Alternating Current
46 Board Physics previous year questions on Alternating Current — options free on every question; 5 include the answer & explanation free, the rest unlock with PYQ Pass.
An alternating voltage \(V(t)=220 \sin 100 \pi t\) volt is applied to a purely resistive load of \(50 \Omega\). The time taken for the current to rise from half of the peak value to the peak value is:
\(3.3 ms\)
\(i(t)={i}_{0}\sin 100\pi t\)
\(\frac{{i}_{0}}{2}={i}_{0}\sin 100\pi {t}_{1}\)
\(\sin 100\pi {t}_{1}=\sin \frac{\pi }{6}\)
\(100\pi {t}_{1}=\frac{\pi }{6}\)
\({t}_{1}=\frac{1}{600}\text{sec}\)
Similarly \({t}_{2}=\frac{1}{200}\text{sec}\)
\(t={t}_{2}−{t}_{1}\)
\(=\frac{1}{200}−\frac{1}{600}\)
\(t=\frac{3−1}{600}=\frac{1}{300}\)
t = 3.3 ms
A bulb is rated \((100\mathrm{W},110\mathrm{V})\). It is operated by current of 1.0 A supplied by a step down transformer. If the input voltage and efficiency of the transformer are 220 V and 0.9 respectively, the input current drawn from the mains is :
\(\frac{5}{9}\mathrm{A}\)
The power output of the transformer is the power consumed by the bulb, which is 100 W. The efficiency of the transformer is given as 0.9.
First, we calculate the power input to the transformer using the efficiency formula:
\(\eta = \frac{P_{out}}{P_{in}}\)
where \(\eta\) is the efficiency, \(P_{out}\) is the output power, and \(P_{in}\) is the input power.
Rearranging for \(P_{in}\):
\(P_{in} = \frac{P_{out}}{\eta}\)
Substituting the given values:
\(P_{in} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\)
Next, we use the input power and input voltage to find the input current. The input voltage is given as 220 V.
Using the power formula:
\(P_{in} = V_{in} \times I_{in}\)
Rearranging for \(I_{in}\):
\(I_{in} = \frac{P_{in}}{V_{in}}\)
Substituting the values:
\(I_{in} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\)
Thus, the input current drawn from the mains is approximately \(\frac{5}{9} \text{ A}\).
Identify the power consumed by the bulb: \(P_{out} = 100 \text{ W}\).
Calculate the input power using the efficiency: \(P_{in} = \frac{P_{out}}{\eta} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\).
Identify the input voltage: \(V_{in}\) = \(220 \text{ V}\).
Use the power formula to find the input current: \(P_{in} = V_{in} \times I_{in}\).
Rearrange and solve for \(I_{in}: I_{in} = \frac{P_{in}}{V_{in}} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\).
final answer: \(\frac{5}{9} \text{ A}\)
An alternating current is given by \(I={I}_{A}\sin \omega t+{I}_{B}\cos \omega t\). The r.m.s current will be
\(\sqrt{\frac{{I}_{A}^{2}+{I}_{B}^{2}}{2}}\)
Given the current as \(I={I}_{A}\sin \omega t+\) \({I}_{B}\cos \omega t\), the r.m.s. current is calculated by integrating the square of the current over time, then taking the square root. i.e.,
\({i}_{ms}=\sqrt{\frac{\int {I}^{2}dt}{\int dt}}\\ {I}_{rms}=\sqrt{\frac{{I}_{A}^{2}+{I}_{B}^{2}}{2}}\)
A coil of resistance \(20\Omega\) and self-inductance 10 mH is connected to an ac source of frequency \(1000/\pi \mathrm{Hz}\). The phase difference between current in the circuit and the source voltage is :
\(45^\circ\)
\({X}_{L}=2\pi (\frac{1000}{\pi })(10\times {10}^{-3})\)3 Sv6Kpe[] \({X}_{L}=20\ \Omega\) Calculate the phase difference (\(ϕ\)) The phase difference (\(ϕ\)) between the current and voltage in an RL circuit is given by the formula \(\tan (ϕ)=\frac{{X}_{L}}{R}\) \(\tan (ϕ)=\frac{20}{20}\) \(\tan (ϕ)=1\)1 Therefore, the phase difference is: \(ϕ=\mathrm{arctan}(1)={45}^{∘}\)or\(\frac{\pi }{4}\)radians.
