roviow constants periodic table par a the capacitance of pontion ofe circuit is t0 de reduced fr

Step 1: Use the formula for calculating capacitance in series: ( frac{C1 times C3}{C1 + C3} =

Roviow constants periodic table par a the capacitance of...

(a) Let, at any moment of time, charge on the plates be $\left(q_{0}-q\right)$ then current through the resistor, $i=-\frac{d\left(q_{0}-q\right)}{d t}$, because the capacitor is discharging. or, $i=\frac{d q}{d t}$ Now, applying loop rule in the circuit, $i R-\frac{q_{0}-q}{C}=0$ or, $\frac{d q}{d t} R=\frac{q_{0}-q}{C}$ or, $\frac{d q}{q_{0}-q}=\frac{1}{R C} d t$ At $t=0, q=0$ and at $t=\tau, q=q$ So, $$ \ln \frac{q_{0}-q}{q_{0}}=\frac{-\tau}{R C} $$ Thus $$ q=q_{0}\left(1-e^{-\tau / R C}\right)=0 \cdot 18 \mathrm{mC} $$ (b) Amount of heat generated $=$ decrement in capacitance energy $$ \begin{gathered} =\frac{1}{2} \frac{q_{0}^{2}}{C}-\frac{1}{2} \frac{\left[q_{0}-q_{0}\left(1-e^{-\tau / R C}\right)\right]^{2}}{C} \\ =\frac{1}{2} \frac{q_{0}^{2}}{C}\left[1-e^{-\frac{2 \tau}{R C}}\right]=82 \mathrm{~mJ} \end{gathered} $$