JEE Advanced · Physics · 18. Capacitance

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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 1 , the energy stored in the capacitor $E_{C}$ and heat dissipated across resistance $E_{D}$ are related by:

A $E_C=E_D$
B $E_C=E_D \ln 2$
C $E_C=\frac{1}{2} E_D$
D $E_C=2 E_D$

Verified Solution

Answer & Solution

Correct Answer

(A) $E_C=E_D$

Step-by-step Solution

Detailed explanation

When switch is closed for a very long time capacitor will get fully charged and charge on capacitor will be

q = C V

Energy stored in capacitor

E_{C} = \frac{1}{2} C V^{2}

.....(i)
Work done by battery

(W) = V q = V C V = C V^{2}

Dissipated across resistance

E_{D}

= (work done by battery) – (energy store)

E_{D} = C V^{2} - \frac{1}{2} C V^{2} = \frac{1}{2} C V^{2}

.....(ii)
From (i) and (ii)

E_{D} = E_{C}

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.

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From JEE Advanced

Let

f (x) = x + \log_{e} x - x \log_{e} x, x \in (0, \infty)

•

Column 1 contains information about zeros of

f (x)

f' (x)

and

f'' (x)

•

Column 2 contains information about the limiting behaviour of

f (x)

f' (x)

and

f'' (x)

at infinity.

•

Column 3 contains information about increasing-decreasing nature of

f (x)

and

f' (x)

Column 1	Column 2	Column 3
(I) $f (x) = 0$ for some $x \in (1, e^{2})$	(i) $\lim_{x \to \infty} f (x) = 0$	(P) $f$ is increasing in $(0, 1)$
(II) $f^{'} (x) = 0$ for some $x i n (1, e)$	(ii) $\lim_{x \to \infty} f (x) = - \infty$	(Q) $f$ is decreasing in $(e, e^{2})$
(III) $f^{'} (x) = 0$ for some $x \in (0, 1)$	(iii) $\lim_{x \to \infty} f^{'} (x) = - \infty$	(R) $f^{'}$ is increasing in $(0, 1)$
(IV) $f^{''} (x) = 0$ for some $x \in (1, e)$	(iv) $\lim_{x \to \infty} f^{''} (x) = 0$	(S) $f^{'}$ is decreasing in $(e, e^{2})$

JEE Advanced 2017 Hard

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