Alternating Current
189 Questions
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th June Morning Shift
If wattless current flows in the AC circuit, then the circuit is :
A.
Purely Resistive circuit
B.
Purely Inductive circuit
C.
LCR series circuit
D.
RC series circuit only
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 24th June Evening Shift
Given below are two statements :
Statement I : The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor.
Statement II : In ac circuit, the average power delivered by the source never becomes zero.
In the light of the above statements, choose the correct answer from the options given below.
A.
Both Statement I and Statement II are true.
B.
Both Statement I and Statement II are false.
C.
Statement I is true but Statement II is false.
D.
Statement I is false but Statement II is true.
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 24th June Morning Shift
A resistance of 40 $\Omega$ is connected to a source of alternating current rated 220 V, 50 Hz. Find the time taken by the current to change from its maximum value to the rms value :
A.
2.5 ms
B.
1.25 ms
C.
2.5 s
D.
0.25 s
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
At very high frequencies, the effective impendence of the given circuit will be ________________ $\Omega$.
Correct Answer: 2
Explanation:
XL = 2$\pi$fL
f is very large
$\therefore$ XL is very large hence open circuit.
${X_C} = {1 \over {2\pi fC}}$
f is very large.
$\therefore$ XC is very small, hence short circuit.
Final circuit
$ \therefore $ ${Z_{eq}} = 1 + {{2 \times 2} \over {2 + 2}} = 2$
f is very large
$\therefore$ XL is very large hence open circuit.
${X_C} = {1 \over {2\pi fC}}$
f is very large.
$\therefore$ XC is very small, hence short circuit.
Final circuit
$ \therefore $ ${Z_{eq}} = 1 + {{2 \times 2} \over {2 + 2}} = 2$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
An ac circuit has an inductor and a resistor resistance R in series, such that XL = 3R. Now, a capacitor is added in series such that XC = 2R. The ratio of new power factor with the old power factor of the circuit is $\sqrt 5 :x$. The value of x is ___________.
Correct Answer: 1
Explanation:
$\cos \phi = {R \over {\sqrt {{R^2} + 3{R^2}} }}$
$ = {1 \over {\sqrt {10} }}$
$\cos \phi ' = {R \over {\sqrt {{R^2} + {R^2}} }}$
$ = {1 \over {\sqrt 2 }}$
${{\cos \phi '} \over {\cos \phi }} = {{\sqrt {10} } \over {\sqrt 2 }} = {{\sqrt 5 } \over 1}$
$\therefore$ x = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
The alternating current is given by $i = \left\{ {\sqrt {42} \sin \left( {{{2\pi } \over T}t} \right) + 10} \right\}A$
The r.m.s. value of of this current is ................. A.
The r.m.s. value of of this current is ................. A.
Correct Answer: 11
Explanation:
$f_{rms}^2 = f_{1\,rms}^2 + f_{2\,rms}^2$
$ = {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}$
$ = 121 \Rightarrow {f_{rms}}$ = 11 A
$ = {\left( {{{\sqrt {42} } \over {\sqrt 2 }}} \right)^2} + {10^2}$
$ = 121 \Rightarrow {f_{rms}}$ = 11 A
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T1 is connected to T2. As the current in R = 6$\Omega$ attains a maximum value of steady state level, T1 is disconnected from T2 and immediately connected to T3. Potential drop across r = 3$\Omega$ resistor immediately after T1 is connected to T3 is __________ V. (Round off to the Nearest Integer)
Correct Answer: 3
Explanation:
What T1 and T2 are connected, then the steady state current in the inductor $I = {6 \over 6} = 1A$
When T1 and T3 are connected then current through inductor remains same. So potential difference across 3$\Omega$
V = Ir = 1 $\times$ 3 = 3 Volt
When T1 and T3 are connected then current through inductor remains same. So potential difference across 3$\Omega$
V = Ir = 1 $\times$ 3 = 3 Volt
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
Two circuits are shown in the figure (a) & (b). At a frequency of ____________ rad/s the average power dissipated in one cycle will be same in both the circuits.
