JEE Advanced PYQ Physics – AC Circuit, LCR Circuit, Quality Factor & Power Factor

Q 1 :

An AC voltage source of variable angular frequency $ω$ and fixed amplitude $V_{0}$ is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When $ω$ is increased [2010]

the bulb glows dimmer
the bulb glows brighter
total impedance of the circuit is unchanged
total impedance of the circuit increases

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(2)

$I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{rms}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}}$

As $ω$ increases, $I_{rms}$ through the bulb increases. Hence the bulb glows brighter.

Q 2 :

Find the time constant (in $μ$ s) for the given RC circuits in the given order respectively. [2006]

[IMAGE 981]

Given: $R_{1} = 1 Ω, R_{2} = 2 Ω, C_{1} = 4 μ F, C_{2} = 2 μ F$

$18, 4, \frac{8}{9}$
$18, \frac{8}{9}, 4$
$4, 18, \frac{8}{9}$
$4, \frac{8}{9}, 18$

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(2)

$Time constant of R - C circuit, τ = R_{eq} C_{eq}$

$(i) R_{1} & R_{2} in series and C_{1} & C_{2} in parallel.$

$τ_{1} = (2 + 1) (2 + 4) = 18 μ s$

$(ii) R_{1} & R_{2} in parallel and C_{1} & C_{2} in series.$

$τ_{2} = (\frac{2 \times 1}{2 + 1}) (\frac{2 \times 4}{2 + 4}) = \frac{8}{9} μ s$

$(iii) R_{1} & R_{2} in parallel and C_{1} & C_{2} in parallel.$

$τ_{3} = (\frac{2 \times 1}{2 + 1}) \times (4 + 2) = 4 μ s$

Q 3 :

When an AC source of emf $e = E_{0} \sin (100 t)$ is connected across a circuit. The phase difference between the emf $e$ and the current $i$ in the circuit is observed to be $π / 4$ , as shown in the diagram. If the circuit consists possibly only of $R - C$ or $R - L$ or $L - C$ in series, find the relationship between the two elements. [2003]

[IMAGE 982]

$R = 1 k Ω, C = 10 μ F$
$R = 1 k Ω, C = 1 μ F$
$R = 1 k Ω, L = 10 H$
$R = 1 k Ω, L = 1 H$

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(1)

$From e = E_{0} \sin (100 t),$ $∴$ $ω = 100$

From the given graph, current leads emf. Hence this is an $R - C$ circuit.

$\tan ϕ = \frac{X_{C}}{R}$

Here, $ϕ = 45 ° = \frac{π}{4}$ $∴$ $X_{C} = R$

$\frac{1}{ω C} = R \Rightarrow R C ω = 1$

or $R C = \frac{1}{100} s^{- 1}$ $∴$ $R = 1 k Ω, C = 10 μ F$

Q 4 :

Consider an LC circuit, with inductance L = 0.1 H and capacitance $C = 10^{- 3} F$ , kept on a plane. The area of the circuit is $1 m^{2}$ . It is placed in a constant magnetic field of strength $B_{0}$ , which is perpendicular to the plane of the circuit. At time $t = 0$ , the magnetic field strength starts increasing linearly as $B = B_{0} + β t$ with $β = 0.04 {Ts}^{- 1} .$ The maximum magnitude of the current in the circuit is _______ $m A$ . [2022]

(4)

[IMAGE 983]

$Induced emf, E = | \frac{d ϕ}{d t} | = \frac{d}{d t} (B A) = \frac{A d}{d t} (B_{0} + β t) = A β = 1 \times 0.04 = 0.04 V$

