39: Alternating Current : Neet-xii-physics

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NEET-XII-Physics

39: Alternating Current

with Solutions - page 5

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Qstn #9
A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for ω = 100 s^-1, 500 s^-1, 1000 s^-1.
Ans : Given:
Inductance of the coil, L = 5.0 mH = 0.005 H
(a) At ω = 100 s^-1:
Reactance of coil `` \left({X}_{L}\right)`` is given by,
X_L = `` \omega L``
Here, `` \omega `` = angular frequency
`` \therefore `` X_L = 100 `` \times `` 0.005 = 0.5 `` \,\mathrm{\,\Omega \,}``
Peak current, I₀ = `` \frac{10}{0.5}=20\,\mathrm{\,A\,}``
(b) At ω = 500 s^-1:
`` \,\mathrm{\,Reactance\,},{X}_{\,\mathrm{\,L\,}}=500\times \frac{5}{1000}``
`` =2.5\,\mathrm{\,\Omega \,}``
`` ``
Peak current, I₀ = `` \frac{10}{2.5}=4\,\mathrm{\,A\,}``
(c) ω = 1000 s^-1:
Reactance, X_L= 1000 `` \times 0.005`` = 5 `` \,\mathrm{\,\Omega \,}``
Peak current, I₀ = `` \frac{10}{5}`` = 2 A
Page No 330:

Qstn #10
A coil has a resistance of 10 Ω and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V,
30πHz. Find the average power consumed in the circuit.
Ans : Given:
Resistance of coil, R = 10 Ω
Inductance of coil, L = 0.4 Henry
Voltage of AC source, Erms = 6.5 V
Frequency of AC source, `` f`` = `` \frac{30}{\,\mathrm{\,\pi \,}}\,\mathrm{\,Hz\,}``
Reactance of resistance-inductance circuit `` \left(Z\right)`` is given by,
Z = `` \sqrt{{R}^{2}+{{X}_{L}}^{2}}``
Here, R = resistance of the circuit
X_L = Reactance of the pure inductive circuit
Z = `` \sqrt{{R}^{2}+{\left(2\,\mathrm{\,\pi \,}fL\right)}^{2}}``
= `` \sqrt{{\left(10\right)}^{2}+{\left(2\times \,\mathrm{\,\pi \,}\times \frac{30}{\,\mathrm{\,\pi \,}}\times 0.4\right)}^{2}}``
Average power consumed in the circuit (P) is given by,
P = E_rmsI_rmscos`` \varphi ``
cos`` \varphi `` = `` \frac{R}{Z}``, I_rms = `` \frac{{E}_{rms}}{Z}``
`` ``
`` \therefore P=6.5\times \frac{6.5}{Z}\times \frac{R}{Z}``
`` \Rightarrow P=\frac{6.5\times 6.5\times 10}{{\left(\sqrt{{R}^{2}+{\left(\omega L\right)}^{2}}\right)}^{2}}``
`` \Rightarrow P=\frac{6.5\times 6.5\times 10}{100+576}``
`` \Rightarrow P=\frac{6.5\times 6.5\times 10}{676}``
`` \Rightarrow P=0.625=\frac{5}{8}\,\mathrm{\,W\,}``
Page No 330:

Qstn #11
A resistor of resistance 100 Ω is connected to an AC source ε = (12 V) sin (250 π s^-1)t. Find the energy dissipated as heat during t = 0 to t = 1.0 ms.
Ans : Given:
Peak voltage of AC source, E₀ = 12 V
Angular frequency, ω = 250 `` \,\mathrm{\,\pi \,}`` s^-1
Resistance of resistor, R = 100 Ω
Energy dissipated as heat `` \left(H\right)`` is given by,
`` H=\frac{{{E}_{\,\mathrm{\,rms\,}}}^{2}}{R}T``
Here, E_rms = RMS value of voltage
R = Resistance of the resistor
T = Temperature
Energy dissipated as heat during t = 0 to t = 1.0 ms,
`` H={\int }_{\mathit{0}}^{t}dH``
`` =\int \frac{{E}_{0}^{2}{\,\mathrm{\,sin\,}}^{2}\,\mathrm{\,\omega \,}t}{R}dt\left(\because {E}_{rms}={E}_{0}\,\mathrm{\,sin\,}\omega t\right)``
`` =\frac{144}{100}{\int }_{0}^{{10}^{-3}}{\,\mathrm{\,sin\,}}^{2}\omega tdt``
`` =1.44{\int }_{0}^{{10}^{-3}}\left(\frac{1-\,\mathrm{\,cos\,}2\omega t}{2}\right)dt``
`` =\frac{1.44}{2}\left[{\int }_{0}^{{10}^{-3}}dt+{\int }_{0}^{{10}^{-3}}\,\mathrm{\,cos\,}2\omega tdt\right]``
`` =0.72\left[{10}^{-3}-{\left\{\frac{\,\mathrm{\,sin\,}2\omega t}{2\omega }\right\}}_{0}^{{10}^{-3}}\right]``
`` =0.72\left[\frac{1}{1000}-\frac{1}{500\,\mathrm{\,\pi \,}}\right]``
`` =0.72\left[\frac{1}{1000}-\frac{2}{1000\,\mathrm{\,\pi \,}}\right]``
`` =\left(\frac{\,\mathrm{\,\pi \,}-2}{1000\,\mathrm{\,\pi \,}}\right)\times 0.72``
`` =0.0002614=2.61\times {10}^{-4}\,\mathrm{\,J\,}``
Page No 330:

Qstn #12
In a series RC circuit with an AC source, R = 300 Ω, C = 25 μF, ε₀ = 50 V and ν = 50/π Hz. Find the peak current and the average power dissipated in the circuit.
Ans : Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε₀ = 50 V
Frequency of the AC source, ν = 50/`` \,\mathrm{\,\pi \,}`` Hz
Capacitive reactance `` \left({X}_{C}\right)`` is given by,
`` {X}_{C}=\frac{1}{\omega C}``
Here, `` \omega `` = angular frequency of AC source
C = capacitive reactance of capacitance
`` \therefore {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {\displaystyle \frac{50}{\,\mathrm{\,\pi \,}}}\times 25\times {10}^{-6}}``
`` \Rightarrow {X}_{C}=\frac{{10}^{4}}{25}\,\mathrm{\,\Omega \,}``
Net reactance of the series RC circuit `` \left(Z\right)`` = `` \sqrt{{R}^{2}+{\left({X}_{C}\right)}^{2}}``
`` \Rightarrow `` Z = `` \sqrt{{\left(300\right)}^{2}+{\left(\frac{{10}^{4}}{25}\right)}^{2}}``
= `` \sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}`` = 500 `` \,\mathrm{\,\Omega \,}``
(a) Peak value of current `` \left({I}_{0}\right)`` is given by,
`` {I}_{0}=\frac{{\epsilon }_{0}}{Z}``
`` \Rightarrow {I}_{0}=\frac{50}{500}=0.1\,\mathrm{\,A\,}``
(b) Average power dissipated in the circuit `` \left(P\right)`` is given by,
P = `` {\epsilon }_{\,\mathrm{\,rms\,}}``I_rms cosϕ.
`` {\epsilon }_{rms}`` = `` \frac{{\epsilon }_{0}}{\sqrt{2}}``
and `` {I}_{\,\mathrm{\,rms\,}}=\frac{{I}_{0}}{\sqrt{2}}``
`` \therefore P=\frac{{E}_{0}}{\sqrt{2}}\times \frac{{I}_{0}}{\sqrt{2}}\times \frac{R}{Z}``
`` \Rightarrow P=\frac{50\times 0.1\times 300}{2\times 500}``
`` \Rightarrow P=\frac{3}{2}=1.5\,\mathrm{\,W\,}``
Page No 330:

Qstn #13
An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage?
Ans : Power consumed by the electric bulb, P = 55 W
Voltage at which the bulb is operated, V= 110 V
Voltage of the line, V = 220 V
Frequency of the source, v = 50 Hz
P =`` \frac{{V}^{2}}{R}``,
where R = resistance of electric bulb
`` \therefore R=\frac{{V}^{2}}{P}``
`` =\frac{110\times 110}{55}=220\,\mathrm{\,\Omega \,}``
If L is the inductance of the coil, then total reactance of the circuit (Z) is given by,
Z = `` \sqrt{{R}^{2}+{\left(\omega L\right)}^{2}}``
=`` \sqrt{{\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}}`` `` \left(\because \omega =2\,\mathrm{\,\pi \,}f\right)``
Here, `` \omega `` = angular frequency of the circuit
Now, current through the bulb, `` I=\frac{V}{Z}``
`` \therefore `` Voltage drop across the bulb, V = `` \frac{V}{Z}\times R``
As per question,
`` 110=\frac{220\times 220}{\sqrt{{\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}}}``
`` 110=\frac{{\left(220\right)}^{2}}{\sqrt{{\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}}}``
`` \Rightarrow 220\times 2=\sqrt{{\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}}``
`` \Rightarrow {\left(220\right)}^{2}+{\left(100\,\mathrm{\,\pi \,}L\right)}^{2}={\left(440\right)}^{2}``
`` \Rightarrow 48400+{10}^{4}{\,\mathrm{\,\pi \,}}^{2}{L}^{2}=193600``
`` \Rightarrow {10}^{4}{\,\mathrm{\,\pi \,}}^{2}{L}^{2}=193600-48400``
`` \Rightarrow {L}^{2}=\frac{145200}{{\,\mathrm{\,\pi \,}}^{2}\times {10}^{4}}``
`` =1.4726``
`` \Rightarrow L=1.2135\cong 1.2\,\mathrm{\,H\,}``
Page No 330:

