3 Marks Question

Question 1013 Marks

An AC source of voltage $\text{V = Vm}\sin\omega\text{t}$ is applied across a series LCR circuit. Draw the phasor diagrams for this circuit, when,

Capacitive impedance exceeds the inductive impedance.
Inductive impedance exceeds capacitive impedance.

Answer

When X_C > X_L; the phasor diagram is shown in fig. (a).

When X_L > X_C; the phasor diagram is shown in fig. (b).

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Question 1023 Marks

Figure shows a typical circuit for a low-pass filter. An AC input V_i= 10mV is applied at the left end and the output V₀ is received at the right end. Find the output voltage for ν = 10kHz, 1.0MHz and 10.0MHz. Note that as the frequency is increased the output decreases and, hence, the name low-pass filter.

Answer

$\text{V}_1=10\times10^{-3}\text{V}$

$\text{R}=1\times10^3\Omega$

$\text{C}=10\times10^{-9}\text{F}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10\times10^3\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-4}}$

$=\frac{10^4}{2\pi}=\frac{5000}{\pi}$

$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$

$=\sqrt{\big(1\times10^3\big)^2+\Big(\frac{5000}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{5000}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5000}{\pi}\Big)^2}}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10^5\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-3}}=\frac{10^3}{2\pi}=\frac{500}{\pi}$

$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{500}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}}\times\frac{500}{\pi}$

$=1.6124\text{V}\approx1.6\text{mV}$

$\text{f}=1\text{MHz}=10^6\text{Hz}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10^6\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-2}}=\frac{10^2}{2\pi}=\frac{50}{\pi}$

$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{50}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}}\times\frac{50}{\pi}$

$\approx1.16\mu\text{V}$

$\text{f}=10\text{MHz}=10^7\text{Hz}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10^7\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-1}}=\frac{10}{2\pi}=\frac{5}{\pi}$

$\text{Z}=\sqrt{\text{R}^2+\text{X}_\text{C}{^2}}$

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{5}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}}\times\frac{5}{\pi}$

$\approx16\mu\text{V}$

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Question 1033 Marks

Prove that the power dissipated in an ideal resistor connected to an ac source is $\frac{\text{V}^2_{\text{eff}}}{\text{R}}.$

Answer

Power in ac circuit, P $=\text{V}_{\text{rms}}\text{i}_{\text{rms}}\cos\varphi$
As rms values of current and voltage are also caled effective values i.e.
$\text{P = V}_{\text{eff}}\text{I}_{\text{eff}}\cos\varphi \ ...(\text{i})$
But $\cos\varphi=\text{power factor}=\frac{\text{R}}{\text{Z}}$
In a purely resistive circuit Z $=\text{R},\cos\varphi=1$
and $\text{i}_{\text{eff}}=\frac{\text{V}_{\text{eff}}}{\text{Z}}=\frac{\text{V}_{\text{eff}}}{\text{R}}$
Substituting these value in (i), we get
$\text{P}=\text{V}_{\text{eff}}.\frac{\text{V}_{\text{eff}}}{\text{R}}\times1=\frac{\text{V}^2_{\text{eff}}}{\text{R}}$

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Question 1043 Marks

An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?

Answer

$\text{E}=12\text{volt}$
$\text{i}^2\text{Rt}=\text{i}^2_\text{rms}\text{RT}$
$\Rightarrow\ \frac{\text{E}^2}{\text{R}^2}=\frac{\text{E}^2_\text{rms}}{\text{R}^2}$
$\Rightarrow\ \text{E}^2=\frac{\text{E}_0^2}{2}$
$\Rightarrow\ \text{E}_0{^2}=2\text{E}^2$
$\Rightarrow\ \text{E}_0{^2}=2\times12^2=2\times144$
$\Rightarrow \text{E}_0=\sqrt{2\times144}=16.97\approx17\text{V}$

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Question 1053 Marks

In the given circuit, the value of resistance effect of the coil L is exactly equal to the resistance R. Bulbs B₁ and B₂ are exactly identical.

