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**Column II**. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ (indicated in circuits) are related as shown in

**Column I**. Match the two

**[JEE 2010]**

**Sol.**(A) RST (B) QRST (C) PQ (D) QRST

**$\omega$**is increased

(A) the bulb glows dimmer

(B) the bulb glows brighter

(C) total impedance of the circuit is unchanged

(D) total impedance of the circuit increases

**[JEE 2010]**

**Sol.**(B)

(A) $I_{R}^{A}>I_{R}^{B}$

$(\mathrm{B}) \quad I_{R}^{A}<I_{R}^{B}$

(C) $V_{c}^{A}>V_{c}^{B}$

(D) $V_{C}^{A}<V_{C}^{B}$

**[JEE 2011]**

**Sol.**(B,C)

$\mathrm{x}_{\mathrm{c}}$ decreases therefore impedence decreases and current increases. $\mathrm{I}_{\mathrm{B}}>\mathrm{I}_{\mathrm{A}}$

As $\mathrm{I}_{\mathrm{B}}$ increases the voltage across ‘R’ increases therefore $\mathrm{V}_{\mathrm{c}}$ decreases.

**[JEE 2011]**

**Sol.**4

$1.25 \mathrm{R}^{2}=\mathrm{R}^{2}+\left(\frac{1}{\omega c}\right)^{2}$

$0.25 \mathrm{R}^{2}=\left(\frac{1}{\omega c}\right)^{2} ; 0.5 \mathrm{R}=\frac{1}{500 \times C} ; \mathrm{C}=\frac{1}{250 R} ; \mathrm{RC}=\frac{1}{250} \mathrm{sec}$

$\tau=4$ millisecond; $\tau=4$

(A) The current through the circuit, I is $0.3 \mathrm{A}$.

(B) The current through the circuit, is $0.3 \sqrt{2 \mathrm{A}}$

(C) The voltage across $100 \Omega$ resistor $=10 \sqrt{2} \mathrm{V}$

(D) The voltage across $50 \Omega$ resistor $=10 \mathrm{V}$

** [JEE 2012]**

**Sol.**(C or A,C)

**Paragraph for Questions 6 and 7**

A thermal power plant produces electric power of 600 kW and 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers’ end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values.

(A) 200 : 1

(B) 150 : 1

(C) 100 : 1

(D) 50 : 1

**[JEE Advance-2013]**

**Sol.**(A)

(A) 20

(B) 30

(C) 40

(D) 50

**[JEE Advance-2013]**

**Sol.**(B)

Current in transmission line $=\frac{\text { Power }}{\text { Voltage }}$

$=\frac{600 \times 10^{3}}{40,000}=150 \mathrm{A}$

Resistance of line $=0.4 \times 20=8 \Omega$

Power loss in line $=\mathrm{i}^{2} \mathrm{R}=(150)^{2} 8$

$=180 \mathrm{KW}$

percentage of power dissipation in during transmission $=\frac{1800 \times 10^{3}}{600 \times 10^{3}} \times 100=30 \%$

(A) Magnitude of the maximum charge on the capacitor before $\mathrm{t}=\frac{7 \pi}{6 \omega}$ is $1 \times 10^{-3} \mathrm{C}$

(B) The current in the left part of the circuit just before $\mathrm{t}=\frac{7 \pi}{6 \omega}$ is clockwise.

(C) Immediately after A is connected to D, the current in $\mathrm{R}$ is $10 \mathrm{A}$

