Question 61:
easy12. A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when: (2013)
Inserting an iron rod into the coil increases its self-inductance $L$. This increases the inductive reactance $X_L = \omega L$ and hence total circuit impedance $Z = \sqrt{R^2 + X_L^2}$. With increased impedance, the circuit current decreases, leading to reduced brightness of the bulb.
Question 62:
easy13. A coil has resistance $30 \Omega$ and inductive reactance $20 \Omega$ at $50 Hz$ frequency. If an AC source, of $200 volt$, $100 Hz$, is connected across the coil, the current in the coil will be: (2011 Mains)
Inductive reactance is $X_L = 2\pi f L \propto f$. At $100 Hz$, $X_L' = 20 \times \frac{100}{50} = 40 \Omega$. Circuit impedance is $Z = \sqrt{R^2 + X_L'^2} = \sqrt{30^2 + 40^2} = 50 \Omega$. Current in the coil is $I = \frac{V}{Z} = \frac{200}{50} = 4.0 A$.
Question 63:
easy14. An AC voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to $3 \Omega$, the phase difference between the applied voltage and the current in the circuit is: (2011 Pre)
Phase difference $\phi$ in a series RL circuit is given by $\tan\phi = \frac{X_L}{R}$. Given $X_L = 3 \Omega$ and $R = 3 \Omega$, $\tan\phi = \frac{3}{3} = 1$. Therefore, the phase difference is $\phi = \frac{\pi}{4}$.
Question 64:
easy15. A series RC circuit is connected to an alternating voltage source. Consider two situations
A. When capacitor is air filled.
B. When capacitor is mica filled.
Current through resistor is i and voltage across capacitor is V then: (2015 Re)
Inserting mica increases capacitance ($C_b > C_a$), so reactance decreases ($X_{Cb} < X_{Ca}$). Total impedance $Z = \sqrt{R^2 + X_C^2}$ decreases, causing current to increase ($i_b > i_a$). Since $V_C = \frac{V}{\sqrt{1 + (R/X_C)^2}}$, as $X_C$ decreases, $V_C$ decreases, hence $V_a > V_b$.
Question 65:
easy17. The time constant of C-R circuit is (1992)
In a series capacitive-resistive (C-R) circuit, charge grows or decays according to $q(t) = q_0 e^{-t/\tau}$. The time constant $\tau$ is defined as the product of capacitance and resistance. Therefore, the time constant is $\tau = CR$.
Question 66:
easy28. An series L-C-R circuit is connected to a source of A.C. current. At resonance, the phase difference between the applied voltage and the current in the circuit, is (1994)
At resonance in a series L-C-R circuit, the inductive and capacitive reactances cancel out ($X_L = X_C$).
The total impedance of the circuit is purely resistive ($Z = R$).
Since the circuit is purely resistive, the current and applied voltage are in phase.
Therefore, the phase difference between the applied voltage and current is zero.
Question 67:
easy29. A condenser of capacity C is charged to a potential difference of $V_1$. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to $V_2$ is: (2010 Mains)
By conservation of energy, the decrease in electrical energy of the capacitor equals the magnetic energy gained by the inductor.
$\frac{1}{2} C V_1^2 - \frac{1}{2} C V_2^2 = \frac{1}{2} L I^2$.
$C(V_1^2 - V_2^2) = L I^2 \implies I^2 = \frac{C(V_1^2 - V_2^2)}{L}$.
Therefore, the current is $I = \left[ \frac{C(V_1^2 - V_2^2)}{L} \right]^{\frac{1}{2}}$.
Question 68:
easy30. A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be: (2006)
The frequency of oscillation for an LC circuit is $f = \frac{1}{2\pi\sqrt{LC}}$.
When the inductance is doubled ($L' = 2L$) and capacitance is quadrupled ($C' = 4C$), the new frequency is $f'$.
$f' = \frac{1}{2\pi\sqrt{(2L)(4C)}} = \frac{1}{2\pi\sqrt{8LC}} = \frac{1}{\sqrt{8}} \cdot \frac{1}{2\pi\sqrt{LC}}$.
Thus, the new frequency is $f' = \frac{f}{2\sqrt{2}}$.
Question 69:
easy20. In a circuit L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by $45^\circ$. The value of C is: (2005)
Current leads voltage by $45^\circ$, so capacitive reactance is greater than inductive reactance.
$\tan \phi = \frac{X_C - X_L}{R} \implies \tan 45^\circ = \frac{X_C - X_L}{R} \implies 1 = \frac{X_C - X_L}{R}$.
$X_C = X_L + R \implies \frac{1}{2\pi f C} = 2\pi f L + R$.
Therefore, $C = \frac{1}{2\pi f(2\pi f L + R)}$.
Question 70:
easyA series LCR circuit with inductance $10 H$, capacitance $10 \mu F$, resistance $50 \Omega$ is connected to an ac source of voltage, $V = 200 \sin(100t)$ volt. If resonant frequency of the LCR circuit is $\nu_0$ and the frequency of the ac source is $\nu$, then:
(2022)
For the AC source $V = 200 \sin(100t)$, angular frequency $\omega = 100 rad/s$.
Source frequency $\nu = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} Hz$.
Resonant frequency $\nu_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10 \times 10 \times 10^{-6}}} = \frac{1}{2\pi\times 10^{-2}} = \frac{50}{\pi} Hz$.
Thus, $\nu_0 = \nu = \frac{50}{\pi} Hz$.