Answer the following questions based on above information:

Which one of the two bulbs lights up earlier, when key K is closed and why?
What will be the comparative brightness of the two bulbs after sometime if the key K is kept closed and why?

Answer

Bulb B₂ lights up earlier. The self-induction effect due to coil L in bulb B₁ arm does not allow the current to attain maximum value immediately on closing the circuit.
Since the resistance effect of the coil L is equal to R and the self-induction effect in coil L will disappear after sometime, the current in both the arms will be equal. Hence, both the bulbs will glow with equal brightness after sometime.

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Question 1063 Marks

Calculate the following:

Impedance of the given ac circuit.
Wattless current of the given ac circuit.

Answer

Potential difference across capacitance, V_C = X_CI

$\therefore$ Capacitive reactance,

$\text{X}_{\text{C}}=\frac{\text{V}_{\text{C}}}{\text{I}}=\frac{40}{2}=20\Omega$

Resistance, $\text{R}=\frac{\text{V}_{\text{R}}}{\text{I}}=\frac{30}{2}=15\Omega$

Impedance, $\text{Z}=\sqrt{\text{R}^2+\text{X}^2_{\text{C}}}=\sqrt{(15)^2+(20)^2}$

$=\sqrt{225+400}=\sqrt{625}\Omega=25\Omega$

The phase lead $(\varphi)$ of current over applied voltage is

$\varphi=\frac{\text{X}_{\text{C}}}{\text{R}}$

Wattless Current, $\text{I}_{\text{wattless}}=\text{I}\sin\varphi=\text{I}\Big(\frac{\text{X}_{\text{C}}}{\text{Z}}\Big)$

$=2\times\frac{20}{25}\text{A}=1.6\text{A}$

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Question 1073 Marks

A coil of inductance 5.0mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for $\omega=100\text{s}^{-1},500\text{s}^{-1},1000\text{s}^{-1}.$

Answer

Inductance $=5.0\text{mH}=0.005\text{H}$

$\omega=100\text{s}^{-1}$

$\text{X}_\text{L}=\omega\text{L}$

$=100\times\frac{5}{1000}=0.5\Omega$

$=\text{i}=\frac{\in_0}{\text{X}_\text{L}}=\frac{10}{0.5}=20\text{A}$

$\omega=500\text{s}^{-1}$

$\text{X}_\text{L}=\omega\text{L}$

$=500\times\frac{5}{1000}=2.5\Omega$

$=\text{i}=\frac{\in_0}{\text{X}_\text{L}}=\frac{10}{2.5}=4\text{A}$

$\omega=1000\text{s}^{-1}$

$\text{X}_\text{L}=\omega\text{L}$

$=1000\times\frac{5}{1000}=5\Omega$

$=\text{i}=\frac{\in_0}{\text{X}_\text{L}}=\frac{10}{5}=2\text{A}$

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Question 1083 Marks

The instantaneous current in an ac circuit is $\text{i}=0.5\sin314\text{t},$ what is (i) rms value and (ii) frequency of the current.

Answer

Given $\text{I}=0.5\sin314\text{t} \ ...(\text{i})$
Standard equation of current is $\text{I = I}_0\sin\omega\text{t} \ ...(\text{ii})$
Comparing (i) and (ii), we get $\text{I}_{0}=0.5\text{A},\omega=314$
Therefore,

rms value $\text{I}_{\text{rms}}=\frac{\text{I}_0}{\sqrt{2}}=\frac{0.5}{\sqrt{2}}\text{A}=0.35\text{A}$
Frequency $\text{v}=\frac{\omega}{2\pi}=\frac{314}{2\times3.14}=50\text{Hz}$

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Question 1093 Marks

The voltage and current in a series AC circuit are given by, $\text{V}=\text{V}_0\cos\omega\text{t}$ and $\text{i}=\text{i}_0\sin\omega\text{t}.$ What is the power dissipated in the circuit?