Qstn #14
In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, ε_rms = 50 V and ν = 50/π Hz. Find
Ans : Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10^-6 F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, ε_rms = 50 V
Frequency of source, f = 50/`` \,\mathrm{\,\pi \,}`` Hz
Reactance of the inductor (X_L) is given by,
X_L= `` \omega L`` = 2`` \,\mathrm{\,\pi \,}``fL
`` \Rightarrow ``X_L = 2`` \times \,\mathrm{\,\pi \,}\times \frac{50}{\,\mathrm{\,\pi \,}}\times 1`` = 100 `` \,\mathrm{\,\Omega \,}``
Reactance of the capacitance `` \left({X}_{C}\right)`` is given by,
`` {X}_{C}=\frac{1}{\omega C}`` = `` \frac{1}{2\,\mathrm{\,\pi \,}fC}``
`` \Rightarrow ``X_C = `` \frac{1}{2\,\mathrm{\,\pi \,}\times {\displaystyle \frac{50}{\,\mathrm{\,\pi \,}}}\times 20\times {10}^{-6}}``
`` \Rightarrow `` X_C = `` 500`` `` \,\mathrm{\,\Omega \,}``

#14-a
the rms current in the circuit and
Ans : Impedance of an LCR circuit `` \left(Z\right)`` is given by,
`` Z=\sqrt{{R}^{2}+{\left({X}_{\,\mathrm{\,C\,}}-{X}_{\,\mathrm{\,L\,}}\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(500-100\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}``
`` \Rightarrow Z=500``
`` ``
RMS value of current `` \left({I}_{rms}\right)`` is given by,
`` {I}_{rms}=\frac{{\epsilon }_{rms}}{Z}``
`` \Rightarrow ``I_rms = `` \frac{50}{500}``
`` \Rightarrow ``I_rms = `` 0.1\,\mathrm{\,A\,}``

#14-b
the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.
Ans : Potential across the capacitor `` \left({V}_{C}\right)`` is given by,
V_C = I_rms × X_C
`` \Rightarrow ``V_C = 0.1 × 500 = 50 V
Potential difference across the resistor `` \left({V}_{R}\right)`` is given by,
V_{R =}I_rms × R
`` \Rightarrow ``V_R = 0.1 × 300 = 30 V
Potential difference across the inductor `` \left({V}_{L}\right)`` is given by,
V_L = I_rms × X_L
`` \Rightarrow `` V_L= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied
Page No 330:

Qstn #15
Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.
Ans : Given:
Resistance in the LCR circuit, R = 300 Ω
Capacitance in the LCR circuit, C = 20 μF = 20 × 10^-6 F
Inductance in the LCR circuit, L = 1 henry
Net impedance of the LCR circuit, Z = 500 ohm
RMS value of voltage, `` {\epsilon }_{\,\mathrm{\,rms\,}}`` = 50 V
RMS value of current, I_rms = 0.1 A
Peak current `` \left({I}_{0}\right)`` is given by,
`` {I}_{0}=\frac{{E}_{\,\mathrm{\,rms\,}}\sqrt{\mathit{2}}}{Z}=\frac{50\times 1.414}{500}=0.1414\,\mathrm{\,A\,}``
Electrical energy stored in capacitor`` \left({U}_{C}\right)`` is given by,
`` {U}_{C}=\frac{1}{2}C{V}^{2}``
`` \Rightarrow {U}_{C}=\frac{1}{2}\times 20\times {10}^{-6}\times 50\times 50``
`` \Rightarrow {U}_{C}=25\times {10}^{-3}\,\mathrm{\,J\,}=25\,\mathrm{\,mJ\,}``
Magnetic field energy stored in the coil `` \left({U}_{L}\right)`` is given by,
`` {U}_{L}=\frac{1}{2}L{I}_{0}^{2}``
`` \Rightarrow {U}_{L}=\frac{1}{2}\times 1\times {\left(0.1414\right)}^{2}``
`` \Rightarrow {U}_{L}\approx 5\times {10}^{-3}\,\mathrm{\,J\,}``
`` \Rightarrow {U}_{L}=5\,\mathrm{\,mJ\,}``
Page No 330:

Qstn #16
An inductance of 2.0 H, a capacitance of 18μF and a resistance of 10 kΩ are connected to an AC source of 20 V with adjustable frequency.
Ans : Given:
Inductance of inductor, L = 2.0 H
Capacitance of capacitor, C = 18 μF
Resistance of resistor, R = 10 kΩ
Voltage of AC source, E = 20 V

#16-a
What frequency should be chosen to maximise the current in the circuit?
Ans : In an LCR circuit, current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
X_L = X_C
`` \Rightarrow \omega L=\frac{1}{\omega C}``
`` \Rightarrow {\omega }^{2}=\frac{1}{LC}=\frac{1}{2\times 18\times {10}^{-6}}``
`` \Rightarrow {\omega }^{2}=\frac{{10}^{6}}{36}``
`` \Rightarrow \omega =\frac{{10}^{3}}{6}``
`` \Rightarrow 2\,\mathrm{\,\pi \,}f=\frac{{10}^{3}}{6}``
`` \Rightarrow f=\frac{1000}{6\times 2\,\mathrm{\,\pi \,}}=26.539\,\mathrm{\,Hz\,}``
`` =27\,\mathrm{\,Hz\,}``

#16-b
What is the value of this maximum current?
Ans : At resonance, reactance is minimum.
Minimum Reactance, Z = R
Maximum current (I) is given by,
`` I=\frac{E}{R}``
`` \Rightarrow I\mathit{=}\frac{20}{10\times {10}^{3}}``
`` \Rightarrow I=\frac{2\,\mathrm{\,A\,}}{{10}^{3}}=2\,\mathrm{\,mA\,}``
Page No 330:

Qstn #17
An inductor-coil, a capacitor and an AC source of rms voltage 24 V are connected in series. When the frequency of the source is varied, a maximum rms current of 6.0 A is observed. If this inductor coil is connected to a battery of emf 12 V and internal resistance 4.0 Ω, what will be the current?
Ans : RMS value of voltage, E_rms = 24 V
Internal resistance of battery, r = 4 Ω
RMS value of current, I_rms = 6 A
Reactance `` \left(R\right)`` is given by,
`` R=\frac{E}{{I}_{\,\mathrm{\,rms\,}}}``
`` \Rightarrow R=\frac{24}{6}=4\,\mathrm{\,\Omega \,}``
Let R' be the total resistance of the circuit. Then,
R' = R + r
`` \Rightarrow `` R' = 4 Ω + 4 Ω
`` \Rightarrow ``R' = 8 Ω
Current, I = `` \frac{12}{8}\,\mathrm{\,A\,}``
= 1.5 A
Page No 330:

Qstn #18
Figure (39-E1) shows a typical circuit for a low-pass filter. An AC input V_i= 10 mV is applied at the left end and the output V₀ is received at the right end. Find the output voltage for ν = 10 k Hz, 1.0 MHz and 10.0 MHz. Note that as the frequency is increased the output decreases and, hence, the name low-pass filter.
Figure
Ans : Here,
Input voltage to the filter, Vi = 10 × 10^-3V
Resistance of the circuit, R = 1 × 10³ Ω
Capacitance of the circuit, C = 10 × 10^-9 F
(a) When frequency, f = 10 kHz
A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
Capacitive reactance `` \left({X}_{C}\right)`` is given by,
`` {X}_{\,\mathrm{\,C\,}}=\frac{\mathit{1}}{\mathit{\omega }\mathit{C}}=\frac{1}{2\,\mathrm{\,\pi \,}fC}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times 10\times {10}^{3}\times 10\times {10}^{-9}}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {10}^{-4}}``
`` \Rightarrow {X}_{C}=\frac{{10}^{4}}{2\,\mathrm{\,\pi \,}}=\frac{5000}{\,\mathrm{\,\pi \,}}\,\mathrm{\,\Omega \,}``
Net impedence of the resistance-capacitance circuit (Z) is given by,
`` Z=\sqrt{{R}^{2}+{X}_{\,\mathrm{\,C\,}}^{2}}``
`` \Rightarrow Z=\sqrt{\left(1+{10}^{3}\right)+{\left(5000/\,\mathrm{\,\pi \,}\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{10}^{6}+{\left(5000/\,\mathrm{\,\pi \,}\right)}^{2}}``
`` ``
Current (I₀) is given by,
`` {I}_{0}=\frac{{V}_{i}}{Z}``
`` \Rightarrow {I}_{0}=\frac{10\times {10}^{-3}}{\sqrt{{10}^{6}+{\left(5000/\,\mathrm{\,\pi \,}\right)}^{2}}}``
`` ``
Output across the capacitor `` \left({V}_{0}\right)`` is given by,
`` {V}_{0}=\frac{{10}^{2}}{\sqrt{{10}^{5}+{\left(50/\,\mathrm{\,\pi \,}\right)}^{2}}}\times \frac{500}{\,\mathrm{\,\pi \,}}``
`` \Rightarrow {V}_{0}=1.6124\,\mathrm{\,V\,}=1.6\,\mathrm{\,mV\,}``
(b)When frequency, f = 1 MHz = `` 1\times ``10⁶ Hz
Capacitive reactance `` \left({X}_{C}\right)`` is given by,
`` {X}_{c}=\frac{1}{\omega C}``
`` \Rightarrow {X}_{C}=\frac{\mathit{1}}{\mathit{2}\pi fC}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {10}^{6}\times {10}^{-9}\times 10}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {10}^{-2}}``
`` \Rightarrow {X}_{C}=\frac{100}{2\,\mathrm{\,\pi \,}}``
`` \Rightarrow {X}_{C}=\frac{500}{\,\mathrm{\,\pi \,}}\,\mathrm{\,\Omega \,}``
`` \,\mathrm{\,Total\,}\,\mathrm{\,impedence\,}\left(Z\right)=\sqrt{{R}^{2}+{{\,\mathrm{\,X\,}}_{\,\mathrm{\,C\,}}}^{2}}``
`` \Rightarrow Z=\sqrt{{\left({10}^{3}\right)}^{2}+{\left(50/\,\mathrm{\,\pi \,}\right)}^{2}}``
`` \,\mathrm{\,Current\,}\left({I}_{0}\right)=\frac{{V}_{\mathit{1}}}{Z}``
`` \Rightarrow {I}_{0}=\frac{10\times {10}^{-3}}{\sqrt{{10}^{6}+{\left(50/\,\mathrm{\,\pi \,}\right)}^{2}}}``
`` \,\mathrm{\,Output\,}\,\mathrm{\,voltage\,}\left({V}_{0}\right)\mathit{=}{I}_{\mathit{0}}{X}_{C}``
`` \Rightarrow {V}_{0}=\frac{{10}^{-2}}{\sqrt{{10}^{5}+{\left(50/\,\mathrm{\,\pi \,}\right)}^{2}}}\times \frac{50}{\,\mathrm{\,\pi \,}}``
`` \Rightarrow {V}_{0}=0.16\,\mathrm{\,mV\,}``
(c) When frequency, f = 10 MHz = 10 × 10⁶ Hz = 10⁷ Hz
Capacitive reactance `` \left({X}_{C}\right)`` is given by,
`` {X}_{c}=\frac{1}{\omega C}``
`` \Rightarrow {X}_{C}=\frac{1}{2\pi fC}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {10}^{7}\times 10\times {10}^{-9}}``
`` \Rightarrow {X}_{C}=\frac{5}{\,\mathrm{\,\pi \,}}\,\mathrm{\,\Omega \,}``
`` \,\mathrm{\,Impedence\,}\left(Z\right)=\sqrt{{R}^{2}+{X}_{c}^{2}}``
`` \Rightarrow Z=\sqrt{{\left({10}^{3}\right)}^{2}+{\left(5/\,\mathrm{\,\pi \,}\right)}^{2}}``
`` \,\mathrm{\,Current\,}\left({I}_{0}\right)=\frac{{V}_{1}}{Z}``
`` \Rightarrow {I}_{0}=\frac{10\times {10}^{-3}}{\sqrt{{10}^{6}+{\left(5/\,\mathrm{\,\pi \,}\right)}^{2}}}``
`` {V}_{0}={I}_{0}{X}_{\,\mathrm{\,C\,}}``
`` \Rightarrow {V}_{0}=\frac{{10}^{-2}}{\sqrt{{10}^{6}+{\left(5/\,\mathrm{\,\pi \,}\right)}^{2}}}\times \frac{5}{\,\mathrm{\,\pi \,}}``
`` \Rightarrow {V}_{o}=16\,\mathrm{\,\mu V\,}``
Page No 331:

Qstn #19
A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary?
Ans : A transformer works on the principle of electromagnetic induction, which is only possible in case of AC.
Hence, when DC (zero frequency) is supplied to it, the primary coil blocks the current supplied to it. Thus, the induced current in the secondary coil is zero. So, the output voltage will be zero.