Alternating Current MCQs for NEET — Physics Questions with Answers

In an RL series circuit, the voltage leads the current by a phase angle φ = tan^(-1)(X_L/R). When R = X_L = 3Ω, φ = 45° = π/4 radians. Therefore, the phase difference between applied voltage and current is π/4.

$I_S$= 2A, $E_S$= 440V, $E_P$=220V 
Efficiency = $\frac{Output power}{input power}=\frac{E_S{I}_S}{E_P{I}_P}$
$\frac{80}{100}$= $\frac{440\times 2}{220\times I_P}$

Solving we get- 

$I_P=5 A$

The rate of doing work is known as the power. It is given that in the ac circuit emf is ‘e’ and the current flowing in the circuit is represented by ‘I’ 
e = E₀ sin $\omega t$t 
I=$I_0$sin($\left(\omega t-\varphi \right)$ 
The average power is given by: 
$P_{avg}$ = W/T 
= $\frac{eI}{T}$
= $\frac{E_0{I}_0\cos \varphi \times T/2}{T}$ 
= $\frac{E_0{I}_0\cos \varphi }{2}$

Given, 
An LCR circuit has a capacitor of capacitance (C) =100 $\mu F$ , $\omega =1000 s^{-1}$ 
In resonance condition, maximum current flows in the circuit. 
Current in LCR series circuit, 
I=$\frac{V}{{\sqrt{R^2+\left(X_L-X_C\right)}}^2}$ 
Where, V is RMS value of current, R is resistance, XL is inductive reactance and XC is capacitive reactance. 
For current to be maximum, denominator should be minimum which can be done, if 
XL=XC 
We know, 
XL =$\omega L and XC =\frac{1}{\omega C}$ 
This happens in resonance state of the circuit i.e., 
ωL=$\frac{1}{\omega C}$ 
Or L =$\frac{1}{{\omega }^{2}C}$ ……..(i) 
Given, ω=1000 s^-1, C = 10 μF= 10×10^-6 F 
Hence, L =$\frac{1}{{\left(1000\right)}^2\times 10\times {10}^{-6}}$ 
= 0.1 H 
= 100 mH

When magnetic flux linked with a coil changes, induced emf is produced in it and the induced current flows through the wire forming the coil. In 1895, Focault experimentally found that these induced currents are set up in the conductor in the form of closed loops. These currents look like eddies or whirlpools and likewise are known as eddy currents. They are also known as Focault’s current. These currents oppose the cause of their origin, therefore, due to eddy currents, a great amount of energy is wasted in the form of heat energy. If the core of the transformer is laminated, then its effect can be minimized.

Given, 
Inductive reactance of coil = 31Ω 
Resistance = 8Ω 
Capacitive reactance = 25 Ω 
Power factor (cos Ï•)is a ratio of resistance and impedance of a.c. circuit. 
Power factor of a.c. circuit is given by: 
Cos Ï• = $\frac{R}{Z}$ ……..(i) 
Where R is resistance employed and z the impedance of the circuit. 
Z = $\sqrt{R^2+{\left(X_L-X_C\right)}^2}$ …………(ii) 
Eqs. (i) and (ii) meet to give, 
Cos Ï• = ${\sqrt{R^2+{\left(X_L-X_C\right)}^2}}^{}$ …………(iii) 

Given R= 8Ω , X_L =31 Ω, X_C =25 Ω 
∴ Cos Ï• = $\frac{8}{\sqrt{8^2+{\left(31-25\right)}^2}}$ 
= $\frac{8}{\sqrt{64+36}}$

 
Hence, Cos Ï• = 0.80

In steady RC circuit connected across DC source , current  is zero as capacitor  behaves as an open circuit. However in the LR circuit, current is present in steady state when connected across the DC source.

According to the NCERT text and Figure 7.2, 'In a pure resistor, the voltage and current are in phase. The minima, zero and maxima occur at the same respective times.' This means their phase difference is zero.

The NCERT states, 'the sum of the instantaneous current values over one complete cycle is zero, and the average current is zero.' This is because the sinusoidal current has symmetrical positive and negative values over a cycle.