Correct Answer: 500
Explanation:
For figure (a)
${P_{avg}} = {{v_{rms}^2} \over R}$
${{v_{rms}^2} \over {{Z^2}}} \times R = {{v_{rms}^2} \over R} \times 1$
${R^2} = {Z^2}$
$25 = {\left( {\sqrt {{{({x_C} - {x_L})}^2} + {5^2}} } \right)^2}$
$ = 25{({x_C} - {x_L})^2} + 25$
${x_C} = {x_L} \Rightarrow {1 \over {\omega C}} = \omega L$
${\omega ^2} = {1 \over {LC}} = {{{{10}^6}} \over {0.1 \times 40}}$
$\omega = 500$
${P_{avg}} = {{v_{rms}^2} \over R}$
${{v_{rms}^2} \over {{Z^2}}} \times R = {{v_{rms}^2} \over R} \times 1$
${R^2} = {Z^2}$
$25 = {\left( {\sqrt {{{({x_C} - {x_L})}^2} + {5^2}} } \right)^2}$
$ = 25{({x_C} - {x_L})^2} + 25$
${x_C} = {x_L} \Rightarrow {1 \over {\omega C}} = \omega L$
${\omega ^2} = {1 \over {LC}} = {{{{10}^6}} \over {0.1 \times 40}}$
$\omega = 500$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
An inductor of 10 mH is connected to a 20V battery through a resistor of 10 k$\Omega$ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 $\mu$s is ${x \over {100}}$ mA. Then x is equal to ___________. (Take e$-$1 = 0.37)
Correct Answer: 74
Explanation:
${I_{\max }} = {V \over R} = {{20V} \over {10K\Omega }} = 2$ mA
For LR - decay circuit
$I = {I_{\max }}{e^{ - Rt/L}}$
$I = 2mA{e^{{{ - 10 \times {{10}^3} \times 1 \times {{10}^{ - 6}}} \over {10 \times {{10}^{ - 3}}}}}}$
I = 2mA e$-$1
I = 2 $\times$ 0.37 mA
$I = {{74} \over {100}}$ mA
x = 74
For LR - decay circuit
$I = {I_{\max }}{e^{ - Rt/L}}$
$I = 2mA{e^{{{ - 10 \times {{10}^3} \times 1 \times {{10}^{ - 6}}} \over {10 \times {{10}^{ - 3}}}}}}$
I = 2mA e$-$1
I = 2 $\times$ 0.37 mA
$I = {{74} \over {100}}$ mA
x = 74
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
A series LCR circuit of R = 5$\Omega$, L = 20 mH and C = 0.5 $\mu$F is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______________ $\times$ 102 W.
Correct Answer: 125
Explanation:
XL = XC (due to resonance)
Z = R so ${i_{rms}} = {V \over Z} = {V \over R}$
${{{V^2}} \over R} = {{250 \times 250} \over 5} = 125 \times {10^2}W$
Z = R so ${i_{rms}} = {V \over Z} = {V \over R}$
${{{V^2}} \over R} = {{250 \times 250} \over 5} = 125 \times {10^2}W$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
In an LCR series circuit, an inductor 30 mH and a resistor 1 $\Omega$ are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45$^\circ$ is ${1 \over x} \times {10^{ - 3}}$ F. Then the value of x is ____________.
Correct Answer: 3
Explanation:
Given,
Inductance, L = 30 mH
Resistance, R = 1 $\Omega$
Angular frequency, $\omega$ = 300 rad/s
We know that in L-C-R circuit, $\tan \phi = {{{X_C} - {X_L}} \over R}$
where, $\phi$ = phase angle = 45$^\circ$
XC = capacitive reactance = ${1 \over {\omega C}}$
XL = inductive reactance = $\omega$L
$\Rightarrow$ $\tan 45^\circ = {{{X_C} - {X_L}} \over R}$
$ \Rightarrow {X_C} - {X_L} = R$ [$\because$ tan 45$^\circ$ = 1]
$ \Rightarrow {1 \over {\omega C}} - \omega L = R \Rightarrow {1 \over {\omega C}} - 300 \times 30 \times {10^{ - 3}} = 1$
$ \Rightarrow {1 \over {\omega C}} = 10 \Rightarrow \omega C = {1 \over {10}}$
$ \Rightarrow C = {1 \over {10\omega }} \Rightarrow C = {1 \over {10 \times 300}}$
$ \Rightarrow C = {1 \over 3} \times {10^{ - 3}}F$ .... (i)
According to question, the value of capacitance is ${1 \over x} \times {10^{ - 3}}F$. So, on comparing it with Eq. (i), we can say x = 3.
