Practice Start Test

Alternating Current

JEE Physics · 99 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be ;
A 50 A
B 45 A
C 25 A
D 20 A
Correct Answer
Option B
Solution

Efficiency n = 0.9 =

PsPp{{{P_s}} \over {{P_p}}}

as

      \,\,\,\,\,\,

P = VI

∴   \therefore\,\,\,

Ps = 0.9 ×\times Pp ⇒\Rightarrow

    \,\,\,\,

Vs Is = 0.9 ×\times Vp Ip ⇒\Rightarrow

    \,\,\,\,

Is =

0.9×2300×5230{{0.9 \times 2300 \times 5} \over {230}}

=

    \,\,\,\,

45 A

Q62
In a series LCRLCR circuit R=200ΩR = 200\Omega and the voltage and the frequency of the main supply is 220V220V and 5050 HzHz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30∘.{30^ \circ }. On taking out the inductor from the circuit the current leads the voltage by 30∘.{30^ \circ }. The power dissipated in the LCRLCR circuit is
A 305305 WW
B 210210 WW
C zerozero WW
D 242242 WW
Correct Answer
Option D
Solution

When capacitance is taken out, the circular is

LR.LR.

∴\therefore

tan⁡ϕ=ωLR\tan \phi = {{\omega L} \over R}
⇒ωL=R tan⁡ϕ\Rightarrow \omega L = R\,\tan \phi
=200×13=2003= 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}

Again, when inductor is taken out, the circuit is

CR.CR.

∴\therefore

tan⁡ϕ=1ωCR\tan \phi = {1 \over {\omega CR}}
⇒1ωc=Rtan⁡ϕ\Rightarrow {1 \over {\omega c}} = R\tan \phi
=200×13=2003= 200 \times {1 \over {\sqrt 3 }} = {{200} \over {\sqrt 3 }}

Now,

Z=R2+(1ωC−ωL)2Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}} - \omega L} \right)}^2}}
=(200)2+(2003−2003)2=200Ω= \sqrt {{{\left( {200} \right)}^2} + {{\left( {{{200} \over {\sqrt 3 }} - {{200} \over {\sqrt 3 }}} \right)}^2}} = 200\Omega

Power dissipated

=VrmsIrmscos⁡ϕ= {V_{rms}}{I_{rms}}\cos \phi
=Vrms.VrmsZ.RZ= {V_{rms}}.{{{V_{rms}}} \over Z}.{R \over Z}
(   \left( {\,\,\,} \right.

as

  cos⁡ϕ=RZ   )\left. {\,\,\cos \phi = {R \over Z}\,\,\,} \right)
=V2rmsRZ2=(220)2×200(200)2= {{{V^2}rmsR} \over {{Z^2}}} = {{{{\left( {220} \right)}^2} \times 200} \over {{{\left( {200} \right)}^2}}}
=220×220200=242 W= {{220 \times 220} \over {200}} = 242\,W
Q63
An ideal capacitor of capacitance 0.2 μF0.2\,\mu F is charged to a potential difference of 1010 V.V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.50.5 mH.mH. The current at a time when the potential difference across the capacitor is 55 V,V, is :
A 0.34  A0.34\,\,A
B 0.25  A0.25\,\,A
C 0.17  A0.17\,\,A
D 0.15  A0.15\,\,A
Correct Answer
Option C
Solution

Capacitance, C = 0.2 μ\muF = 0.2 ×\times 10−-6 F Inductance, L = 0.5 m H = 0.5 ×\times 10−-3 H Let, current = I.

Using energy conservation, UE + 0 = UE' + Ub' ⇒\Rightarrow

    \,\,\,\,
12{1 \over 2}

cv2 + 0 =

12{1 \over 2}

c

v12v_1^2

+

12{1 \over 2}

LI2 ⇒\Rightarrow

    \,\,\,\,
12{1 \over 2}

×\times 0.2 ×\times 10−-6 ×\times 102 =

12×{1 \over 2} \times

0.2 ×\times 10−-6 ×\times 52 +

12{1 \over 2}

×\times 0.5 ×\times 10−-3 ×\times I2 By solving this, I =

3×\sqrt 3 \times

10−-1 A = 0.17 A.

Q64
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
Correct Answer
Option B
Solution

For wattless current to flow in AC circuit the circuit will be Purely Inductive circuit.

Q65
An alternating voltage v(t) = 220 sin 100 π\pi t volt is applied to a purely resistance load of 50Ω\Omega . The time taken for the current to rise from half of the peak value to the peak value is :
A 5 ms
B 2.2 ms
C 3.3 ms
D 7.2 ms
Correct Answer
Option C
Solution

In an AC resistive circuit, current and voltage are in phase. So,

I=VR⇒I=22050sin⁡(100πt)I = {V \over R} \Rightarrow I = {{220} \over {50}}\sin (100\pi t)

..... (i) ∴\therefore Time period of one complete cycle of current is

T=2πω=2π100π=150sT = {{2\pi } \over \omega } = {{2\pi } \over {100\pi }} = {1 \over {50}}s

So, current reaches its maximum value at

t1=T4=1200s{t_1} = {T \over 4} = {1 \over {200}}s

When current is half of its maximum value, then from Eq.(i), we have

I=Imax⁡2=Imax⁡sin⁡(100πt2)I = {{{I_{\max }}} \over 2} = {I_{\max }}\sin (100\pi {t_2})
⇒sin⁡(100πt2)=12⇒100πt2=5π6\Rightarrow \sin (100\pi {t_2}) = {1 \over 2} \Rightarrow 100\pi {t_2} = {{5\pi } \over 6}