So, the circuit can be drawn as

$By KVL, E = L \frac{d i}{d t} + \frac{q}{C}$

$\Rightarrow L \frac{d i}{d t} = E - \frac{q}{C}$

Since $i = \frac{d q}{d t}$ ,

$\Rightarrow \frac{d^{2} q}{d t^{2}} = - \frac{1}{L C} (q - C E)$

Comparing it with the equation of SHM, we get

$q = C E + A \sin (ω t + ϕ), where ω = \frac{1}{\sqrt{L C}}$

So, $i = A ω \cos (ω t + ϕ)$

At $t = 0$ , $q = 0$ and $i = 0$

So, $O = C E + A \sin ϕ \Rightarrow A \sin ϕ = - C E$ $\dots (i)$

$O = \cos ϕ \Rightarrow ϕ = \frac{π}{2}$ $\dots (i i)$

From (i) and (ii), we get

$A = - C E$

So, $i = - C E ω \cos (ω t + \frac{π}{2}) = C E ω \sin ω t$

Therefore, $i_{\max} = C E ω = 10^{- 3} \times 0.04 \times \frac{1}{\sqrt{0.1 \times 10^{- 3}}} = 4 mA$

Q 5 :

Two inductors $L_{1}$ (inductance 1 mH, internal resistance $3 Ω$ ) and $L_{2}$ (inductance 2 mH, internal resistance $4 Ω$ ), and a resistor R (resistance $12 Ω$ ) are all connected in parallel across a 5 V battery. The circuit is switched on at time $t = 0$ . The ratio of the maximum to the minimum current $(\frac{I_{\max}}{I_{\min}})$ drawn from the battery is ______ [2016]

(8)

[IMAGE 984]

$At t = 0, I_{\min} = \frac{5}{12}$

At $t = \infty$ ,

$I_{\max} = \frac{5}{R_{eq}} = \frac{5}{3 / 2} = \frac{10}{3}$

$[\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12} = \frac{8}{12}]$

$∴$ $\frac{I_{\max}}{I_{\min}} = \frac{10}{3} \times \frac{12}{5} = 8$

Q 6 :

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $C μ F$ across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take $r m s$ values of the voltages. The magnitude of the phase angle (in degrees) between the current and the supply voltage is $φ$ .

Assume, $π \sqrt{3} \approx 5 .$

Q. The value of C is ______. [2021]

(100)

[IMAGE 985]

From $V_{RMS} = \sqrt{V_{C}^{2} + V_{R}^{2}},$

$\Rightarrow V_{C}^{2} + 100^{2} = 200^{2},$

or, $V_{C}^{2} + 10000 = 40000$

$∴$ $V_{C} = 100 \sqrt{3} V$ $\dots (i)$

$\tan ϕ = \frac{V_{C}}{V_{R}} = \frac{100 \sqrt{3}}{100} = \sqrt{3},$

$∴$ $ϕ = 60 °$ $\dots (i i)$

Power consumed, $P = I_{rms} V_{rms} \cos ϕ = \frac{1}{2} \frac{V_{rms}^{2}}{Z}$

$\Rightarrow 500 = \frac{200^{2}}{Z} \times \frac{1}{2}$

$∴$ $Z = 40 Ω$ $\dots (i i i)$

$\cos ϕ = \frac{R}{Z} \Rightarrow \frac{1}{2} = \frac{R}{40}$

$∴$ $R = 20 Ω$

And, $X_{C} = \sqrt{Z^{2} - R^{2}} = \sqrt{40^{2} - 20^{2}} = 20 \sqrt{3} Ω$

$X_{C} = \frac{1}{C ω} \Rightarrow 20 \sqrt{3} = \frac{1}{C 2 π f}$

$∴$ $C = \frac{1}{2 π f (20 \sqrt{3})} = \frac{1}{20 π \sqrt{3} \times 100} = 10^{- 4} F = 100 μ F$

Q 7 :

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $C μ F$ across a 200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take $r m s$ values of the voltages. The magnitude of the phase angle (in degrees) between the current and the supply voltage is $φ$ .

Assume, $π \sqrt{3} \approx 5 .$

Q. The value of $φ$ is ______. [2021]

(60)

[IMAGE 986]

