NEET Practice Questions (MCQs) with Answers & Solutions

The AC voltage can be represented as $V(t) = V_0 ext{sin}( heta)$ where $V_0$ is the peak voltage. For a frequency $f = 50 ext{ Hz}$, the angular frequency $ heta = 2 ext{Ï€}f = 100 ext{Ï€} ext{ rad/s}$. The voltage changes from its peak value to zero in a quarter cycle, which is $T/4$ where $T = 1/f$. Therefore, $T = 1/50 = 0.02 ext{ s}$ and $T/4 = 0.02/4 = 0.005 ext{ s} = 5 imes 10^{-3} ext{ s}.$

The given voltage is $ ext{ε} = 80 ext{ sin } 100 ext{π} t $ and the impedance is $ 20 ext{ Ω}$. The peak value of voltage $V_0 = 80$ V. The effective (RMS) value of voltage is $V_{ ext{rms}} = V_0/ ext{√2} = 80/ ext{√2} = 56.57$ V. Using Ohm's law, the effective current $I_{ ext{rms}} = V_{ ext{rms}}/Z = 56.57/20 = 2.828$ A.

A choke coil is designed to have high inductance and low resistance. The high inductance allows it to effectively block high-frequency AC signals, while the low resistance minimizes power loss in the form of heat.

with help pf phasor $ \upsilon = \sqrt {\upsilon _R^2 + \upsilon _C^2 } $

The average power in an AC circuit is given by the formula: \[ P_{avg} = \frac{E_0 I_0}{2} \cos \varphi \] where \( E_0 \) is the peak emf, \( I_0 \) is the peak current, and \( \varphi \) is the phase difference between the emf and the current. This formula is derived from the general power formula for AC circuits and accounts for the phase difference.

The impedance \( Z \) of a series circuit containing resistance \( R \) and inductance \( L \) is given by: \[ Z = \sqrt{R^2 + (\omega L)^2} \] where \( \omega = 2\pi f \) is the angular frequency. Substituting \( \omega \), the impedance becomes: \[ Z = \sqrt{R^2 + (2\pi fL)^2} \]

$ I = { \epsilon _0 \over \sqrt { R^2 + \omega^2 L^2 } } = { \epsilon _0 \over R \sqrt {1 + ({ \omega L \over R }) ^2 }} = { I _0 \over \sqrt {1 + ({ \omega L \over R }) ^2 }}$

$ I_{rms } = { \upsilon _ {rms } \over |z| } =300 /30 = 10 A $

In an inductive circuit, the current \( I \) is inversely proportional to the frequency \( f \) of the AC supply, according to the formula \( I = \frac{V}{\omega L} \). Thus, as frequency increases, the current decreases. In a capacitive circuit, the current is directly proportional to the frequency, according to \( I = \omega CV \). Hence, as frequency increases, the current increases. Therefore, the effect of increasing frequency is a decrease in current for an inductive circuit and an increase in current for a capacitive circuit.