Inductance, L = 30 mH
Resistance, R = 1 $\Omega$
Angular frequency, $\omega$ = 300 rad/s
We know that in L-C-R circuit, $\tan \phi = {{{X_C} - {X_L}} \over R}$
where, $\phi$ = phase angle = 45$^\circ$
XC = capacitive reactance = ${1 \over {\omega C}}$
XL = inductive reactance = $\omega$L
$\Rightarrow$ $\tan 45^\circ = {{{X_C} - {X_L}} \over R}$
$ \Rightarrow {X_C} - {X_L} = R$ [$\because$ tan 45$^\circ$ = 1]
$ \Rightarrow {1 \over {\omega C}} - \omega L = R \Rightarrow {1 \over {\omega C}} - 300 \times 30 \times {10^{ - 3}} = 1$
$ \Rightarrow {1 \over {\omega C}} = 10 \Rightarrow \omega C = {1 \over {10}}$
$ \Rightarrow C = {1 \over {10\omega }} \Rightarrow C = {1 \over {10 \times 300}}$
$ \Rightarrow C = {1 \over 3} \times {10^{ - 3}}F$ .... (i)
According to question, the value of capacitance is ${1 \over x} \times {10^{ - 3}}F$. So, on comparing it with Eq. (i), we can say x = 3.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8$\Omega$, L = 24 mH and C = 60 $\mu$F. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ____________.
Correct Answer: 4
Explanation:
At resonance power (P)
$P = {{{{({V_{rms}})}^2}} \over R}$
$ \therefore $ $P = {{{{(250/\sqrt 2 )}^2}} \over 8}$
$ \Rightarrow $ P = 3906.25 w
$ \Rightarrow $ P $ \cong $ 4 Kw
$P = {{{{({V_{rms}})}^2}} \over R}$
$ \therefore $ $P = {{{{(250/\sqrt 2 )}^2}} \over 8}$
$ \Rightarrow $ P = 3906.25 w
$ \Rightarrow $ P $ \cong $ 4 Kw
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be __________.
Correct Answer: 283
Explanation:
Quality factor = ${{{X_L}} \over R} = {{\omega L} \over R}$
$Q = {1 \over {\sqrt {LC} }}{L \over R}$
$Q = \left( {{1 \over {\sqrt C }}} \right){{\sqrt L } \over R}$
$Q = {{XL} \over R} = {{\omega L} \over R} = {1 \over {\sqrt {LC} }}{L \over R} = {1 \over R}{{\sqrt L } \over {\sqrt C }}$
$Q' = {{\sqrt {2L} } \over {\left( {{R \over 2}} \right)\sqrt C }} = 2\sqrt 2 Q$
Q' = 282.84
$Q = {1 \over {\sqrt {LC} }}{L \over R}$
$Q = \left( {{1 \over {\sqrt C }}} \right){{\sqrt L } \over R}$
$Q = {{XL} \over R} = {{\omega L} \over R} = {1 \over {\sqrt {LC} }}{L \over R} = {1 \over R}{{\sqrt L } \over {\sqrt C }}$
$Q' = {{\sqrt {2L} } \over {\left( {{R \over 2}} \right)\sqrt C }} = 2\sqrt 2 Q$
Q' = 282.84
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 $\mu$F is used in the resonant circuit. The self inductance of coil necessary for resonance is __________ $\times$ 10$-$8 H.
Correct Answer: 10
Explanation:
$\lambda$ = 960 m
C = 2.56 $\mu$F = 2.56 $\times$ 10$-$6 F
c = 3 $\times$ 108 m/s
L = ?
Now at resonance, ${\omega _0} = {1 \over {\sqrt {LC} }}$
[Resonant frequency]
$2\pi {f_0} = {1 \over {\sqrt {LC} }}$
On substituting ${f_0} = {c \over \lambda }$, we have $2\pi {c \over \lambda } = {1 \over {\sqrt {LC} }}$
Squaring both sides : $4{\pi ^2}{{{c^2}} \over {{\lambda ^2}}} = {1 \over {LC}}$
$ = {{4 \times 10 \times {{(3 \times {{10}^8})}^2}} \over {{{(960)}^2}}} = {1 \over {L \times 2.56 \times {{10}^{ - 6}}}}$
$ \Rightarrow {1 \over L} = {{4 \times 10 \times 9 \times {{10}^{16}} \times 2.56 \times {{10}^{ - 6}}} \over {960 \times 960}}$
$ \Rightarrow L = 10 \times {10^{ - 8}}$ H
C = 2.56 $\mu$F = 2.56 $\times$ 10$-$6 F
c = 3 $\times$ 108 m/s
L = ?