Answer

Power $=\text{I}_\text{rms}\text{E}_\text{rms}\cos\phi$
$\phi=\frac{\pi}{2}$ so $\text{P}=0.$

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Question 1103 Marks

What is the power dissipated in an ac circuit in which voltage and current are given by $\text{V}=230\sin(\omega\text{t}+\frac{\pi}{2})$ and $\text{I}=10\sin\omega\text{t}?$

Answer

Power dissipated P $=\frac{1}{2}\text{V}_0\text{I}_0\cos\varphi$
Here, $\text{V}_0=230\text{V},\text{I}_0=10\text{A},\varphi=\frac{\pi}{2}$
$\therefore\text{P}=\frac{1}{2}\times230\times10\cos\frac{\pi}{2}=0$

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Question 1113 Marks

When a capacitor is connected in series LR circuit, the alternating current flowing in the circuit increases. Explain why.

Answer

Impedance of series LR circuit
$\text{Z}_1=\sqrt{\text{R}^2+\text{X}^2_{\text{L}}}$
When capacitor is also connected in circuit,
Then impedance
$\text{Z}_{\text{L}}=\sqrt{\text{R}^2+(\text{X}_{\text{L}}-\text{X}_{\text{C}})^2}$
Clearly impedance of circuit decreases (Z₂ < Z₁), so the value of current $\text{I}=\frac{\text{v}}{\text{z}}\propto\frac{1}{\text{z}}$ in the circuit increases.

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Question 1123 Marks

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.

Answer

$\text{R}=300\Omega$
$\text{C}=20\mu\text{F}=20\times10^{-6}\text{F}$
$\text{L}=1\text{H},\text{Z}=500$ $(\text{from}\ 14)$
$\in_0=50\text{V},\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{50}{500}=0.1\text{A}$
Electric Energy stored in Capacitor $=\Big(\frac{1}{2}\Big)\text{CV}^2$
$=\Big(\frac{1}{2}\Big)\times20\times10^{-6}\times50\times50$
$=25\times10^{-3}\text{J}=25\text{mJ}$
Magnetic field energy stored in the coil $=\Big(\frac{1}{2}\Big)\text{LI}_0{^2}$
$=\Big(\frac{1}{2}\Big)\times1\times(0.1)^{2}$
$=5\times10^{-3}\text{J}=5\text{mJ}$

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Question 1133 Marks

In the given diagram, a coil B is connected to low voltage bulb L and placed parallel to another coil A as shown. Explain the following observations:

Bulb lights.
Bulb gets dimmer if the coil B moves upwards.

Answer

Bulb lights up due to induced current in B because of change in flux linked with it as a consequence of continuous variation of magnitude of alternative current flowing in A.
When coil B moves upward, the magnetic flux linked with B decreases and hence lesser current is induced in B.

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Question 1143 Marks

Figure shows a light bulb (B) and iron cored inductor connected to a dc battery through a switch (S).

What will one observe when switch (S) is closed?
How will the glow of the bulb change when the battery is replaced by an ac source of rms voltage equal to the voltage of dc battery? Justify your answer in each case.

Answer

When switch S is closed, the bulb will give full brightness slowly, because inductor opposes the rise of current in the circuit depending on the value of ratio $\frac{\text{L}}{\text{R}}.$

(L = inductance, R = resistance of bulb).

When battery is replaced by an ac source, the inductor offers reactance $(\omega\text{L})$ so impedance of circuit increases and the bulb will glow with less brightness.

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Question 1153 Marks

In the given circuit, the potential difference across the inductor L and resistor R are 200V and 150V respectively and the rms. value of current is 5A. Calculate (i) the impedance of the circuit and (ii) the phase angle between the voltage and the current.

