A \(160 - \mu F\)capacitor charged to \(450\;V\)is discharged through a \(31.2 - k\Omega \)resistor. (a) Find the time constant.(b) Calculate the temperature increase of the resistor, given that its mass is \(2.50\;g\)and its specific heat is \(1.67\frac{{kJ}}{{kg{ \cdot ^\circ }C}}\), noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance seem significant?

RC time constant: This measurement informs us how long it will take a cap to charge to a particular voltage level.

Anything that prevents current flow is referred to as "resistance" in this context.

Capacitor=\(160 - \mu F\).

Resistor=\(31.2 - k\Omega \).

Mass = \(2.50g\).

Specific heat=\(1.67\frac{{kJ}}{{kg{ \cdot ^\circ }C}}\).

(a)

Let us calculate the time constant.

\(\begin{align}{}\tau & = RC\\ & = 31.2 \cdot {10^3} \cdot 160 \cdot {10^{ - 6}}\\\tau & = 4.99\;s\end{align}\)

Hence, the time constant is \(4.99\;s\).

(b)

Let us calculate the energy in the circuit.

\(\begin{align}{}E& = \frac{1}{2}C{U^2}\\ & = \frac{1}{2}160 \cdot {10^{ - 6}} \cdot {450^2}\\& = 16.2\;J\end{align}\)

And all of that energy is transferred into an increase in temperature.

\(\begin{align}{}\Delta T& = \frac{E}{{m{c_p}}}\\ & = \frac{{16.2}}{{2.5 \cdot 1.67}}\\\Delta T& = {3.9^\circ }C\end{align}\)

Hence, the total energy in the circuit is \(\Delta T = {3.9^\circ }C\).

(c)

Let us calculate the new resistance.

\(\begin{align}{}{R_n} = R + ( - 0.025 \cdot \Delta T)\\ = 31.2 - (0.025 \cdot 3.9)\\{R_n} = 31.1{\rm{ }}k\Omega \end{align}\)

Hence, the new resistance is \(31.1{\rm{ }}k\Omega \).

(d)

Change in the resistance does not seem significant.