Now at resonance, ${\omega _0} = {1 \over {\sqrt {LC} }}$
[Resonant frequency]
$2\pi {f_0} = {1 \over {\sqrt {LC} }}$
On substituting ${f_0} = {c \over \lambda }$, we have $2\pi {c \over \lambda } = {1 \over {\sqrt {LC} }}$
Squaring both sides : $4{\pi ^2}{{{c^2}} \over {{\lambda ^2}}} = {1 \over {LC}}$
$ = {{4 \times 10 \times {{(3 \times {{10}^8})}^2}} \over {{{(960)}^2}}} = {1 \over {L \times 2.56 \times {{10}^{ - 6}}}}$
$ \Rightarrow {1 \over L} = {{4 \times 10 \times 9 \times {{10}^{16}} \times 2.56 \times {{10}^{ - 6}}} \over {960 \times 960}}$
$ \Rightarrow L = 10 \times {10^{ - 8}}$ H
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
A series L-C-R circuit is designed to resonate at an angular frequency $\omega$0 = 105 rad/s. The circuit draws 16W power from 120V source at resonance. The value of resistance 'R' in the circuit is _________ $\Omega$.
Correct Answer: 900
Explanation:
Given, angular frequency at resonance, $\omega$0 = 105 rads$-$1
Power drawn from circuit, P = 16 W
and supply voltage, V = 120 V
Let resistance of circuit = R.
As, $P = {V^2}/R$
$ \Rightarrow R = {V^2}/P = {{120 \times 100} \over {16}}$
$ = 30 \times 30 = 900\,\Omega $
Power drawn from circuit, P = 16 W
and supply voltage, V = 120 V
Let resistance of circuit = R.
As, $P = {V^2}/R$
$ \Rightarrow R = {V^2}/P = {{120 \times 100} \over {16}}$
$ = 30 \times 30 = 900\,\Omega $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
A common transistor radio set requires 12 V (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V (A.C.) on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are ___________.
Correct Answer: 440
Explanation:
In a transformer,
${{{N_p}} \over {{N_s}}} = {{{V_p}} \over {{V_s}}}$
where, Np = number of turns in primary circuit, Ns = number of turns in secondary circuit = 24, Vp = potential of primary circuit = 220 V and Vs = potential of secondary circuit = 12 V
$ \Rightarrow {{{N_p}} \over {24}} = {{220} \over {12}}$
$ \Rightarrow {N_p} = 440$
${{{N_p}} \over {{N_s}}} = {{{V_p}} \over {{V_s}}}$
where, Np = number of turns in primary circuit, Ns = number of turns in secondary circuit = 24, Vp = potential of primary circuit = 220 V and Vs = potential of secondary circuit = 12 V
$ \Rightarrow {{{N_p}} \over {24}} = {{220} \over {12}}$
$ \Rightarrow {N_p} = 440$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
A resonance circuit having inductance and resistance 2 $\times$ 10$-$4 H and 6.28$\Omega$ respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is ___________. [$\pi$ = 3.14]
Correct Answer: 2000
Explanation:
Given, L = 2 $\times$ 10$-$4 H, R = 6.28 $\Omega$, f0 = 10 MHz = 10 $\times$ 106 Hz
$\therefore$ Quality factor $ = {\omega _0}{L \over R} = 2\pi {f_0}{L \over R}$
$ = 2\pi \times 10 \times {10^6} \times {{2 \times {{10}^{ - 4}}} \over {6.28}}$
$ = 2 \times {10^3} = 2000$
$\therefore$ Quality factor $ = {\omega _0}{L \over R} = 2\pi {f_0}{L \over R}$
$ = 2\pi \times 10 \times {10^6} \times {{2 \times {{10}^{ - 4}}} \over {6.28}}$
$ = 2 \times {10^3} = 2000$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
For the given circuit the current i through the battery when the key in closed and the steady state has been reached is __________.
A.
6 A
B.
25 A
C.
10 A
D.
0 A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
In an ac circuit, an inductor, a capacitor and a resistor are connected in series with XL = R = XC. Impedance of this circuit is :
A.
2R2
B.
Zero
C.
R
D.
R$\sqrt 2 $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Evening Shift
In the given circuit the AC source has $\omega$ = 100 rad s-1. Considering the inductor and capacitor to be ideal, what will be the current I flowing through the circuit?
A.
5.9 A
B.
3.16 A
C.
0.94 A
D.
6 A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k$\Omega$, an inductor of inductive reactance XL = 250 $\pi$$\Omega$ and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take $\pi$2 = 10)
A.
4 $\mu$F
B.
25 $\mu$F
C.
400 $\mu$F
D.
40 $\mu$F
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Evening Shift
A 100$\Omega$ resistance, a 0.1 $\mu$F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.
A.
0.70 H
B.
70.3 mH
C.
7.03 $\times$ 10$-$5 H
D.
70.3 H
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
A 0.07 H inductor and a 12$\Omega$ resistor are connected in series to a 220V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $\pi$ as ${{22} \over 7}$]
A.
8.8 A and ${\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$
B.
88 A and ${\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$
C.
0.88 A and ${\tan ^{ - 1}}\left( {{{11} \over 6}} \right)$
D.
8.8 A and ${\tan ^{ - 1}}\left( {{{6} \over 11}} \right)$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
A 10 $\Omega$ resistance is connected across 220V $-$ 50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is :
A.
2.5 ms
B.
1.5 ms
C.
3.0 ms
D.
4.5 ms
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
In a circuit consisting of a capacitance and a generator with alternating emf Eg = Eg0 sin$\omega$t, VC and IC are the voltage and current. Correct phasor diagram for such circuit is
A.
B.
C.
D.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Match List - I with List - II
Choose the correct answer from the options given below
| List - I | List - II | ||
|---|---|---|---|
| (a) | $\omega L > {1 \over {\omega C}}$ | (i) | Current is in phase with emf |
| (b) | $\omega L = {1 \over {\omega C}}$ | (ii) | Current lags behind the applied emf |
| (c) | $\omega L < {1 \over {\omega C}}$ | (iii) | Maximum current occurs |
| (d) | Resonant frequency | (iv) | Current leads the emf |
Choose the correct answer from the options given below
A.
a(ii), b(i), c(iv), d(iii)
B.
a(ii), b(i), c(iii), d(iv)
C.
a(iii), b(i), c(iv), d(ii)
D.
a(iv), b(iii), c(ii), d(i)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
For a series LCR circuit with R = 100 $\Omega$, L = 0.5 mH and C = 0.1 pF connected across 220V$-$50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :
A.
0$^\circ$, resistive circuit
B.
$ \approx $ 90$^\circ$, predominantly inductive circuit
C.
0$^\circ$, resonance circuit
D.
$ \approx $ 90$^\circ$, predominantly capacitive circuit
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
AC voltage V(t) = 20 sin$\omega$t of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2. The amplitude of the oscillating displacement current for the applied AC voltage is _________. [Take $\varepsilon $0 = 8.85 $\times$ 10$-$12 F/m]
A.
55.58 $\mu$A
B.
21.14 $\mu$A
C.
27.79 $\mu$A
D.
83.37 $\mu$A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
In a series LCR circuit, the inductive reactance (XL) is 10$\Omega$ and the capacitive reactance (XC) is 4$\Omega$. The resistance (R) in the circuit is 6$\Omega$. The power factor of the circuit is :
A.
${1 \over 2}$
B.
${{\sqrt 3 } \over 2}$
C.
${1 \over {\sqrt 2 }}$
D.
${1 \over {2\sqrt 2 }}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :
A.
The bandwidth of resonance circuit will increase.
B.
The resonance frequency will increase.
C.
The quality factor will increase.
D.
The quality factor and the resonance frequency will remain constant.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :
A.
2.5 ms
B.
25 ms
C.
2.5 s
D.
0.25 ms
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
Match List - I with List - II
Choose the most appropriate answer from the options given below :
| List - I | List - II | ||
|---|---|---|---|
| (a) | Phase difference between current and voltage in a purely resistive AC circuit | (i) | ${\pi \over 2}$; current leads voltage |
| (b) | Phase difference between current and voltage in a pure inductive AC circuit | (ii) | zero |
| (c) | Phase difference between current and voltage in a pure capacitive AC circuit | (iii) | ${\pi \over 2}$; current lags voltage |
| (d) | Phase difference between current and voltage in an LCR series circuit | (iv) | ${\tan ^{ - 1}}\left( {{{{X_C} - {X_L}} \over R}} \right)$ |
Choose the most appropriate answer from the options given below :
A.
