Q.A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.9.

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The ratio of the turns per unit length in two inductors is 2 : 1. Then the ratio of the impedances produced by the inductors is (A) 2 (B) 4 (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Inductance (and hence inductive impedance at fixed frequency) scales as the square of the turn density. Answer: 4.

   Concept and Intuition

   For a solenoid-type inductor, $L = \mu_0 n^2 A l$, where $n = N/l$ is the number of turns per unit length. This comes from $B = \mu_0 n I$ inside the coil combined with flux linkage through all $N=nl$ turns — the $n$ appears twice (once from the field, once from the number of turns linking that field), giving the $n^2$ dependence. Since inductive reactance $X_L = \omega L$ at a fixed angular frequency $\omega$, the impedance ratio mirrors the inductance ratio.

   Step-by-Step Solution

   1. $L \propto n^2$ (for identical core, area, and length).
   2. $Z = X_L = \omega L \propto n^2$ (same $\omega$ for both). …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.A resistor of 450 $\Omega$ and an inductor are connected in series to an ac source of frequency $\dfrac{75}{\pi}$ Hz. If the power factor of the circuit is 0.6, then the inductance connected in the circuit is (A) 6 mH (B) 4 H (C) 4 mH (D) 6 H

   ›Reveal solutionSolution

   This tests series $RL$ ac-circuit relations: power factor gives impedance, impedance gives reactance, and reactance gives inductance. The answer is $L=4$ H.

   Concept and Intuition

   In a series $RL$ circuit, the resistor and inductor's voltage phasors are perpendicular (resistor voltage in phase with current, inductor voltage $90^\circ$ ahead), so the impedance triangle has $R$ as the base and $X_L$ as the perpendicular side, with $Z$ the hypotenuse. The power factor $\cos\phi=R/Z$ tells us how "resistive" the circuit looks overall; once we know $\cos\phi$ and $R$, geometry pins down $Z$ and hence $X_L$, and dividing $X_L$ by the angular frequency gives $L$.

   Step-by-Step Solution

   1. Power factor: $\cos\phi=\dfrac{R}{Z}=0.6\Rightarrow Z=\dfrac{R}{0.6}=\dfrac{450}{0.6}=750\ \Omega$.
   2. Impedance relation: $Z^2=R^2+X_L^2\Rightarrow X_L=\sqrt{Z^2-R^2}=\sqrt{750^2-450^2}$. $750^2=562500,\quad 450^2=202500,\quad X_L=\sqrt{360000}=600\ \Omega$ …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.If an inductor of inductance $0.5\ \mu\mathrm{H}$ is connected to an ac source of frequency $70\ \mathrm{MHz}$ and voltage $3.3\ \mathrm{V}$, then the current through the inductor is (A) 5 mA (B) 7.5 mA (C) 15 mA (D) 30 mA

   ›Reveal solutionSolution

   The inductive reactance $X_L = 2\pi fL$ at $70\,\mathrm{MHz}$ for a $0.5\,\mu\mathrm{H}$ inductor works out to about $220\,\Omega$; dividing the given voltage by this reactance gives a current of $15\,\mathrm{mA}$.

   Concept and Intuition

   An ideal inductor connected to an AC source opposes the current with an impedance called inductive reactance, $X_L = \omega L = 2\pi f L$, which increases with frequency. The current amplitude (or rms current, matching whichever the voltage is quoted as) is then simply Ohm's-law-like: $I = V/X_L$.

   Step-by-Step Solution

   1. $X_L = 2\pi f L = 2\pi \times (70\times10^6\,\mathrm{Hz}) \times (0.5\times10^{-6}\,\mathrm{H})$.
   2. $f L = 70\times10^6 \times 0.5\times10^{-6} = 35$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.In the figure, if A & B are identical bulbs, which bulb glows brighter [FIGURE] (an AC circuit with two parallel branches between a common pair of rails, one branch containing a 100 mH inductor in series with bulb A, the other branch containing a 10 pF capacitor in series with bulb B; both branches are connected across an AC source) (A) A (B) B (C) Both with equal brightness (D) Both do not glow

   ›Reveal solutionSolution

   The brightness of each bulb depends on the current through it, which is determined by the impedance of its branch at the AC source frequency. Since the inductor and capacitor have opposite frequency-dependent reactances, one branch will have much lower impedance than the other, making one bulb glow brighter. The correct option is (A).

   1. Understand the circuit and the goal

      Bulbs A and B are identical, so their brightness is directly proportional to the power dissipated, which depends on the square of the current through them. The two branches are in parallel across the same AC source. The current in each branch is limited by the total impedance of that branch:

      • Branch A: inductor (100 mH) in series with bulb A.
      • Branch B: capacitor (10 pF) in series with bulb B. The bulbs themselves have some resistance $R$ (same for both). The source is AC, but its frequency is not given — this is the key puzzle.

   2. Recall the frequency dependence of reactances

      • Inductive reactance: $X_L = 2\pi f L$ — increases with frequency.
      • Capacitive reactance: $X_C = \frac{1}{2\pi f C}$ — decreases with frequency. For a typical AC mains frequency (e.g., 50 Hz or 60 Hz), let’s see what happens.

   3. Estimate the reactances at a typical mains frequency

      Take $f = 50 \, \text{Hz}$ (common in many countries).

      • $L = 100 \, \text{mH} = 0.1 \, \text{H}$

   $X_L = 2\pi (50)(0.1) = 10\pi \approx 31.4 \, \Omega$

   • $C = 10 \, \text{pF} = 10 \times 10^{-12} \, \text{F}$

   $X_C = \frac{1}{2\pi (50)(10 \times 10^{-12})} = \frac{1}{2\pi \times 5 \times 10^{-10}} \approx \frac{1}{3.14 \times 10^{-9}} \approx 3.18 \times 10^8 \, \Omega$

   That’s 318 million ohms — an enormous impedance. Even if the bulb’s resistance is, say, 100 Ω, the capacitor branch is essentially an open circuit.

