JEE Advanced · Physics · 18. Capacitance

Paragraph:

Consider a simple $R C$ circuit as shown in Figure $1$.

Process 1: In the circuit the switch $S$ is closed at $t=0$ and the capacitor is fully charged to voltage $V_{0}$ (i.e., charging continues for time $T>>R C$ ). In the process some dissipation $\left(E_{D}\right)$ occurs across the resistance $R$. The amount of energy finally stored in the fully charged capacitor is $E_{C}$.

Process 2: In a different process the voltage is first set to $\frac{V_{0}}{3}$ and maintained for a charging time $T>>R C$. Then the voltage is raised to $\frac{2 V_{0}}{3}$ without discharging the capacitor and again maintained for a time $T>>R C$. The process is repeated one more time by raising the voltage to $V_{0}$ and the capacitor is charged to the same final voltage $V_{0}$ as in Process 1.

These two processes are depicted in Figure $2 .$

Question:

In Process 2, total energy dissipated across the resistance $E_{D}$ is:

A $E_D=\frac{1}{2} C V_0^2$
B $E_D=3\left(\frac{1}{2} C V_0^2\right)$
C $E_D=\frac{1}{3}\left(\frac{1}{2} C V_0^2\right)$
D $E_D=3 C V_0^2$

Verified Solution

Answer & Solution

Correct Answer

(C) $E_D=\frac{1}{3}\left(\frac{1}{2} C V_0^2\right)$

Step-by-step Solution

Detailed explanation

For process (i)
Charge on capacitor

= \frac{C V_{0}}{3}

Energy stored in capacitor

= \frac{1}{2} C \frac{V_{0}^{2}}{9} = \frac{C V_{0}^{2}}{18}

Work done by battery

= \frac{C V_{0}}{3} \times \frac{V}{3} = \frac{C V_{0}^{2}}{9}

Heat loss

= \frac{C V_{0}^{2}}{9} - \frac{C V_{0}^{2}}{18} = \frac{C V_{0}^{2}}{18}

For process (ii)
Charge on capacitor

= \frac{2 C V_{0}}{3}

Extra charge flow through battery

= \frac{C V_{0}}{3}

Work done by battery:

\frac{C V_{0}}{3} . \frac{2 V_{0}}{3} = \frac{2 C V_{0}^{2}}{9}

Final energy store in capacitor:

\frac{1}{2} C {(\frac{2 V_{0}}{3})}^{2} = \frac{4 C V_{0}^{2}}{18}

Energy store in process 2:

\frac{4 C V_{0}^{2}}{18} - \frac{C V_{0}^{2}}{18} = \frac{3 C V_{0}^{2}}{18}

Heat loss in process (ii) = work done by battery in process (ii) – energy store in capacitor process (ii)

= \frac{2 C V_{0}^{2}}{9} - \frac{3 C V_{0}^{2}}{18} = \frac{C V_{0}^{2}}{18}

For process (iii)
Charge on capacitor

= C V_{0}

Extra charge flow through battery:

C V_{0} - \frac{2 C V_{0}}{3} = \frac{C V_{0}}{3}

Work done by battery in this process:

(\frac{C V_{0}}{3}) (V_{0}) = \frac{C V_{0}^{2}}{3}

Find energy store in capacitor:

\frac{1}{2} C V_{0}^{2}

Energy stored in this process:

\frac{1}{2} C V_{0}^{2} - \frac{4 C V_{0}^{2}}{18} = \frac{5 C V_{0}^{2}}{18}

Heat loss in process (iii):

\frac{C V_{0}^{2}}{3} - \frac{5 C V_{0}^{2}}{18} = \frac{C V_{0}^{2}}{18}

Now total heat loss

(E_{D}) : \frac{C V_{0}^{2}}{18} + \frac{C V_{0}^{2}}{18} + \frac{C V_{0}^{2}}{18} = \frac{C V_{0}^{2}}{6}

Final energy store in capacitor:

\frac{1}{2} C V_{0}^{2}

So we can say that

E_{D} = \frac{1}{3} (\frac{1}{2} C V_{0}^{2})

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

	List-I		List-II
(I)	$10^{- 3} kg$ of water $100 ° C$ is converted to steam at the same temperature, at a pressure of $10^{5} Pa$ . The volume of the system changes from $10^{- 6} m^{3}$ to $10^{- 3} m^{3}$ in the process. Latent heat of water $= 2250 kJ {kg}^{- 1}$	(P)	$2 kJ$
(II)	$0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 K$ undergoes an isobaric expansion to volume $3 V$ . Assume $8.0 J {mol}^{- 1} K^{- 1}$	(Q)	$7 kJ$
(III)	One mole of a monoatomic ideal gas is compressed adiabatically from volume $V = \frac{1}{3} m^{3}$ and pressure $2 kPa$ to volume $\frac{V}{8}$ .	(R)	$4 kJ$
(IV)	Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 kJ$ of heat and undergoes isobaric expansion.	(S)	$5 kJ$
		(T)	$3 kJ$

Which one of the following options is correct?

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