From $V_{RMS} = \sqrt{V_{C}^{2} + V_{R}^{2}},$

$\Rightarrow V_{C}^{2} + 100^{2} = 200^{2}$

or, $V_{C}^{2} + 10000 = 40000$

$∴$ $V_{C} = 100 \sqrt{3} V$ $\dots (i)$

$\tan ϕ = \frac{V_{C}}{V_{R}} = \frac{100 \sqrt{3}}{100} = \sqrt{3}$

$∴$ $ϕ = 60 °$ $\dots (i i)$

Power consumed, $P = I_{rms} V_{rms} \cos ϕ = \frac{1}{2} \frac{V_{rms}^{2}}{Z}$

$\Rightarrow 500 = \frac{200^{2}}{Z} \times \frac{1}{2}$

$∴$ $Z = 40 Ω$ $\dots (i i i)$

$\cos ϕ = \frac{R}{Z} \Rightarrow \frac{1}{2} = \frac{R}{40},$

$∴$ $R = 20 Ω$

And, $X_{C} = \sqrt{Z^{2} - R^{2}} = \sqrt{40^{2} - 20^{2}} = 20 \sqrt{3} Ω$

$X_{C} = \frac{1}{C ω} \Rightarrow 20 \sqrt{3} = \frac{1}{C 2 π f}$

$∴$ $C = \frac{1}{2 π f (20 \sqrt{3})} = \frac{1}{20 π \sqrt{3} \times 100} = 10^{- 4} F = 100 μ F$

Q 8 :

The inductors of two $L R$ circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced $E M F$ in the inductors by the time the currents reach their steady-state values is ________ mJ. [2020]

[IMAGE 987]

(55)

$Given: Mutual inductance, M = 5 mH$

$L_{1} = 10 mH, V_{1} = 5 V, L_{2} = 20 mH, V_{2} = 20 V$

$I_{1} = \frac{V_{1}}{R_{1}} = \frac{5}{5} = 1 A, I_{2} = \frac{V_{2}}{R_{2}} = \frac{20}{10} = 2 A$

After both the switches are closed simultaneously, the total work done by the batteries against the induced emf equals the increase in magnetic energy.

$∴$ $W = Δ U = \frac{1}{2} L_{1} I_{1}^{2} + \frac{1}{2} L_{2} I_{2}^{2} + M I_{1} I_{2}$

$= \frac{1}{2} \times (10 \times 10^{- 3}) \times 1^{2} + \frac{1}{2} \times (20 \times 10^{- 3}) \times 2^{2} + (5 \times 10^{- 3}) \times 1 \times 2$

$= (5 + 40 + 10) \times 10^{- 3} J$

$∴$ $W = 55 mJ$

Q 9 :

In the figure below, the switches $S_{1}$ and $S_{2}$ are closed simultaneously at $t = 0$ and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current $I$ in the middle wire reaches its maximum magnitude $I_{\max}$ at time $t = τ$ . Which of the following statements is (are) true? [2018]

[IMAGE 988]

$I_{\max} = \frac{V}{2 R}$
$I_{\max} = \frac{V}{4 R}$
$τ = \frac{L}{R} \ln 2$
$τ = \frac{2 L}{R} \ln 2$

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Select one or more options

(2, 4)

$Here I + I_{2} = I_{1}$ $∴$ $I = I_{1} - I_{2}$

$∴$ $I = \frac{V}{R} [1 - e^{- \frac{R t}{2 L}}] - \frac{V}{R} [1 - e^{- \frac{R t}{L}}]$

$\Rightarrow I = \frac{V}{R} [e^{- \frac{R t}{L}} - e^{- \frac{R t}{2 L}}]$ $\dots (i)$

For $I_{\max}$ , $\frac{d I}{d t} = 0$

[IMAGE 989]

$∴$ $\frac{V}{R} [- \frac{R}{L} e^{- \frac{R t}{L}} - (- \frac{R}{2 L}) e^{- \frac{R t}{2 L}}] = 0$

$∴$ $e^{- \frac{R t}{2 L}} = \frac{1}{2}$ $\Rightarrow (\frac{R}{2 L}) t = \ln 2$ $∴$ $t = \frac{2 L}{R} \ln 2$

This is the time when $I$ is maximum.

