Question 51:
easyWhich of the following AC circuit will not dissipate energy? (Symbols have their usual meaning)
In a purely reactive circuit like a series LC circuit, the average power dissipated is zero because the phase difference between the current and voltage is \( 90^\circ \), making the power factor \( \cos\phi = 0 \).
Question 52:
easyConsider a series \(LCR\) circuit in which reactance and resistance are \(100\ \Omega\) each. When the circuit is connected to ac source \(220\text{ V}\), \(50\text{ Hz}\), then current drawn from the source is
Here, \(R = 100\ \Omega\) and net reactance \(X = 100\ \Omega\). Total impedance is \(Z = \sqrt{R^2 + X^2} = 100\sqrt{2}\ \Omega\). The current \(I = \frac{V}{Z} = \frac{220}{100\sqrt{2}} = 1.1\sqrt{2}\text{ A}\).
Question 53:
easy7. A small signal voltage $V(t) = V_0 \sin \omega t$ is applied across an ideal capacitor C : (2016 – I)
For an ideal capacitor, the phase difference between current and voltage is $\phi = 90^\circ$ (current leads voltage). The average power consumed over a full cycle is $P_{avg} = V_{rms} I_{rms} \cos(90^\circ) = 0$. Since average power is zero, the capacitor does not consume any energy from the voltage source over a complete cycle.
Question 54:
easy8. In an AC circuit an alternating voltage $E = 200\sqrt{2} \sin 100t$ volts is connected to a capacitor of capacity $1 \mu F$. The r.m.s value of the current in the circuit is: (2011 Pre)
Peak voltage is $E_0 = 200\sqrt{2} V$, so $E_{rms} = \frac{E_0}{\sqrt{2}} = 200 V$. Capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = 10^4 \Omega$. The rms current is $I_{rms} = \frac{E_{rms}}{X_C} = \frac{200}{10^4} = 20 \times 10^{-3} A = 20 mA$.
Question 55:
easy1. The peak voltage of the ac source is equal to (2022)
The root-mean-square (rms) voltage for a sinusoidal alternating voltage is given by $V_{rms} = \frac{V_0}{\sqrt{2}}$. Rearranging for peak voltage $V_0$, we get $V_0 = \sqrt{2} V_{rms}$. Hence, the peak voltage is $\sqrt{2}$ times the rms value of the ac source.
Question 56:
easy3. In an A.C. circuit, $I_{rms}$ and $I_0$ are related as (1994)
For an alternating current $I = I_0 \sin \omega t$, the rms current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_0^T I^2 dt}$. Evaluating this integral over one complete cycle gives $I_{rms} = \frac{I_0}{\sqrt{2}}$. Therefore, the relation between rms and peak current is $I_{rms} = I_0/\sqrt{2}$.
Question 57:
easy5. A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by $V = V_0 \sin \omega t$. The displacement current between the plates of the capacitor, would then be given by: (2021)
The displacement current between the capacitor plates is equal to the conduction current, $I_d = \frac{dq}{dt}$. Since charge on the capacitor is $q = C V = C V_0 \sin \omega t$, we differentiate with respect to time. $I_d = \frac{d}{dt}(C V_0 \sin \omega t) = V_0 \omega C \cos \omega t$.
Question 58:
easy6. A $40 \mu F$ capacitor is connected to a $200 V, 50 Hz$ ac supply. The rms value of the current in the circuit is, nearly: (2020)
Capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$. RMS current is given by $I_{rms} = \frac{V_{rms}}{X_C} = V_{rms} \times 2\pi f C$. Substituting the values: $I_{rms} = 200 \times 2\pi \times 50 \times 40 \times 10^{-6} = 0.8\pi \approx 2.51 A \approx 2.5 A$.
Question 59:
easy9. A capacitor of capacity C and reactance X if capacitance and frequency become double then reactance will be: (2001)
Capacitive reactance is $X = \frac{1}{2\pi f C}$. When capacitance becomes $2C$ and frequency becomes $2f$, the new reactance is $X' = \frac{1}{2\pi(2f)(2C)} = \frac{1}{4(2\pi f C)} = \frac{X}{4}$.
Question 60:
easy11. A resistance R draws power P when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes Z, the power drawn will be: (2015)
Initially, power dissipated in resistance is $P = \frac{V_{rms}^2}{R}$. When inductance is connected in series, current is $I_{rms} = \frac{V_{rms}}{Z}$ and power factor is $\cos\phi = \frac{R}{Z}$. New power drawn is $P' = V_{rms} I_{rms} \cos\phi = V_{rms} \left(\frac{V_{rms}}{Z}\right) \left(\frac{R}{Z}\right) = \frac{V_{rms}^2 R}{Z^2} = P \left(\frac{R}{Z}\right)^2$.