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Alternating Current

NEET Physics · 95 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
An alternating emf E=440sin⁡100πt\mathrm{E}=440 \sin 100 \pi \mathrm{t} is applied to a circuit containing an inductance of 2πH\dfrac{\sqrt{2}}{\pi} \mathrm{H}. If an a.c. ammeter is connected in the circuit, its reading will be :
A 4.4 A
B 1.55 A
C 2.2 A
D 3.11 A
Correct Answer
Option C
Solution
I=VωLI = {V \over {\omega L}}
=440100π×2π=44102= {{440} \over {100\pi \times {{\sqrt 2 } \over \pi }}} = {{44} \over {10\sqrt 2 }}
⇒Irms=I2=4420=2.2 A\Rightarrow {I_{rms}} = {I \over {\sqrt 2 }} = {{44} \over {20}} = 2.2\,A
Q62
A coil of inductance 1 H and resistance 100 Ω100 \,\Omega is connected to a battery of 6 V. Determine approximately : (a) The time elapsed before the current acquires half of its steady - state value. (b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given ln⁡2=0.693,e−3/2=0.25\ln 2=0.693, \mathrm{e}^{-3 / 2}=0.25)
A t = 10 ms; U = 2 mJ
B t = 10 ms; U = 1 mJ
C t = 7 ms; U = 1 mJ
D t = 7 ms; U = 2 mJ
Correct Answer
Option C
Solution
i(t)=VR(1−e−Rt/L)i(t) = {V \over R}(1 - {e^{ - Rt/L}})

...... (1)

LR=1100s⇒LR=10 ms{L \over R} = {1 \over {100}}s \Rightarrow {L \over R} = 10\,ms

...... (2)

V2R=VR(1−e−Rt/L){V \over {2R}} = {V \over R}(1 - {e^{ - Rt/L}})
⇒e−Rt/L=12⇒t=LRln⁡2=6.93 ms\Rightarrow {e^{ - Rt/L}} = {1 \over 2} \Rightarrow t = {L \over R}\ln 2 = 6.93\,ms
U=12Li2=12[1−e−15/10]2[6100]2U = {1 \over 2}L{i^2} = {1 \over 2}{[1 - {e^{ - 15/10}}]^2}{\left[ {{6 \over {100}}} \right]^2}
=12[1−0.25]2×36×10−4= {1 \over 2}{[1 - 0.25]^2} \times 36 \times {10^{ - 4}}
=1 mJ= 1\,mJ
Q63
In a series LR circuit with XL=R\mathrm{X_L=R}, power factor P1. If a capacitor of capacitance C with XC=XL\mathrm{X_C=X_L} is added to the circuit the power factor becomes P2. The ratio of P1 to P2 will be :
A 1 : 2\sqrt2
B 1 : 3
C 1 : 2
D 1 : 1
Correct Answer
Option A
Solution
XL=R{X_L} = R
⇒P1=RXL2+R2=12\Rightarrow {P_1} = {R \over {\sqrt {X_L^2 + {R^2}} }} = {1 \over {\sqrt 2 }}

Now,

XL=XC=R{X_L} = {X_C} = R
⇒P2=RR2+(XL−XC)2=1\Rightarrow {P_2} = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }} = 1
⇒P1P2=12\Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}
Q64
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

Q65
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 }}
Q66
A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μ\muF and R = 100 Ω\Omega. The phase difference (Φ\Phi ) between the applied voltage and resultant current will be :
A tan−-1(0.17)
B tan−-1(9.46)
C tan−-1(0.30)
D tan−-1(13.33)
Correct Answer
Option A
Solution
XL=3000×10×10−3=30 Ω{X_L} = 3000 \times 10 \times {10^{ - 3}} = 30\,\Omega
XC=13000×25×106=403 Ω{X_C} = {1 \over {3000 \times 25}} \times {10^6} = {{40} \over 3}\,\Omega

So

XL−XC=30−403=503 Ω{X_L} - {X_C} = 30 - {{40} \over 3} = {{50} \over 3}\,\Omega
tan⁡θ=XL−XCR=50/3100=16\tan \theta = {{{X_L} - {X_C}} \over R} = {{50/3} \over {100}} = {1 \over 6}

So

θ=tan⁡−1(0.17)\theta = {\tan ^{ - 1}}(0.17)
Q67
In series RLC resonator, if the self inductance and capacitance become double, the new resonant frequency (f2) and new quality factor (Q2) will be : (f1 = original resonant frequency, Q1 = original quality factor)
A f2=f12{f_2} = {{{f_1}} \over 2} and Q2=Q1{Q_2} = {Q_1}
B f2=f1{f_2} = {f_1} and Q2=Q1Q2{Q_2} = {{{Q_1}} \over {{Q_2}}}
C f2=2f1{f_2} = 2{f_1} and Q2=Q1{Q_2} = {Q_1}
D f2=f1{f_2} = {f_1} and Q2=2Q1{Q_2} = 2{Q_1}
Correct Answer
Option A
Solution

We know, Quality factor (Q factor)

Q1=w1Δw{Q_1} = {{{w_1}} \over {\Delta w}}
=1LC×LR= {1 \over {\sqrt {LC} }} \times {L \over R}
=1RLC= {1 \over R}\sqrt {{L \over C}}

Now, when

L′=2LL' = 2L

and

C′=2CC' = 2C

then

Q2=1R2L2C=1RLC=Q1{Q_2} = {1 \over R}\sqrt {{{2L} \over {2C}}} = {1 \over R}\sqrt {{L \over C}} = {Q_1}

∴\therefore Q2 remains same as Q1. Also, as

w1=1LC{w_1} = {1 \over {\sqrt {LC} }}
⇒2πf1=1LC\Rightarrow 2\pi {f_1} = {1 \over {\sqrt {LC} }}
⇒f1=12πLC\Rightarrow {f_1} = {1 \over {2\pi \sqrt {LC} }}

∴\therefore When

L′=2LL' = 2L

and

C′=2CC' = 2C

then new resonating frequency

f2=12π2L×2C=12π×2LC=12×f1{f_2} = {1 \over {2\pi \sqrt {2L \times 2C} }} = {1 \over {2\pi \times 2\sqrt {LC} }} = {1 \over 2} \times {f_1}
Q68
To increase the resonant frequency in series LCR circuit,
A source frequency should be increased.
B another resistance should be added in series with the first resistance.
C another capacitor should be added in series with the first capacitor.
D the source frequency should be decreased.
Correct Answer
Option C
Solution

Resonant frequency

=1LC=ω0= {1 \over {\sqrt {LC} }} = {\omega _0}

⇒\Rightarrow If we decrease C, ω\omega0 would increase ⇒\Rightarrow Another capacitor should be added in series.

Q69
A direct current of 4 A4 \mathrm{~A} and an alternating current of peak value 4 A4 \mathrm{~A} flow through resistance of 3 Ω3\, \Omega and 2 Ω2\,\Omega respectively. The ratio of heat produced in the two resistances in same interval of time will be :
A 3 : 2
B 3 : 1
C 3 : 4
D 4 : 3
Correct Answer
Option B
Solution

Ratio =

i12R1(i22)2R2=42×3(42)2×2{{i_1^2{R_1}} \over {{{\left( {{{{i_2}} \over {\sqrt 2 }}} \right)}^2}{R_2}}} = {{{4^2} \times 3} \over {{{\left( {{4 \over {\sqrt 2 }}} \right)}^2} \times 2}}

⇒\Rightarrow Ratio = 3 : 1

Q70
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

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