(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
B.
(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
C.
(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
D.
(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Evening Shift
What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?
A.
Both, inducting reactance and current will be doubled.
B.
Inductive reactance will be doubled and current will be halved.
C.
Both, inductive reactance and current will be halved.
D.
Inductive reactance will be halved and current will be doubled.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
An AC current is given by I = I1 sin$\omega$t + I2 cos$\omega$t. A hot wire ammeter will give a reading :
A.
${{{I_1} + {I_2}} \over {\sqrt 2 }}$
B.
$\sqrt {{{I_1^2 - I_2^2} \over 2}} $
C.
$\sqrt {{{I_1^2 + I_2^2} \over 2}} $
D.
${{{I_1} + {I_2}} \over {2\sqrt 2 }}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
For the given circuit, comment on the type of transformer used.
A.
Auxiliary transformer
B.
Step down transformer
C.
Step-up transformer
D.
Auto transformer
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
An RC circuit as shown in the figure is driven by a AC source generating a square wave. The output wave pattern monitored by CRO would look close to :
A.
B.
C.
D.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
Find the peak current and resonant frequency of the following circuit (as shown in figure).
A.
2 A and 100 Hz
B.
2 A and 50 Hz
C.
0.2 A and 100 Hz
D.
0.2 A and 50 Hz
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
An alternating current is given by the equation i = i1 sin $\omega$t + i2 cos $\omega$t. The rms current will be :
A.
${1 \over {\sqrt 2 }}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}$
B.
${1 \over {\sqrt 2 }}({i_1} + {i_2})$
C.
${1 \over {\sqrt 2 }}{({i_1} + {i_2})^2}$
D.
${1 \over 2}{\left( {i_1^2 + i_2^2} \right)^{{1 \over 2}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
An LCR circuit contains resistance of 110$\Omega$ and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45$^\circ$. If on the other hand, only inductor is removed the current leads by 45$^\circ$ with the applied voltage. The rms current flowing in the circuit will be :
A.
1 A
B.
2.5 A
C.
2 A
D.
1.5 A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
Match List I with List II.
Choose the correct answer from the options given below :
| List I | List II | ||
|---|---|---|---|
| (a) | Rectifier | (i) | Used either for stepping up or stepping down the a.c. voltage |
| (b) | Stabilizer | (ii) | Used to convert a.c. voltage into d.c. voltage |
| (c) | Transformer | (iii) | Used to remove any ripple in the rectified output voltage |
| (d) | Filter | (iv) | Used for constant output voltage even when the input voltage or load current change |
Choose the correct answer from the options given below :
A.
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
B.
(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
C.
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
D.
(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
The angular frequency of alternating current in a L-C-R circuit is 100 rad/s. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.
A.
0.8 H and 150 $\mu$F
B.
0.8 H and 250 $\mu$F
C.
1.33 H and 150 $\mu$F
D.
1.33 H and 250 $\mu$F
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
The current (i) at time t = 0 and t = $\infty $ respectively for the given circuit is :
A.
${{18E} \over {55}},{{5E} \over {18}}$
B.
${{5E} \over {18}},{{18E} \over {55}}$
C.
${{5E} \over {18}},{{10E} \over {33}}$
D.
${{10E} \over {33}},{{5E} \over {18}}$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
In a series LR circuit, power of 400W is dissipated from a source of 250 V, 50 Hz. The power factor
of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in
series to the L and R. Taking the value of C as $\left( {{n \over {3\pi }}} \right)$ $\mu $F, then value of n is __________.
Correct Answer: 400
Explanation:
Given, power factor of LR circuit,
cos $\phi $ = 0.8 = ${R \over {\sqrt {{R^2} + X_L^2} }}$ = ${R \over Z}$
We know,
Power, P = ${{V_{rms}^2} \over {{Z^2}}} \times R$
$ \Rightarrow $ 400 = ${{{{\left( {250} \right)}^2} \times 0.8Z} \over {{Z^2}}}$
$ \Rightarrow $ Z = 125
$ \therefore $ R = 0.8 $ \times $ 125 = 100 $\Omega $
As Z2 = $X_L^2 + {R^2}$
$ \Rightarrow $ ${\left( {125} \right)^2} = X_L^2 + {\left( {100} \right)^2}$
$ \Rightarrow $ XL = 75
In 2nd case given.
