Practice Start Test

Alternating Current

JEE Physics · 99 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
A circuit has a resistance of 1212 ohmohm and an impedance of 1515 ohmohm. The power factor of the circuit will be
A 0.40.4
B 0.80.8
C 0.1250.125
D 1.251.25
Correct Answer
Option B
Solution

Power factor

=cos⁡ϕ=RZ=1215=45=0.8= \cos \phi = {R \over Z} = {{12} \over {15}} = {4 \over 5} = 0.8
Q12
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.
Correct Answer
Option C
Solution
X=∣XC−XL∣X = |{X_C} - {X_L}|

So, it can be zero if

XC=XL{X_C} = {X_L}

And, average power in ac circuit can be zero.

Q13
A resistance of 40 Ω\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
Correct Answer
Option A
Solution
I=I0cos⁡(ωt)I = {I_0}\cos (\omega t)

say ⇒\Rightarrow At maximum

ωt1=0\omega {t_1} = 0

or

t1=0{t_1} = 0

Then at rms value

I=I0/2I = {I_0}/\sqrt 2
⇒ωt2=π/4\Rightarrow \omega {t_2} = \pi /4
⇒ω(t2−t1)=π/4\Rightarrow \omega ({t_2} - {t_1}) = \pi /4
Δt=π4ω=πT4×2π\Delta t = {\pi \over {4\omega }} = {{\pi T} \over {4 \times 2\pi }}
=1400s= {1 \over {400}}s

or 2.5 ms

Q14
When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on :
A Electromagnetic induction
B Resonance in ac circuits
C Mutual induction in ac circuits
D Interference of electromagnetic waves
Correct Answer
Option B
Solution

Metal detector works on the principle of resonance in ac circuits.

Q15
In a series LRL R circuit XL=RX_{L}=R and power factor of the circuit is P1P_{1}. When capacitor with capacitance CC such that XL=XCX_{L}=X_{C} is put in series, the power factor becomes P2P_{2}. The ratio P1P2\dfrac{P_{1}}{P_{2}} is:
A 12\dfrac{1}{2}
B 12\dfrac{1}{\sqrt{2}}
C 32\dfrac{\sqrt{3}}{\sqrt{2}}
D 2 : 1
Correct Answer
Option B
Solution
P1=cos⁡ϕ=12(XL=R){P_1} = \cos \phi = {1 \over {\sqrt 2 }}({X_L} = R)
P2=cos⁡ϕ′=1{P_2} = \cos \phi ' = 1

(will become resonance circuit) So,

P1P2=12{{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}
Q16
A coil of self inductance 10 mH and resistance 0.1 Ω\Omega is connected through a switch to a battery of internal resistance 0.9 Ω\Omega . After the switch is closed, the time taken for the current to attain 80% of the saturation value is: [take ln 5 = 1.6]
A 0.324 s
B 0.002 s
C 0.103 s
D 0.016 s
Correct Answer
Option D
Solution

L = 10 × 10–3 H, r1 = 0.1

Ω\Omega
i=ε{1−e−1/2}i = \varepsilon \left\{ {1 - {e^{ - 1/2}}} \right\}
isaturation=ε{i_{saturation}}{\rm{ }} = {\rm{ }}\varepsilon
80%isaturation=0.8ε80\% {\rm{ }}{i_{saturation}}{\rm{ }} = {\rm{ }}0.8{\rm{ }}\varepsilon

0.8 = 1 - e-t/2 ; e-t/2 = 0.2 et/L = 5 t = L ln 5 = 10 × 10–3 × 1.6 = 16 × 10–3

Q17
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
Correct Answer
Option A
Solution
I=I0sin⁡ωtI = {I_0}\sin \omega t

⇒\Rightarrow

IM2=IMsin⁡ωt{{{I_M}} \over {\sqrt 2 }} = {I_M}\sin \omega t

⇒\Rightarrow

ωt=π4\omega t = {\pi \over 4}

⇒\Rightarrow

t=π4ωt = {\pi \over {4\omega }}
=π4(2πf)= {\pi \over {4\left( {2\pi f} \right)}}

⇒\Rightarrow t =

18×30=1400{1 \over {8 \times 30}} = {1 \over {400}}

= 2.5 ms

Q18
In a series LCR circuit, the inductive reactance (XL) is 10Ω\Omega and the capacitive reactance (XC) is 4Ω\Omega. The resistance (R) in the circuit is 6Ω\Omega. The power factor of the circuit is :
A 12{1 \over 2}
B 32{{\sqrt 3 } \over 2}
C 12{1 \over {\sqrt 2 }}
D 122{1 \over {2\sqrt 2 }}
Correct Answer
Option C
Solution