Putting this value of time in Eq.(i),

Further $I_{\max} = \frac{V}{R} [e^{- \frac{R}{L} (\frac{2 L}{R} \ln 2)} - e^{- \frac{R}{2 L} (\frac{2 L}{R} \ln 2)}]$

$\Rightarrow I_{\max} = \frac{V}{R} [\frac{1}{4} - \frac{1}{2}]$

Taking the magnitude,

$I_{\max} = \frac{V}{R} [\frac{1}{2} - \frac{1}{4}] = \frac{V}{4 R}$

Q 10 :

In the circuit shown, $L = 1 μ H$ , $C = 1 μ F$ and $R = 1 k Ω$ . They are connected in series with an a.c. source $V = V_{0} \sin ω t$ as shown. Which of the following options is/are correct? [2017]

[IMAGE 990]

The current will be in phase with the voltage if $ω = 10^{4} rad s^{- 1} .$
The frequency at which the current will be in phase with the voltage is independent of R.
At $ω ~ 0$ , the current flowing through the circuit becomes nearly zero.
At $ω ≫ 10^{6} rad s^{- 1}$ , the circuit behaves like a capacitor.

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Select one or more options

(2, 3)

The frequency at which the current is in phase with the voltage is the resonance frequency.

$ω_{r} = \frac{1}{\sqrt{L C}} = \frac{1}{{(10^{- 6} \times 10^{- 6})}^{1 / 2}} = 10^{6} {rad s}^{- 1}$

This frequency is independent of $' R'$ .

At $ω \approx 0$ , the current is $i = \frac{V}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$

i.e., the current through the circuit nearly becomes zero.

If $ω ≫ ω_{r},$ $X_{L} > X_{C}$ so the circuit behaves like an inductor.

Q 11 :

In the given circuit, the AC source has $ω = 100 rad / s .$ Considering the inductor and capacitor to be ideal, the correct choice(s) is (are) [2012]

[IMAGE 991]

The current through the circuit, $I$ is 0.3 A
The current through the circuit, $I$ is $0.3 \sqrt{2} A$
The voltage across $100 Ω$ resistor is $10 \sqrt{2} V$
The voltage across $50 Ω$ resistor is $10 V$

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Select one or more options

(1, 3)

$Impedance across A B, RC part of the circuit.$

$Z_{1} = \sqrt{X_{C}^{2} + R_{1}^{2}} = \sqrt{{(\frac{1}{ω C})}^{2} + R_{1}^{2}}$

$= \sqrt{{(100)}^{2} + {(100)}^{2}} = 100 \sqrt{2}$

[IMAGE 992]

$∴$ $I_{1} = \frac{V}{Z_{1}} = \frac{20}{100 \sqrt{2}}$ $(Current leads emf by ϕ_{1})$

where $\cos ϕ_{1} = \frac{R}{Z_{1}} = \frac{100}{100 \sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow ϕ = 45 °$

$Impedance across C D, LR part of the circuit.$

$Z_{2} = \sqrt{X_{L}^{2} + R_{2}^{2}} = \sqrt{{(ω L)}^{2} + R_{2}^{2}}$

$= \sqrt{{(0.5 \times 100)}^{2} + {(50)}^{2}} = 50 \sqrt{2} Ω$

[IMAGE 993]

$∴$ $I_{2} = \frac{V}{Z_{2}} = \frac{20}{50 \sqrt{2}}$ $(Current lags emf by ϕ_{2})$

where $\cos ϕ_{2} = \frac{R}{Z_{2}} = \frac{50}{50 \sqrt{2}} = \frac{1}{\sqrt{2}}$ $\Rightarrow ϕ_{2} = 45 °$

$∴$ $Current I from the circuit$

$I = \frac{20}{100 \sqrt{2}} + \frac{20}{50 \sqrt{2}} = I_{1} + I_{2} = 0.3 A$

Q 12 :

The circuit shown in the figure contains an inductor L, a capacitor $C_{0}$ , a resistor $R_{0}$ and an ideal battery. The circuit also contains two keys $K_{1}$ and $K_{2}$ . Initially, both the keys are open and there is no charge on the capacitor. At an instant, key $K_{1}$ is closed and immediately after this the current in $R_{0}$ is found to be $I_{1}$ . After a long time, the current attains a steady state value $I_{2}$ . Thereafter, $K_{2}$ is closed and simultaneously $K_{1}$ is opened and the voltage across $C_{0}$ oscillates with an amplitude $V_{0}$ and angular frequency $ω_{0}$ .