So, instantaneous time at which current is half of maximum value is

t2=1120s{t_2} = {1 \over {120}}s

Hence, time duration in which current reaches half of its maximum value after reaching maximum value is

Δt=t2−t1=1120−1200=1300s=3.3\Delta t = {t_2} - {t_1} = {1 \over {120}} - {1 \over {200}} = {1 \over {300}}s = 3.3

ms

Q66
An AC circuit has R= 100 Ω\Omega , C = 2 μ\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
Correct Answer
Option B
Solution
Q=1RLCQ = {1 \over R}\sqrt {{L \over C}}

=

110080×10−32×10−6{1 \over {100}}\sqrt {{{80 \times {{10}^{ - 3}}} \over {2 \times {{10}^{ - 6}}}}}

=

110040×103{1 \over {100}}\sqrt {40 \times {{10}^3}}

=

200100{{200} \over {100}}

= 2

Q67
In series LCR circuit, the capacitance is changed from CC to 4C4 C. To keep the resonance frequency unchanged, the new inductance should be:
A increased by 2 L2 \mathrm{~L}
B reduced by 14 L\dfrac{1}{4} \mathrm{~L}
C reduced by 34 L\dfrac{3}{4} \mathrm{~L}
D increased to 4 L4 \mathrm{~L}
Correct Answer
Option C
Solution

The resonance frequency

f0f_0

of an LCR circuit is given by:

f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}

where:

LL

is the inductance.

CC

is the capacitance. To keep the resonance frequency unchanged when the capacitance is changed from

CC

to

4C4C

, we must adjust the inductance

LL

to a new value

L′L'

such that:

12πLC=12πL′⋅4C\frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{L' \cdot 4C}}

Now solving for

L′L'

:

LC=4CL′\sqrt{LC} = \sqrt{4CL'}

Squaring both sides, we have:

LC=4CL′LC = 4CL'

Dividing both sides by

4C4C

, we get:

L4=L′\frac{L}{4} = L'

Thus, the new inductance

L′L'

is one fourth of the original inductance

LL

. Therefore, to achieve the same resonance frequency with the capacitance increased to

4C4C

, the inductance should be reduced to a quarter of its initial value:

L′=L4L' = \frac{L}{4}

This means we have reduced the inductance by

34L\frac{3}{4}L

, so the correct answer is: Option C: reduced by

34L\frac{3}{4}L
Q68
A 750 Hz, 20 V (rms) source is connected to a resistance of 100 Ω\Omega , an inductance of 0.1803 H and a capacitance of 10 μ\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
Correct Answer
Option A
Solution

f = 750 Hz, Vrms = 20 V, R = 100

Ω\Omega

, L = 0.1803 H, C = 10μ\mu F, S = 2 J/°C |Z| =

R2+(XL−XC)2\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}

=

R2+(ωL−1ωC)2\sqrt {{R^2} + {{\left( {\omega L - {1 \over {\omega C}}} \right)}^2}}

=

R2+(2πfL−12πfC)2\sqrt {{R^2} + {{\left( {2\pi fL - {1 \over {2\pi fC}}} \right)}^2}}

=

(100)2+(2×3.14×750×0.1803−12×3.14×750×10−5)2\sqrt {{{(100)}^2} + {{\left( {2 \times 3.14 \times 750 \times 0.1803 - {1 \over {2 \times 3.14 \times 750 \times {{10}^{ - 5}}}}} \right)}^2}}

= 834

Ω\Omega

In AC, power (P) = irmsVrms cos ϕ\phi and irms =

Vrms∣Z∣{{{V_{rms}}} \over {\left| Z \right|}}

Power factor (cos ϕ\phi) =

R∣Z∣{R \over {\left| Z \right|}}

∴\therefore P =

Vrms∣Z∣.Vrms.R∣Z∣{{{V_{rms}}} \over {\left| Z \right|}}.{V_{rms}}.{R \over {\left| Z \right|}}

=

(Vrms∣Z∣)2R{\left( {{{{V_{rms}}} \over {\left| Z \right|}}} \right)^2}R

=

(20834)2×100{\left( {{{20} \over {834}}} \right)^2} \times 100

= 0.0575 J/S Also, H = Pt = S

Δ\Delta

θ\theta ⇒\Rightarrow t =

2(10)0.0575{{2\left( {10} \right)} \over {0.0575}}

= 348 sec

Q69
An inductance coil has a reactance of 100 Ω\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
Correct Answer
Option D
Solution

L-R circuit : tan 45o =

XLR{{{X_L}} \over R}

⇒\Rightarrow 1 =

XLR{{{X_L}} \over R}

⇒\Rightarrow XL = R Now Z =

R2+XL2\sqrt {{R^2} + X_L^2}

or Z =

XL2+XL2=2XL2\sqrt {X_L^2 + X_L^2} = \sqrt {2X_L^2}

=

2XL\sqrt 2 {X_L}

⇒\Rightarrow 100 =

2XL\sqrt 2 {X_L}

XL =

1002{{100} \over {\sqrt 2 }}

⇒\Rightarrow

ωL\omega L

=

1002{{100} \over {\sqrt 2 }}

⇒\Rightarrow L =

1002×2×3.14×1000{{100} \over {\sqrt 2 \times 2 \times 3.14 \times 1000}}

= 1.1 ×\times 10–2 H

Q70
An LCR circuit contains resistance of 110Ω\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
Correct Answer
Option C
Solution

Since ϕ\phi remain same, circuit is in resonance ∴\therefore

Irms=vrmsz{I_{rms}} = {{{v_{rms}}} \over z}

=

vrmsR{{{v_{rms}}} \over R}
=220110= {{220} \over {110}}

⇒\Rightarrow

Irms=2A{I_{rms}} = 2A
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →