A charged capacitor is discharged through a resistance. The time constant of the circuit is $\eta $. Then the value of time constant for the power dissipated through the resistance will be:
A. $\eta $
B. $2\eta $
C. $\dfrac{\eta }{2}$
D. Zero

Hint: When a charged capacitor is connected through a circuit that includes a resistor then, the electric current flows into the circuit and gets dissipated in the form of heat generated in the resistor.
One time constant is defined as the time taken to discharge a completely charged capacitor to get discharged to approximately 37% of the initial charge on the capacitor.

Complete step by step solution:
The time constant for the given $RC$ circuit is $\eta $
The electric current in the circuit during discharging of the capacitor is given by
$i={{i}_{o}}{{e}^{-\dfrac{t}{\eta }}}$
Where,
$i=$Electric current in the circuit
${{i}_{o}}=$Maximum electric current in the circuit
$t=$Time spent after switch is closed
$\eta =$Time constant of the RC circuit
If $i$ is electric current flowing in the resistor of resistance $R$
Then power generated in the resistance is given by formula,
$P={{i}^{2}}R$
Putting the value of electric current in the circuit, we get
\[\begin{align}
  & P={{\left( {{i}_{o}}{{e}^{-\dfrac{t}{\eta }}} \right)}^{2}}R \\
 & =\left( i_{o}^{2}R \right){{e}^{-\dfrac{2t}{\eta }}}
\end{align}\]
Putting $i_{0}^{2}R={{P}_{o}}$ the equation becomes,
$P={{P}_{o}}{{e}^{-\dfrac{t}{\left( \dfrac{\eta }{2} \right)}}}$
Putting $\left( \dfrac{\eta }{2} \right)=\eta '$
Then power consumed can be given as,
$P={{P}_{o}}{{e}^{-\dfrac{t}{\eta '}}}$

The time constant of power consumed in resistance is $\dfrac{\eta }{2}$.

Note: During discharging of the capacitor, the charge on the capacitor after time equal to one time constant is equal to 37% of the initial charge.
During charging of the capacitor, the charge on the capacitor after time equal to one time constant is equal to 63% of the maximum charge.