[IMAGE 994]

Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [2024]

List-I List-II

(P) The value of $I_{1}$ in Ampere is (1) 0

(Q) The value of $I_{2}$ in Ampere is (2) 2

(R) The value of $ω_{0}$ in kilo-radians/s is (3) 4

(S) The value of $V_{0}$ in Volt is (4) 20

(5) 200

P → 1; Q → 3; R → 2; S → 5
P → 1; Q → 2; R → 3; S → 5
P → 1; Q → 3; R → 2; S → 4
P → 2; Q → 5; R → 3; S → 4

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(1)

[IMAGE 995]

$(P) When key K_{1} is closed, current in R_{0} is I_{1}$

At $t = 0$ , the circuit is as follows.

[IMAGE 996]

$I_{1} = 0$

$(Q) After a long time, the inductor behaves as a wire, so$

[IMAGE 997]

$I_{2} = \frac{V}{R} = \frac{20}{5} = 4 A$

$(R) When K_{2} is closed and K_{1} is open,$

[IMAGE 998]

$ω_{0} = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{25 \times 10^{- 3} \times 10 \times 10^{- 6}}} = \frac{1}{5 \times 10^{- 4}}$

$∴$ $ω_{0} = 2 \times 10^{3} rad/s$ $∴$ $ω_{0} = 2 kilo-radian/s$

$(S) When K_{2} is closed and K_{1} is open,$

[IMAGE 999]

$\frac{1}{2} L I_{2}^{2} = \frac{1}{2} C V_{0}^{2}$

$\frac{1}{2} (25 \times 10^{- 3}) {(4)}^{2} = \frac{1}{2} (10 \times 10^{- 6}) V_{0}^{2}$

$V_{0}^{2} = 2500 \times 16$ or, $V_{0} = 50 \times 4 = 200 V$

Q 13 :

A series LCR circuit is connected to a $45 \sin (ω t)$ Volt source. The resonant angular frequency of the circuit is $10^{5} rad s^{- 1}$ and current amplitude at resonance is $I_{0}$ . When the angular frequency of the source is $ω = 8 \times 10^{4} rad s^{- 1},$ the current amplitude in the circuit is $0.05 I_{0}$ . If $L = 50 mH$ , match each entry in List-I with appropriate value from List-II and choose the correct option. [2023]

List-I List-II

(P) $I_{0} in mA$ (1) 44.4

(Q) The quality factor of the circuit (2) 18

(R) The bandwidth of the circuit in $rad s^{- 1}$ (3) 400

(S) The peak power dissipated at resonance in Watt (4) 2250

(5) 500

P → 2, Q → 3, R → 5, S → 1
P → 3, Q → 1, R → 4, S → 2
P → 4, Q → 5, R → 3, S → 1
P → 4, Q → 2, R → 1, S → 5

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(2)

As, $V = V_{0} \sin ω t$ and the resonant angular frequency is

$ω_{0} = \frac{1}{\sqrt{L C}} \Rightarrow ω_{0} = 10^{5} rad/s = \frac{1}{\sqrt{L C}} \Rightarrow C = \frac{1}{L ω_{0}^{2}} = \frac{1}{5 \times 10^{- 2} \times 10^{10}} = 2 \times 10^{- 9} F$

$[∵ L = 50 mH = 5 \times 10^{- 2} H]$

$I_{0} = \frac{V_{0}}{R} = \frac{45}{R}$ $[∵ V = 45 \sin ω t is given]$

$ω = 8 \times 10^{4} rad/s = 0.8 ω_{0}$

$I = 0.05 I_{0} = \frac{I_{0}}{20} \Rightarrow Z = 20 R$

$X_{L} = L ω = 8 \times 10^{4} \times 5 \times 10^{- 2} = 4 k Ω$

$X_{C} = \frac{1}{C ω} = \frac{1}{8 \times 10^{4} \times 2 \times 10^{- 9}} = \frac{1}{16} \times 10^{5} Ω = \frac{25}{4} k Ω$

$Z^{2} = R^{2} + {(X_{C} - X_{L})}^{2}$ or, $400 R^{2} = R^{2} + {(\frac{9}{4} k Ω)}^{2}$