Given : XL = 10

Ω\Omega

XC = 4

Ω\Omega

R = 6

Ω\Omega

∴\therefore Power factor = cosθ\theta =

RZ{R \over Z}
=RR2+(XL−XC)2= {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }}
=662+(10−4)2= {6 \over {\sqrt {{6^2} + {{(10 - 4)}^2}} }}
=662=12= {6 \over {6\sqrt 2 }} = {1 \over {\sqrt 2 }}
Q19
An alternating current is given by I=IAsin⁡ωt+IBcos⁡ωt\mathrm{I}=\mathrm{I}_{\mathrm{A}} \sin \omega \mathrm{t}+\mathrm{I}_{\mathrm{B}} \cos \omega \mathrm{t}. The r.m.s current will be
A IA2+IB22\dfrac{\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}}{2}
B IA2+IB22\sqrt{\dfrac{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}{2}}
C ∣IA+IB∣2\dfrac{\left|\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}\right|}{\sqrt{2}}
D IA2+IB2\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}
Correct Answer
Option B
Solution
Irms=IA2+IB22I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}

To determine the root mean square (r.m.s) value of the alternating current given by

I(t)=IAsin⁡(ωt)+IBcos⁡(ωt),I(t) = I_A \sin (\omega t) + I_B \cos (\omega t),

we start by squaring the current:

I(t)2=IA2sin⁡2(ωt)+IB2cos⁡2(ωt)+2IAIBsin⁡(ωt)cos⁡(ωt).I(t)^2 = I_A^2 \sin^2 (\omega t) + I_B^2 \cos^2 (\omega t) + 2 I_A I_B \sin (\omega t) \cos (\omega t).

The r.m.s value is defined as the square root of the average of this squared current over one complete period.

When averaging over a full cycle, we utilize the known averages:

⟨sin⁡2(ωt)⟩=12,⟨cos⁡2(ωt)⟩=12,⟨sin⁡(ωt)cos⁡(ωt)⟩=0.\langle \sin^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \cos^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \sin (\omega t) \cos (\omega t) \rangle = 0.

Thus, the time-averaged square of the current becomes:

⟨I(t)2⟩=IA212+IB212=IA2+IB22.\langle I(t)^2 \rangle = I_A^2 \frac{1}{2} + I_B^2 \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}.

Taking the square root gives the r.m.s current:

Irms=IA2+IB22.I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}.

This corresponds to Option B.

Q20
An alternating voltage of amplitude 40 V40 \mathrm{~V} and frequency 4 kHz4 \mathrm{~kHz} is applied directly across the capacitor of 12μF12 \mu \mathrm{F}. The maximum displacement current between the plates of the capacitor is nearly :
A 10 A
B 8 A
C 13 A
D 12 A
Correct Answer
Option D
Solution

Let's calculate the maximum displacement current between the plates of the capacitor when an alternating voltage is applied directly across it.

The formula for the capacitive reactance

XCX_C

of a capacitor is given by: XC=12πfCX_C = \dfrac{1}{2\pi fC} where

ff

is the frequency of the alternating voltage, and

CC

is the capacitance of the capacitor. Given: The amplitude of the voltage (

VmaxV_{max}

) =

4040

V Frequency (

ff

) =

40004000

Hz (or

4 kHz4 \mathrm{~kHz}

) Capacitance (

CC

) =

12μF=12×10−6 F12 \mu \mathrm{F} = 12 \times 10^{-6} \mathrm{~F}

First, we calculate

XCX_C

: XC=12π(4000)(12×10−6)≈12π⋅4⋅12⋅10−3≈196π⋅10−3X_C = \dfrac{1}{2\pi(4000)(12 \times 10^{-6})} \approx \dfrac{1}{2\pi \cdot 4 \cdot 12 \cdot 10^{-3}} \approx \dfrac{1}{96\pi \cdot 10^{-3}} XC≈13.14⋅96⋅10−3≈10.30144≈3.316ΩX_C \approx \dfrac{1}{3.14 \cdot 96 \cdot 10^{-3}} \approx \dfrac{1}{0.30144} \approx 3.316 \Omega Now, the maximum current (

ImaxI_{max}

) in the circuit can be calculated using Ohm's law, considering the maximum voltage across the capacitor and its capacitive reactance: Imax=VmaxXCI_{max} = \dfrac{V_{max}}{X_C} Imax=403.316≈12.06 AI_{max} = \dfrac{40}{3.316} \approx 12.06 \mathrm{~A} Hence, the maximum displacement current between the plates of the capacitor is nearly: Option D: 12 A

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