Answer

Voltage applied $\text{V}=\sqrt{\text{V}^2_\text{L}+\text{V}^2_\text{R}}=\sqrt{(200)^2+(150^2)}=250\text{V}$

Impedance of circuit, $\text{Z}=\frac{\text{V}}{\text{I}}=\frac{250}{5}=50\Omega$

Phase angle between voltage and current

$\tan\varphi=\frac{\text{x}_{\text{L}}}{\text{R}}=\frac{\text{V}_{\text{L}}}{\text{V}_{\text{R}}}=\frac{200}{150}=\frac{4}{3}$

$\varphi=\tan^{-1}\Big(\frac{4}{3}\Big)=53^{\circ}$

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Question 1163 Marks

Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.

Answer

A capacitor does not allow flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency; it is given by $\frac{1}{\omega\text{C}}$.

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Question 1173 Marks

An inductor-coil, a capacitor and an AC source of rms voltage 24V are connected in series. When the frequency of the source is varied, a maximum rms current of 6.0A is observed. If this inductor coil is connected to a battery of emf 12V and internal resistance $4.0\Omega,$ what will be the current?

Answer

$\text{E}_\text{rms}=24\text{V}$
$\text{r}=4\Omega,\text{I}_\text{rms}=6\text{A}$
$\text{R}=\frac{\text{E}}{\text{I}}=\frac{24}{6}=4\Omega$
Internal Resistance $=4\Omega$
Hence net resistance $=4+4=8\Omega$
$\therefore$ Current $=\frac{12}{8}=1.5\text{A}$

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Question 1183 Marks

In a series LCR circuit, $\text{R = 1k}\Omega,\text{C}=2\mu\text{F}$ and voltage across R is 100V. The resonant frequency of the circuit $\omega$ is 200 rad s^-1. Calculate the value of voltage across L at resonance.

Answer

Current flowing in the circuit is
$\text{I}=\frac{\text{V}_{\text{R}}}{\text{R}}=\frac{100}{1000}=0.1\text{A}$
$\therefore$ Also at resonance, $\omega\text{L}=\frac{1}{\omega\text{C}}$
$\omega\text{L}=\frac{1}{200\times2\times10^{-6}}=2500\Omega$
$\therefore$ Volage across, $\text{L = I}(\omega\text{L})=0.1\times2500=250\text{V}$

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Question 1193 Marks

Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?

Answer

For a Direct Current (DC),
1 Ampere = 1 Coulomb/Sec
Direction of AC changes with the frequency of source with the source frequency and the attractive force would average to zero. Thus, the AC ampere must be defined in terms of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of AC.
So, r.m.s. value of AC is equal to that value of DC, which when passed through a resistance for a given time will produce the same amount of heat as produced by the alternating current when passed through the same resistance for same time.

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Question 1203 Marks

In series LCR circuit, the plot of I_max vs $\omega$ is shown in Fig. Find the bandwidth and mark in the figure.

Answer

According top the given diagram,
Bandwidth $\Delta\omega=\omega_2-\omega_1$

where $\omega_1$ and $\omega_2$ correspond to frequencies at which magnitude of current is $\frac{1}{\sqrt{2}}$ times of maximum value.
i.e., $\text{I}_\text{rms}=\frac{\text{I}_\text{max}}{\sqrt{2}}=\frac{1\text{A}}{\sqrt{2}}\approx0.7\text{A}$
From the graph these frequencies are $\omega_1=0.8\frac{\text{rad}}{\text{s}}$ and $\omega_2=1.2\frac{\text{red}}{\text{s}}.$
Thus, bandwidth, $\Delta\omega=\omega_1-\omega_2=1.2-0.8=0.4\frac{\text{rad}}{\text{s}}.$

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Question 1213 Marks

Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.

Answer

The inductive reactance is given by $\text{X}_\text{L}=2\pi\text{fL}$, X_L is proportional to the frequency and current is inversely proportional to the reactance. An inductor opposes the flow of current through it by developing a back emf according to Lenz’s law. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice-versa.
Since, the induced emf is proportional to the rate of change of current.