Power factor = 1
that means XL = XC (Resonance condition)
XL = ${1 \over {{\omega _c}}}$
$ \Rightarrow $ 75 = ${1 \over {\left( {2\pi F} \right)C}}$
$ \Rightarrow $ C = ${1 \over {\left( {2\pi \times 50} \right)75}}$
Also given, C = $\left( {{n \over {3\pi }}} \right)$ $\mu $F
$ \therefore $ ${1 \over {\left( {2\pi \times 50} \right)75}} = {{n \times {{10}^{ - 6}}} \over {3\pi }}$
$ \Rightarrow $ n = 400
cos $\phi $ = 0.8 = ${R \over {\sqrt {{R^2} + X_L^2} }}$ = ${R \over Z}$
We know,
Power, P = ${{V_{rms}^2} \over {{Z^2}}} \times R$
$ \Rightarrow $ 400 = ${{{{\left( {250} \right)}^2} \times 0.8Z} \over {{Z^2}}}$
$ \Rightarrow $ Z = 125
$ \therefore $ R = 0.8 $ \times $ 125 = 100 $\Omega $
As Z2 = $X_L^2 + {R^2}$
$ \Rightarrow $ ${\left( {125} \right)^2} = X_L^2 + {\left( {100} \right)^2}$
$ \Rightarrow $ XL = 75
In 2nd case given.
Power factor = 1
that means XL = XC (Resonance condition)
XL = ${1 \over {{\omega _c}}}$
$ \Rightarrow $ 75 = ${1 \over {\left( {2\pi F} \right)C}}$
$ \Rightarrow $ C = ${1 \over {\left( {2\pi \times 50} \right)75}}$
Also given, C = $\left( {{n \over {3\pi }}} \right)$ $\mu $F
$ \therefore $ ${1 \over {\left( {2\pi \times 50} \right)75}} = {{n \times {{10}^{ - 6}}} \over {3\pi }}$
$ \Rightarrow $ n = 400
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
An AC circuit has R= 100 $\Omega $, C = 2 $\mu $F and L = 80 mH, connected in series. The quality factor of the
circuit is :
A.
20
B.
2
C.
0.5
D.
400
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then
the time at which the energy stored in the inductor reaches $\left( {{1 \over n}} \right)$ times of its maximum value, is :
A.
${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)$
B.
${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)$
C.
${L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)$
D.
${L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
A 750 Hz, 20 V (rms) source is connected to a
resistance of 100 $\Omega $, an inductance of 0.1803 H
and a capacitance of 10 $\mu $F all in series. The
time in which the resistance (heat capacity
2 J/oC) will get heated by 10oC. (assume no loss
of heat to the surroudnings) is close to :
A.
348 s
B.
418 s
C.
245 s
D.
365 s
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
An inductance coil has a reactance of 100 $\Omega $.
When an AC signal of frequency 1000 Hz is
applied to the coil, the applied voltage leads
the current by 45o. The self-inductance of the
coil is
A.
6.7 $ \times $ 10–7 H
B.
1.1 $ \times $ 10–1 H
C.
5.5 $ \times $ 10–5 H
D.
1.1 $ \times $ 10–2 H
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
In LC circuit the inductance L = 40 mH and
capacitance C = 100 $\mu $F. If a voltage
V(t) = 10sin(314t) is applied to the circuit, the
current in the circuit is given as :
capacitance C = 100 $\mu $F. If a voltage
V(t) = 10sin(314t) is applied to the circuit, the
current in the circuit is given as :
A.
0.52 cos 314 t
B.
5.2 cos 314 t
C.
0.52 sin 314 t
D.
10 cos 314 t
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 $\Omega $
resistor. The ratio of the currents at time t = $\infty $ and at t = 40 s is close to : (Take e2 = 7.389)
A.
1.06
B.
0.84
C.
1.15
D.
1.46
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be:
A.
L $ \leftrightarrow $ k, C $ \leftrightarrow $ b, R $ \leftrightarrow $ m
B.
L $ \leftrightarrow $ m, C $ \leftrightarrow $ k, R $ \leftrightarrow $ b
C.
L $ \leftrightarrow $ m, C $ \leftrightarrow $ ${1 \over k}$, R $ \leftrightarrow $ b
D.
L $ \leftrightarrow $ ${1 \over b}$, C $ \leftrightarrow $ ${1 \over m}$, R $ \leftrightarrow $ ${1 \over k}$