Q. At time t = 0, a battery of 10V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4V?
[Take : $\ell \mathrm{n} 5=1.6, \ell \mathrm{n} 3$ = 1.1]
[IIT-JEE 2010]
[IIT-JEE 2010]
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Sol. 2
$Q=Q_{0}\left(1-e^{-t / R C}\right)$
$16 \mu \mathrm{Q}=40 \mu \mathrm{Q}\left(1-\mathrm{e}^{-1 / 4}\right)$
t = 2 sec
Q. A 2 $\mu \mathrm{F}$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is
(A) 0%
(B) 20%
(C) 75%
(D) 80%
[IIT-JEE 2011]
(A) 0%
(B) 20%
(C) 75%
(D) 80%
[IIT-JEE 2011]
Q. In the given circuit, a charge of $+80 \mu C$ is given to the upper plate of the $4 \mu \mathrm{F}$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mu \mathrm{F}$ capacitor is –
$(\mathrm{A})+32 \mu \mathrm{C}$
(B) $+40 \mu \mathrm{C}$
(C) $+48 \mu \mathrm{C}$
$(\mathrm{D})+80 \mu \mathrm{C}$
[IIT-JEE 2012]
$(\mathrm{A})+32 \mu \mathrm{C}$
(B) $+40 \mu \mathrm{C}$
(C) $+48 \mu \mathrm{C}$
$(\mathrm{D})+80 \mu \mathrm{C}$
[IIT-JEE 2012]
Q. In the circuit shown in the figure, there are two parallel plate capacitors each of the capacitance C. The switch $\mathrm{S}_{1}$ is pressed first to fully charge the capacitor $\mathrm{C}_{1}$ and then released. The switch $\mathrm{S}_{2}$ is then pressed to charge the capacitor $\mathrm{C}_{2}$. After some time, $\mathrm{S}_{2}$ is released and then $\mathrm{S}_{3}$ is pressed, After some time,
(A) the charge on the upper plate of $C_{1}$ is $2 C V_{0}$
(B) the charge on the upper plate of $\mathrm{C}_{1}$ is $\mathrm{CV}_{0}$
(C) the charge on the upper plate of $\mathrm{C}_{2}$ is $0 .$
(D) the charge on the upper plate of $\mathrm{C}_{2}$ is $-\mathrm{CV}_{0}$
[IIT-JEE 2013]
(A) the charge on the upper plate of $C_{1}$ is $2 C V_{0}$
(B) the charge on the upper plate of $\mathrm{C}_{1}$ is $\mathrm{CV}_{0}$
(C) the charge on the upper plate of $\mathrm{C}_{2}$ is $0 .$
(D) the charge on the upper plate of $\mathrm{C}_{2}$ is $-\mathrm{CV}_{0}$
[IIT-JEE 2013]
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Sol. (B,D) Before $\mathrm{S}_{3}$ is pressed
After $\mathrm{S}_{3}$ is pressed
Q. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is $\mathrm{C}_{1}$. When the capacitor is charged, the plate area covered by the dielectric gets charge $\mathrm{Q}_{1}$ and the rest of the area gets charge $Q_{2}$. The electric field in the dielectric is $\mathbf{E}_{1}$ and that in the other portion is $\mathrm{E}_{2}$. Choose the correct option/options, ignoring edge effects.
(A) $\frac{E_{1}}{E_{2}}=1$
(B) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{\mathrm{K}}$
(C) $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{3}{\mathrm{K}}$
(D) $\frac{\mathrm{C}}{\mathrm{C}_{1}}=\frac{2+\mathrm{K}}{\mathrm{K}}$
[IIT-JEE 2014]
(A) $\frac{E_{1}}{E_{2}}=1$
(B) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{\mathrm{K}}$
(C) $\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{3}{\mathrm{K}}$
(D) $\frac{\mathrm{C}}{\mathrm{C}_{1}}=\frac{2+\mathrm{K}}{\mathrm{K}}$
[IIT-JEE 2014]
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Sol. (A,D)
$\mathrm{C}_{1}=\frac{\mathrm{K} \varepsilon_{0}(\mathrm{A} / 3)}{\mathrm{d}}(\mathrm{With} \text { dielectric })$
$\&$ let $\mathrm{C}_{2}=\frac{\epsilon_{0}(2 \mathrm{A} / 3)}{\mathrm{d}}(\text { without dielectric })$
$\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{A} / 3}{\mathrm{d}}+\frac{\varepsilon_{0} 2 \mathrm{A} / 3}{\mathrm{d}}=\frac{\varepsilon_{0} \mathrm{A} / 3}{\mathrm{d}}[\mathrm{K}+2]$
$\therefore \frac{\mathrm{C}}{\mathrm{C}_{1}}=\frac{\mathrm{K}+2}{\mathrm{K}}$
As potential difference is same and gap is same.
