Physics MCQs for NEET — Practice Questions with Answers

The power factor of an L-C-R circuit is given by cosφ, where φ is the phase angle between voltage and current. The given potential differences across R, C, and L are 80V, 40V, and 100V respectively. Using these values, cosφ = 80/√(80^2 + 40^2 + 100^2) = 0.8. Therefore, the power factor is 0.8.

impedance of the R-C circuit, Z 
= $\sqrt{R^2+X_C^2}$ where, R =100$\Omega$ and X_C = 100$\Omega$ 
 
= 100$\sqrt{2}\Omega$ 
The peak value of the current, 
$I_{max}=\frac{V_{max}}{Z}=\frac{220\sqrt{2}}{100\sqrt{2}}=2.2 A$

In a series R-C circuit, the voltage drop across the capacitor is inversely proportional to the capacitance. When the capacitor is mica-filled, its capacitance increases due to the higher dielectric constant of mica compared to air. Therefore, the voltage drop across the mica-filled capacitor will be lower than the air-filled capacitor for the same current.

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.