(D) $\mathrm{Q}=2 \times 10^{-3} \mathrm{C}$

**[JEE Advance-2014]**

**Sol.**(C,D)

Current $\mathrm{I}=\mathrm{I}_{0} \cos (\omega \mathrm{t})$

$\frac{\mathrm{d} q}{\mathrm{dt}}=\mathrm{I}_{0} \cos (\omega \mathrm{t})$

$\Rightarrow \mathrm{q}=\frac{\mathrm{I}_{0}}{\omega} \sin (\omega \mathrm{t})$

$\Rightarrow \mathrm{q}=\frac{1}{500} \sin (\omega \mathrm{t})$

$\Rightarrow \mathrm{q}=\left(2 \times 10^{-3}\right) \sin (\omega \mathrm{t})$

So, maximum charge $=2 \times 10^{-3} \mathrm{C}$

immediately before $\mathrm{t}=\frac{7 \pi}{6 \omega}$

Current in left part just before $\mathrm{t}=\frac{7 \pi}{6 \omega}$

$\mathrm{I}=\mathrm{I}_{0} \cos \left(\omega \times \frac{7 \pi}{6 \omega}\right)=-\frac{\mathrm{I}_{0} \sqrt{3}}{2}$

Since current is negative hence current will be anticlockwise.

immediately after $\mathrm{t}=\frac{7 \pi}{6 \omega}$

$\mathrm{q}=\left(2 \times 10^{-3}\right) \sin \left(\omega \times \frac{7 \pi}{6 \omega}\right)$

$\mathrm{q}=\left(2 \times 10^{-3}\right) \sin \left(\omega \times \frac{7 \pi}{6 \omega}\right)$

$=-1 \times 10^{-3} \mathrm{C}$

Current in $10 \Omega$ resistance,

$\mathrm{I}=\frac{100}{10}=10 \mathrm{A}$

At steady state, potential difference of capaictor is same as of battery,

So, final charge is

$\mathrm{Q}_{\mathrm{f}}=\mathrm{C} \varepsilon=(20 \mu \mathrm{F})(50 \mathrm{V})=+1 \times 10^{-3} \mathrm{C}$

change in charge $=+10^{-3}-\left(-10^{-3}\right)=2 \times 10^{-3} \mathrm{C}$

(A) The frequency at which the current will be in phase with the voltage is independent of R.

(B) At $\omega \sim 0$ the current flowing through the circuit becomes nearly zero

(C) At $\omega>>10^{6}$ rad.s $^{-1},$ the circuit behaves like a capacitor.

(D) The current will be in phase with the voltage if $\omega=10^{4} \mathrm{rad.s}^{-1}$

**[JEE Advance-2017]**

**Sol.**(A,B)

$\omega \mathrm{t} \mathrm{V}_{\mathrm{Y}}=\mathrm{V}_{0} \sin \left(\omega \mathrm{t}+\frac{2 \pi}{3}\right)$ and $\mathrm{V}_{\mathrm{Z}}=\mathrm{V}_{0} \sin \left(\omega \mathrm{t}+\frac{4 \pi}{3}\right)$

An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be:-

(A) $\mathrm{V}_{\mathrm{XY}}^{\mathrm{rms}}=\mathrm{V}_{0}$

(B) $\quad \mathrm{V}_{\mathrm{YZ}}^{\mathrm{ms}}=\mathrm{V}_{0} \sqrt{\frac{1}{2}}$

(C) Independent of the choice of the two terminals

(D) $\quad V_{\mathrm{XY}}^{\mathrm{ms}}=\mathrm{V}_{0} \sqrt{\frac{3}{2}}$

**[JEE Advance-2017]**

**Sol.**(C,D)

Potential difference between $X \& Y=V_{X}-V_{Y}$

Potential difference between $\mathrm{Y} \& \mathrm{Z}=\mathrm{V}_{\mathrm{Y}}-\mathrm{V}_{7}$

Phasor of the voltages :

$\therefore \mathrm{V}_{\mathrm{X}}-\mathrm{V}_{\mathrm{Y}}=\sqrt{3} \mathrm{V}_{0}$

$\mathrm{V}_{\mathrm{XY}}^{\mathrm{ms}}=\frac{\sqrt{3} \mathrm{V}_{0}}{\sqrt{2}}$

similarly $\mathrm{V}_{\mathrm{YZ}}^{\mathrm{ms}}=\frac{\sqrt{3} \mathrm{V}_{0}}{\sqrt{2}}$

Also difference is independent of choice of two terminals.