$\therefore \mathrm{E}_{4}=\mathrm{E}_{2}$
$\therefore \frac{E_{1}}{E_{2}}=1$
$\mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{V}, \mathrm{Q}_{2}=\mathrm{C}_{2} \mathrm{V}$
$\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{K}}{2}$
Q. In the circuit shown below, the key is pressed at time t = 0. Which of the following statement(s) is(are) true?
(A) The voltmeter displays –5V as soon as the key is pressed, and displays +5V after a long time
(B) The voltmeter will display 0 V at time t = ln 2 seconds
(C) The current in the ammeter becomes 1/e of the initial value after 1 second
(D) The current in the ammeter becomes zero after a long time
[JEE Advanced 2016]
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Sol. (A,B,C,D) (A) At t = 0, capacitor acts as short-circuit $\therefore$
At $\mathrm{t} \rightarrow \infty,$ capacitor acts as open circuit $\&$ no current flows through voltmeter.
$\therefore$
$\mathrm{q}_{\mathrm{x}}=2 \mathrm{CV}\left(1-\mathrm{e}^{-\mathrm{t} / 2 \mathrm{CR}}\right) \quad \mathrm{x}=\frac{\mathrm{V}}{\mathrm{R}} \mathrm{e}^{-\mathrm{t} / 2 \mathrm{CR}}$
$\mathrm{q}_{\mathrm{y}}=\mathrm{CV}\left(1-\mathrm{e}^{-t / 2 \mathrm{CR}}\right) \quad \mathrm{y}=\frac{\mathrm{V}}{2 \mathrm{R}} \mathrm{e}^{-\mathrm{t} / 2 \mathrm{CR}}$
$\Delta \mathrm{V}=-\mathrm{y} 2 \mathrm{R}+\frac{\mathrm{q}_{\mathrm{x}}}{2 \mathrm{C}}$
$=\mathrm{V}\left[1-2 \mathrm{e}^{-\mathrm{t} / 2 \mathrm{CR}}\right]=0$
(C) $\tau=1$ sec
So by $\mathrm{i}=\mathrm{i}_{0} \mathrm{e}^{-t / \tau}$ current at $\mathrm{t}=1$ sec is $=\mathrm{i}_{0} / \mathrm{e}$
(D) After long time no current flows since both capacitor & voltmeter does not allow.
Paragraph–1 Consider a simple RC 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 $\left.\mathrm{V}_{0} \text { (i.e., charging continues for time } \mathrm{T}>>\mathrm{RC}\right)$. In the process some dissipation $\left(\mathrm{E}_{\mathrm{D}}\right)$ occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is $\mathrm{E}_{\mathrm{c}}$. Process 2 : In a different process the voltage is first set to $\frac{\mathrm{v}_{0}}{3}$ and maintained for a charging time T >> RC. Then the voltage is raised to $\frac{2 v_{0}}{3}$ without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to $\mathrm{V}_{0}$ and the capacitor is charged to the same final voltage $\mathrm{V}_{0}$ as in Process 1. These two processes are depicted in Figure 2.
Q. In Process 1, the energy stored in the capacitor $\mathrm{E}_{\mathrm{C}}$ and heat dissipated across resistance $\mathrm{E}_{\mathrm{D}}$ are related by :-
(A) $\mathrm{E}_{\mathrm{C}}=\mathrm{E}_{\mathrm{D}}$
(B) $\mathrm{E}_{\mathrm{c}}=2 \mathrm{E}_{\mathrm{D}}$
(C) $\mathrm{E}_{\mathrm{C}}=\frac{1}{2} \mathrm{E}_{\mathrm{D}}$
(D) $\mathrm{E}_{\mathrm{C}}=\mathrm{E}_{\mathrm{D}} \ln 2$
[JEE Advanced 2017]
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Sol. (A)
When switch is closed for a very long time capacitor will get fully charged & charge on capacitor will be q = CV
Energy stored in capacitor $\mathrm{E}_{\mathrm{C}}=\frac{1}{2} \mathrm{CV}^{2} \quad \ldots$ (i)
Work done by battery $(\mathrm{W})=\mathrm{Vq}=\mathrm{VCV}=\mathrm{CV}^{2}$
dissipated across resistance $\in_{\mathrm{D}}=(\text { work done by battery })-$ (energy store)
$\mathrm{E}_{\mathrm{D}}=\mathrm{CV}^{2}-\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \mathrm{CV}^{2}$ …(ii)
from (i) $\&($ ii )
$\mathrm{E}_{\mathrm{D}}=\mathrm{E}_{\mathrm{C}}$
