A resistor with \(R = 850\;\Omega \) is connected to the plates of a charged capacitor with capacitance \(C = 4.62\;{\rm{\mu F}}\). Just before the connection is made, the charge on the capacitor is \(6.90\;{\rm{mC}}\).

(a) What is the energy initially stored in the capacitor?

(b) What is the electrical power dissipated in the resistor just after the connection is made?

(c) What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part (a)?

Resistance of the resistor:

\(R = 850\;\Omega \)

Capacitance of the capacitor:

\(\begin{aligned}C = 4.62\;{\rm{\mu F}}\\ = {\rm{4}}{\rm{.62}} \cdot \left( {1\;{\rm{\mu F}} \times \frac{{1\;{\rm{F}}}}{{{{10}^6}\;{\rm{\mu F}}}}} \right)\\ = 4.62 \times {10^6}\;{\rm{F}}\end{aligned}\)

Initial charge of the capacitor:

\(\begin{aligned}{Q_0} = 6.90\;{\rm{mC}}\\ = 6.90 \cdot \left( {1\;{\rm{mC}} \times \frac{{1\;{\rm{C}}}}{{1000\;{\rm{mC}}}}} \right)\\ = 6.90 \times {10^{ - 3}}\;{\rm{C}}\end{aligned}\)

The energy stored in a capacitor of capacitance \(C\) and holding a charge \(Q\) is

\(E = \frac{{{Q^2}}}{{2C}}\) .....(i)

From equation (i), the initial energy stored in the capacitor is

\(\begin{aligned}E = \frac{{Q_0^2}}{{2C}}\\ = \frac{{{{\left( {6.90 \times {{10}^{ - 3}}\;{\rm{C}}} \right)}^2}}}{{2 \times 4.62 \times {{10}^6}\;{\rm{F}}}}\\ = 5.15 \cdot \left( {1\;{{{{\rm{C}}^2}} \mathord{\left/

{\vphantom {{{{\rm{C}}^2}} {\rm{F}}}} \right.

\kern-\nulldelimiterspace} {\rm{F}}} \times \frac{{1\;{\rm{J}}}}{{1\;{{{{\rm{C}}^2}} \mathord{\left/

{\vphantom {{{{\rm{C}}^2}} {\rm{F}}}} \right.

\kern-\nulldelimiterspace} {\rm{F}}}}}} \right)\\ = 5.15\;{\rm{J}}\end{aligned}\)

Thus, the required energy is \(5.15\;{\rm{J}}\).