$∴$ $R = \frac{\frac{9}{4} k Ω}{\sqrt{399}} = \frac{9}{80} k Ω = \frac{900}{8} Ω$

$I_{0} = \frac{V_{0}}{R} = \frac{45 \times 8}{900} = \frac{8}{20} A \approx 0.4 A = 400 mA$

Quality factor,

$Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{8}{900} \sqrt{\frac{5 \times 10^{- 2}}{2 \times 10^{- 9}}} = \frac{8}{900} \sqrt{25 \times 10^{6}}$

$= \frac{8}{900} \times 5000 = 44.4$

$Q = \frac{ω_{0}}{Δ ω}$

$∴$ $Bandwidth, Δ ω = \frac{ω_{0}}{Q} = \frac{10^{5}}{44.4} = 2250.0 rad/s$

Peak power dissipated, $P_{\max} = I_{0}^{2} R = \frac{45^{2}}{R^{2}} \times R = \frac{45^{2}}{R}$

$= \frac{45^{2}}{900} \times 8 = 18.4 W \approx 18 W$

Topic Question Set

Q 1 :

An AC voltage source of variable angular frequency $ω$ and fixed amplitude $V_{0}$ is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When $ω$ is increased [2010]

Q 2 :

Find the time constant (in $μ$ s) for the given RC circuits in the given order respectively. [2006]

[IMAGE 981]

Given: $R_{1} = 1 Ω, R_{2} = 2 Ω, C_{1} = 4 μ F, C_{2} = 2 μ F$

Q 10 :

In the circuit shown, $L = 1 μ H$ , $C = 1 μ F$ and $R = 1 k Ω$ . They are connected in series with an a.c. source $V = V_{0} \sin ω t$ as shown. Which of the following options is/are correct? [2017]

[IMAGE 990]

Q 11 :

In the given circuit, the AC source has $ω = 100 rad / s .$ Considering the inductor and capacitor to be ideal, the correct choice(s) is (are) [2012]

[IMAGE 991]

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	List-I		List-II
(P)	The value of $I_{1}$ in Ampere is	(1)	0
(Q)	The value of $I_{2}$ in Ampere is	(2)	2
(R)	The value of $ω_{0}$ in kilo-radians/s is	(3)	4
(S)	The value of $V_{0}$ in Volt is	(4)	20
		(5)	200

	List-I		List-II
(P)	$I_{0} in mA$	(1)	44.4
(Q)	The quality factor of the circuit	(2)	18
(R)	The bandwidth of the circuit in $rad s^{- 1}$	(3)	400
(S)	The peak power dissipated at resonance in Watt	(4)	2250
		(5)	500

Topic Question Set

Q 1 : An AC voltage source of variable angular frequency ω and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased [2010]

Q 2 : Find the time constant (in μs) for the given RC circuits in the given order respectively. [2006] [IMAGE 981] Given: R1=1Ω, R2=2Ω, C1=4μF, C2=2μF

Q 10 : In the circuit shown, L=1 μH, C=1 μF and R=1 kΩ. They are connected in series with an a.c. source V=V0sinωt as shown. Which of the following options is/are correct? [2017] [IMAGE 990]

Q 11 : In the given circuit, the AC source has ω=100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are) [2012] [IMAGE 991]

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Q 1 :

An AC voltage source of variable angular frequency $ω$ and fixed amplitude $V_{0}$ is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When $ω$ is increased [2010]

Q 2 :

Find the time constant (in $μ$ s) for the given RC circuits in the given order respectively. [2006]

[IMAGE 981]

Given: $R_{1} = 1 Ω, R_{2} = 2 Ω, C_{1} = 4 μ F, C_{2} = 2 μ F$

Q 10 :

In the circuit shown, $L = 1 μ H$ , $C = 1 μ F$ and $R = 1 k Ω$ . They are connected in series with an a.c. source $V = V_{0} \sin ω t$ as shown. Which of the following options is/are correct? [2017]

[IMAGE 990]

Q 11 :

In the given circuit, the AC source has $ω = 100 rad / s .$ Considering the inductor and capacitor to be ideal, the correct choice(s) is (are) [2012]

[IMAGE 991]