is,

$$ C_{\mathrm{eq}}=\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)^{-1}=\left(\frac{1}{150 \mathrm{n} \mathrm{F}}+\frac{1}{120 \mathrm{nF}}\right)^{-1}=66.67 \mathrm{nF} $$

the charge on each capacitor remains the same becuase they

connected in series. Hence, the charge on each capacitor is,

$$ Q=C V=(66.67 \mathrm{nF})(48 \mathrm{~V})=3200 \mathrm{nC} $$

the charge on the \(150 \mathrm{nFcapacitor}\) is,

\(Q_{150}=3200 \mathrm{nC}\)

the charge on the \(120 \mathrm{nFcapacitor}\) is,

\(Q_{120}=3200 \mathrm{nC}\)

the total energy stored in the network is,

$$ U=\frac{Q^{2}}{2 C}=\frac{\left(3200 \times 10^{-9} \mathrm{C}\right)^{2}}{2\left(66.67 \times 10^{-9} \mathrm{~F}\right)}=76.8 \times 10^{-6} \mathrm{~J} $$

the energy stored in the \(150 \mathrm{nF}\) capacitor is,

$$ U=\frac{Q_{100}{ }^{2}}{2 C}=\frac{\left(3200 \times 10^{-9} \mathrm{C}\right)^{2}}{2\left(150 \times 10^{-9} \mathrm{~F}\right)}=34.1 \times 10^{-6} \mathrm{~J} $$

the energy stored in the \(120 \mathrm{nF}\) capacitor is,

$$ U=\frac{Q_{100}{ }^{2}}{2 C}=\frac{\left(3200 \times 10^{-9} \mathrm{C}\right)^{2}}{2\left(120 \times 10^{-9} \mathrm{~F}\right)}=42.67 \times 10^{-6} \mathrm{~J} $$

the potential difference across the \(150 \mathrm{nF}\) capacitor is,

$$ V_{100}=\frac{Q_{100}}{C_{100}}=\frac{3200 \times 10^{-9} \mathrm{C}}{150 \times 10^{-9} \mathrm{~F}}=21.33 \mathrm{~V} $$

the potential difference across the \(120 \mathrm{nF}\) capacitor is,

$$ V_{120}=\frac{Q_{120}}{C_{120}}=\frac{3200 \times 10^{-9} \mathrm{C}}{120 \times 10^{-9} \mathrm{~F}}=26.67 \mathrm{~V} $$