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Question 1223 Marks

An AC voltage V = V_m is applied across a:

Series RC circuit in which capacitive reactance is ‘a’ times the resistance in the circuit.
Series RL circuit in which inductive reactance is ‘b’ times the resistance in the circuit.

Find the value of power factor of the circuit in each case.

Answer

Power factor $\cos\varphi=\frac{\text{R}}{\text{Z}},$ when $\text{Z}=\sqrt{\text{R}^2+\text{X}^2}$

$\text{X = X}_{\text{C}} = \text{aR},$

$\therefore\text{Z}=\sqrt{\text{R}^2+(\text{aR})^2}=\text{R}\sqrt{1+\text{a}^2}$

$\therefore\cos\varphi=\frac{\text{R}}{\text{R}\sqrt{1+\text{a}^2}}=\frac{1}{\sqrt{1+\text{a}^2}}$

$\text{X = X}_{\text{L}}=\text{bR},$

$\therefore\text{Z}=\sqrt{\text{R}^2+(\text{bR})^2}=\text{R}\sqrt{1+\text{b}^2}$

$\therefore\cos\varphi=\frac{\text{R}}{\text{R}\sqrt{1+\text{b}^2}}=\frac{1}{\sqrt{1+\text{b}^2}}$

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Question 1233 Marks

An inductor 200mH, a capacitor $100\mu\text{F}$ and a resistor $10\Omega$ are connected in series to an a.c. source of 100V, having variable frequency.

At what frequency of the applied voltage will the power factor of the circuit be 1?
What will be the current amplitude at this frequency?
Calculate the Q-factor of the circuit.

Answer

Since the power factor $\varphi=1$ it means the phase difference between voltage and the current is zero.

This is possible when

$\text{wL}=\frac{1}{\omega\text{C}}$

$\omega^2=\frac{1}{\text{LC}}\Rightarrow\text{v}=\frac{1}{2\pi\sqrt{\text{LC}}}=35.58\text{Hz}$

Current Amplitude, $\text{I}=\frac{\text{V}_{\text{eff}}}{\text{Z}}=\frac{\text{V}_{\text{eff}}}{\text{R}}=\frac{100}{10}=10\text{A}$ $(\therefore\text{Z = R})$
Quality factor, $\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{}\text{C}}=4.47$

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Question 1243 Marks

Draw the effective equivalent circuit of the circuit shown in Fig, at very high frequencies and find the effective impedance.

Answer

At high frequencies, capacitor ≈ short circuit (low reactance) and inductor ≈ open circuit (high reactance). Therefore, the equivalent circuit Z ≈ R₁ + R₃ as shown in the figure given below

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Question 1253 Marks

Find the time required for a 50Hz alternating current to change its value from zero to the rms value.

Answer

$\text{f}=50\text{Hz}$
$=\text{I}=\text{I}_0\sin\omega\text{t}$
Peak value $\text{I}=\frac{\text{I}_0}{\sqrt{2}}$
$\frac{\text{I}_0}{\sqrt{2}}=\text{I}_0\sin\omega\text{t}$
$\Rightarrow\ \frac{1}{\sqrt{2}}=\sin\omega\text{t}=\sin\frac{\pi}{4}$
$\Rightarrow\ \frac{\pi}{4}=\omega\text{t}$
Or $\text{t}=\frac{\pi}{400}$
$=\frac{\pi}{4\times2\pi\text{f}}=\frac{1}{8\text{f}}$
$=\frac{1}{8\times50}=0.0025\text{s}=2.5\text{ms}$

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Question 1263 Marks

A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220V DC supply, what will be the voltage across the secondary?

Answer

Transformer works upon the principle of induction which is only possible in case of AC.
Hence when DC is supplied to it, the primary coil blocks the Current supplied to it and hence induced current supplied to it and hence induced Current in the secondary coil is zero.

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Physics 3 Marks Question