Q. In Process 2, total energy dissipated across the resistance $E_{D}$ is :-
(A) $\mathrm{E}_{\mathrm{D}}=\frac{1}{3}\left(\frac{1}{2} \mathrm{CV}_{0}^{2}\right)$
(B) $\mathrm{E}_{\mathrm{D}}=3\left(\frac{1}{2} \mathrm{CV}_{0}^{2}\right)$
(C) $\mathrm{E}_{\mathrm{D}}=\frac{1}{2} \mathrm{CV}_{0}^{2}$
(D) $\mathrm{E}_{\mathrm{D}}=3 \mathrm{CV}_{0}^{2}$
[JEE Advanced 2017]
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Sol. (A) For process (1) Charge on capacitor $=\frac{\mathrm{CV}_{0}}{3}$ energy stored in capacitor $=\frac{1}{2} \mathrm{C} \frac{\mathrm{V}_{0}^{2}}{9}=\frac{\mathrm{CV}_{0}^{2}}{18}$ work done by battery $=\frac{\mathrm{CV}_{0}}{3} \times \frac{\mathrm{V}}{3}=\frac{\mathrm{CV}_{0}^{2}}{9}$ Heat loss $=\frac{\mathrm{CV}_{0}^{2}}{9}-\frac{\mathrm{CV}_{0}^{2}}{18}=\frac{\mathrm{CV}_{0}^{2}}{18}$ For process (2) Charge on capacitor $=\frac{2 \mathrm{CV}_{0}}{3}$ Extra charge flow through battery $=\frac{\mathrm{CV}_{0}}{3}$ Work done by battery $: \frac{\mathrm{CV}_{0}}{3} \cdot \frac{2 \mathrm{V}_{0}}{3}=\frac{2 \mathrm{CV}_{0}^{2}}{9}$ Final energy store in capacitor : $\frac{1}{2} \mathrm{C}\left(\frac{2 \mathrm{V}_{0}}{3}\right)^{2}=\frac{4 \mathrm{CV}_{0}^{2}}{18}$ energy store in process $2: \frac{4 \mathrm{CV}_{0}^{2}}{18}-\frac{\mathrm{CV}_{0}^{2}}{18}=\frac{3 \mathrm{CV}_{0}^{2}}{18}$ Heat loss in process (2) = work done by battery in process (2) – energy store in capacitor process (2) $=\frac{2 \mathrm{CV}_{0}^{2}}{9}-\frac{3 \mathrm{CV}_{0}^{2}}{18}=\frac{\mathrm{CV}_{0}^{2}}{18}$ For process (3) Charge on capacitor $=\mathrm{CV}_{0}$ extra charge flow through battery : $\mathrm{CV}_{0}-\frac{2 \mathrm{CV}_{0}}{3}=\frac{\mathrm{CV}_{0}}{3}$ work done by battery in this process : $\left(\frac{\mathrm{CV}_{0}}{3}\right)\left(\mathrm{V}_{0}\right)=\frac{\mathrm{C} \mathrm{V}_{0}^{2}}{3}$ find energy store in capacitor $: \frac{1}{2} \mathrm{CV}_{0}^{2}$ energy stored in this process : $\frac{1}{2} \mathrm{CV}_{0}^{2}-\frac{4 \mathrm{CV}_{0}^{2}}{18}=\frac{5 \mathrm{CV}_{0}^{2}}{18}$ heat loss in process ( 3)$: \frac{\mathrm{CV}_{0}^{2}}{3}-\frac{5 \mathrm{CV}_{0}^{2}}{18}=\frac{\mathrm{CV}_{0}^{2}}{18}$ Now total heat loss $\left(\mathrm{E}_{\mathrm{D}}\right): \frac{\mathrm{CV}_{0}^{2}}{18}+\frac{\mathrm{CV}_{0}^{2}}{18}+\frac{\mathrm{CV}_{0}^{2}}{18}=\frac{\mathrm{CV}_{0}^{2}}{6}$ final energy store in capacitor : $\frac{1}{2} \mathrm{CV}_{0}^{2}$ so we can say that $\mathrm{E}_{\mathrm{D}}=\frac{1}{3}\left(\frac{1}{2} \mathrm{CV}_{0}^{2}\right)$
Q. Three identical capacitors $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ have a capacitance of $1.0 \mu \mathrm{F}$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $\mathrm{C}_{1}$ is then filled completely with a dielectric material of relative permittivity $\epsilon_{\mathrm{r}}$. The cell electromotive force (emf) $\mathrm{V}_{0}=8 \mathrm{V}$ . First the switch $\mathrm{S}_{1}$ is closed while the switch $\mathrm{S}_{2}$ is kept open. When the capacitor $\mathrm{C}_{3}$ is fully charged, $\mathrm{S}_{1}$ is opened and $\mathrm{S}_{2}$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $\mathrm{C}_{3}$ is found to be $5 \mu \mathrm{C}$. The value of $\in_{\mathbf{I}}$.
[JEE Advanced 2018]
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Sol. 1.5
Applying loop rule
$\frac{5}{1}-\frac{3}{\epsilon_{\mathrm{r}}}-\frac{3}{1}=0 \quad \frac{3}{\epsilon_{\mathrm{r}}}=2$
$\epsilon_{\mathrm{r}}=1.50$
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