A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit. The average power consumed in the circuit over a cycle is
\(\frac{{i}_{0}{v}_{0}}{2}\cos ϕ\)
To find the average power consumed in the circuit, we begin with the instantaneous power, which is the product of the voltage
\(v(t)\)and the current\(i(t)\)
:
\[p(t)=v(t) \cdot i(t)=\left(v_0 \sin (\omega t)\right) \cdot\left(i_0 \sin (\omega t+\phi)\right)\] \[\sin A \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]\] \[\sin (\omega t) \sin (\omega t+\phi)=\frac{1}{2}[\cos (-\phi)-\cos (2 \omega t+\phi)]\] \[\cos (-\phi)=\cos (\phi)\], thus:
\[p(t)=v_0 i_0 \frac{1}{2}[\cos (\phi)-\cos (2 \omega t+\phi)]\]To find the average power over a cycle, integrate this expression over one period
\[T=\frac{2 \pi}{\omega}\], then divide by
\[T\]:
\[P_{\mathrm{avg}}=\frac{1}{T} \int_0^T v_0 i_0 \frac{1}{2}[\cos (\phi)-\cos (2 \omega t+\phi)] d t\]This separates into two integrals:
\[P_{\mathrm{avg}}=\frac{v_0 i_0}{2 T}\left[\int_0^T \cos (\phi) d t-\int_0^T \cos (2 \omega t+\phi) d t\right]\] \[P_{\mathrm{avg}}=\frac{v_0 i_0}{2 T}\left[\int_0^T \cos (\phi) d t-\int_0^T \cos (2 \omega t+\phi) d t\right]\]To an ac power supply of \(220\mathrm{V}\) at \(50\mathrm{Hz}\) , a resistor of \(20\Omega\), a capacitor of reactance \(25\Omega\) and an inductor of reactance \(45\Omega\) are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively -
[NEET 2025]
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Which of the following quantity/quantities remains same in primary andsecondary coils of an ideal transformer ?
Current, Voltage, Power, Magnetic flux
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The reactance of a capacitor of capacitance \(C\) connected to an ac source of frequency \(\omega\) is ' X '. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become :
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The reactance of a capacitor of capacitance \(C\) connected to an ac source of frequency \(\omega\) is ' X '. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become :
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A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit. The average power consumed in the circuit over a cycle is
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A resistor and an ideal inductor are connected in series to a \(100\sqrt{2}\mathrm{V}\), 50 Hz ac source. When a voltmeter is connected across the resistor or the inductor, it shows the same reading. The reading of the voltmeter is :
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An AC voltage \(V=20 \sin 200 \pi t\) is applied to a series LCR circuit which drives a current \(I=10 \sin \left(200 \pi t+\frac{\pi}{3}\right)\). The average power dissipated is:
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A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit. The average power consumed in the circuit over a cycle is
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Which of the following quantity/quantities remains same in primary andsecondary coils of an ideal transformer ?
Current, Voltage, Power, Magnetic flux
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The reactance of a capacitor of capacitance \(C\) connected to an ac source of frequency \(\omega\) is ' X '. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become :
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A resistor and an ideal inductor are connected in series to a \(100\sqrt{2}\mathrm{V}\), 50 Hz ac source. When a voltmeter is connected across the resistor or the inductor, it shows the same reading. The reading of the voltmeter is :
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A bulb is rated \((100\mathrm{W},110\mathrm{V})\). It is operated by current of 1.0 A supplied by a step down transformer. If the input voltage and efficiency of the transformer are 220 V and 0.9 respectively, the input current drawn from the mains is :
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The reactance of a capacitor of capacitance \(C\) connected to an ac source of frequency \(\omega\) is ' X '. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become :
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A coil of resistance \(20\Omega\) and self-inductance 10 mH is connected to an ac source of frequency \(1000/\pi \mathrm{Hz}\). The phase difference between current in the circuit and the source voltage is :
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The reactance of a capacitor of capacitance \(C\) connected to an ac source of frequency \(\omega\) is ' X '. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become :
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In a series LCR circuit, inductance L = 10 mH and capacitance C = 10 nF. The angular frequency of the source when current has maximum amplitude in the circuit is
(Shift - II Memory Based)
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A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit. The average power consumed in the circuit over a cycle is
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In a series LCR circuit, the maximum amplitude of current is \({I}_{0}\) when the resistance is \(R\). What will be the maximum amplitude of current if the resistor is replaced by a resistor of resistance \(R\mathrm{/}2\)?
(Shift II Memory Based)
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The reactance of a capacitor of capacitance \(C\) connected to an ac source of frequency \(\omega\) is ' X '. If the capacitance of the capacitor is doubled and the frequency of the source is tripled, the reactance will become :
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A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is \(I_0\). If the value of resistance \(R\) becomes twice of its initial value then amplitude of current at resonance will be
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Given below are two statements :
Statement I: In an LCR series circuit, current is maximum at resonance.
Statement II: Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source.
In the light of the above statements, choose the correct from the options given below:
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A resistor and an ideal inductor are connected in series to a \(100\sqrt{2}\mathrm{V}\), 50 Hz ac source. When a voltmeter is connected across the resistor or the inductor, it shows the same reading. The reading of the voltmeter is :
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\text { If } I=I_A \sin \omega t+I_B \cos \omega t \text {, then find } r m s \text { value of current. }
(Shift I - Memory Based)
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A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit. The average power consumed in the circuit over a cycle is
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A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit. The average power consumed in the circuit over a cycle is
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An ac voltage \(\mathrm{v}=\mathrm{v}_0\) sin \(\omega \mathrm{t}\) is applied to a series combination of a resistor \(R\) and an element \(X\). The instantaneous current in the circuit is \(\mathrm{I}=\mathrm{I}_0 \sin \left(\omega \mathrm{t}+\frac{\pi}{4}\right)\). Then which of the following is correct ?
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\text { If } I=I_A \sin \omega t+I_B \cos \omega t \text {, then find } r m s \text { value of current. }
(Shift I - Memory Based)
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An alternating current is given by \(I={I}_{A}\sin \omega t+{I}_{B}\cos \omega t\). The r.m.s current will be
[JEE Main 2025, 24 Jan (Shift 1)]
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A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit. The average power consumed in the circuit over a cycle is
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A voltage \(v={v}_{0}\sin \omega t\) applied to a circuit drives a current \(i={i}_{0}\sin (\omega t+ϕ)\) in the circuit. The average power consumed in the circuit over a cycle is
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An alternating voltage \(V(t)=220 \sin 100 \pi t\) volt is applied to a purely resistive load of \(50 \Omega\). The time taken for the current to rise from half of the peak value to the peak value is:
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A coil of resistance \(20\Omega\) and self-inductance 10 mH is connected to an ac source of frequency \(1000/\pi \mathrm{Hz}\). The phase difference between current in the circuit and the source voltage is :
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Which of the following quantity/quantities remains same in primary andsecondary coils of an ideal transformer ?
Current, Voltage, Power, Magnetic flux
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The primary and secondary coils of a transformer have 500 turns and 5000 turns respectively. The primary coil is connected to an ac source of \(220\mathrm{V}-50\mathrm{Hz}\). The output across the secondary coil is :
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A bulb is rated \((100\mathrm{W},110\mathrm{V})\). It is operated by current of 1.0 A supplied by a step down transformer. If the input voltage and efficiency of the transformer are 220 V and 0.9 respectively, the input current drawn from the mains is :
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The amplitude of the charge oscillating in a circuit decreases exponentially as \(Q={Q}_{0}{e}^{-Rt/2L}\), where \({\mathrm{Q}}_{0}\) is the charge at \(t=0\mathrm{s}\). The time at which charge amplitude decreases to \(0.50{Q}_{0}\) is nearly :
[Given that \(\mathrm{R}=1.5\Omega ,\mathrm{L}=12\mathrm{mH},\ln (2)=0.693\) ]
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A 100 \(\Omega\) resistance, a \(0.1\mu F\) capacitor and an inductor are connected in series across a \(250\mathrm{V}\) supply at variable frequency. Calculate the value of inductance of the inductor at which resonance will occur. Given that the resonant frequency is \(60\mathrm{Hz}\).
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An inductor, a capacitor and a resistor are connected in series across an ac source of voltage. If the frequency of the source is decreased gradually, the reactance of :
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An LCR circuit contains resistance of \(110\Omega\) and a supply of \(220\mathrm{V}\) at \(300\mathrm{rad}/s\) 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:
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Which of the following statements about a series LCR circuit connected to an ac source is correct?
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A step down transformer connected to an ac mains supply of \( 220 \mathrm{~V} \) is made to operate at \( 11 \mathrm{~V}, 44 \mathrm{~W} \) lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
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