   4. Compare the branch impedances

      • Branch A impedance: $Z_A = \sqrt{R^2 + X_L^2}$. With $X_L \approx 31.4 \, \Omega$ and $R$ maybe 100 Ω, $Z_A$ is on the order of 100–200 Ω.
      • Branch B impedance: $Z_B = \sqrt{R^2 + X_C^2} \approx X_C$ (since $X_C$ is huge), so $Z_B \approx 3 \times 10^8 \, \Omega$. The current in branch B is therefore microscopic compared to branch A. Bulb B will not glow at all, while bulb A glows normally.

   5. What if the frequency is extremely high?

      If the source were at a very high frequency (e.g., radio frequencies), $X_L$ would become large and $X_C$ small, possibly reversing the situation. But the problem is from a typical physics exam context, where the AC source is assumed to be standard mains (50/60 Hz). No special frequency is mentioned, so we assume ordinary conditions.

   6. Check the possibility of resonance …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.An ac voltage of 10 sin $\omega t$ volt is applied to a pure inductor of inductance 10 H. The current through the inductor in ampere is (A) $\dfrac{1}{\omega}\sin\left(\omega t-\dfrac{\pi}{2}\right)$ (B) $\omega\sin\left(\omega t-\dfrac{\pi}{2}\right)$ (C) $\dfrac{1}{\omega^2}\sin\left(\omega t-\dfrac{\pi}{2}\right)$ (D) $\omega^2\sin\left(\omega t-\dfrac{\pi}{2}\right)$

   ›Reveal solutionSolution

   This tests the standard AC-inductor relation: current lags voltage by $90°$ and has

   amplitude $V_0/(\omega L)$. Answer: $\dfrac1\omega\sin(\omega t-\pi/2)$.

   Concept and Intuition

   For a purely inductive AC circuit, the induced back-EMF opposes the rate of change of

   current, which mathematically results in the current lagging the applied voltage by a

   quarter cycle ($\pi/2$ radians). The peak current is found by dividing the peak

   voltage by the inductive reactance $X_L=\omega L$, analogous to Ohm's law for AC.

   Step-by-Step Solution

   1. Given voltage $v=10\sin\omega t$ volts, so $V_0=10$ V.
   2. Inductive reactance: $X_L=\omega L=\omega(10)=10\omega\ \Omega$.
   3. Peak current: $I_0=\dfrac{V_0}{X_L}=\dfrac{10}{10\omega}=\dfrac1\omega$.
   4. For a pure inductor, current lags voltage by $\pi/2$: …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
   Q.An inductor of $200\ \mu H$ is connected to an alternating source of frequency 50 Hz. Then the inductive reactance is (A) $6.28\times10^{-2}\ \Omega$ (B) $1.64\times10^{-2}\ \Omega$ (C) $3.14\times10^{-2}\ \Omega$ (D) $4.82\times10^{-2}\ \Omega$

   ›Reveal solutionSolution

   Direct application of the inductive-reactance formula $X_L=2\pi f L$.

   Concept and Intuition

   Inductive reactance measures how strongly an inductor opposes a changing AC current; it grows in proportion to both the frequency of the source and the inductance value, $X_L=\omega L=2\pi f L$.

   Step-by-Step Solution

   1. $L=200\ \mu H = 2\times10^{-4}$ H, $f=50$ Hz.
   2. $X_L=2\pi\times50\times2\times10^{-4}=2\pi\times0.01=0.0628\ \Omega$.
   3. $X_L=6.28\times10^{-2}\ \Omega$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.A coil of inductance $0.1$ H and resistance $110\ \Omega$ is connected to a source of $110$ V and $350$ Hz. The phase difference between the voltage maximum and the current maximum is (A) $\tan^{-1}(1.5)$ (B) $\tan^{-1}(0.5)$ (C) $\tan^{-1}(1.73)$ (D) $\tan^{-1}(2)$

   ›Reveal solutionSolution

   This tests the phase relationship in a series RL AC circuit; computing $X_L$ and dividing by $R$ gives $\tan\phi \approx 2$.

   Concept and Intuition

   In a pure resistor, voltage and current are in phase. In a pure inductor, the voltage leads the current by $90^\circ$ because the inductor opposes changes in current. In a series RL circuit, the net phase difference between the applied voltage and the resulting current is somewhere between $0^\circ$ and $90^\circ$, set by the ratio of the inductive reactance to the resistance: $\tan\phi = X_L/R$.

   Step-by-Step Solution

   1. $\omega = 2\pi f = 2\pi(350) \approx 2199.1$ rad/s.
   2. $X_L = \omega L = 2199.1 \times 0.1 \approx 219.9\ \Omega$. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.A 70 mH inductor is connected to 220 V, 50 Hz AC supply. The rms value of the current in the circuit is (A) $\frac{100}{\sqrt{2}\pi}$ A (B) 10 A (C) $\frac{50}{\pi}$ A (D) $\frac{10\sqrt{2}}{\pi}$ A

   ›Reveal solutionSolution

   The inductor's reactance is $X_L=2\pi fL=7\pi\ \Omega\approx22\ \Omega$; dividing the rms voltage

   by this gives $I_{rms}\approx10$ A.

   Concept and Intuition

   A pure inductor opposes changes in current, and in an AC circuit this opposition manifests as the

   inductive reactance $X_L=\omega L=2\pi f L$, which plays the same role as resistance does in Ohm's

   law for the rms quantities: $I_{rms}=V_{rms}/X_L$ (with a $90^\circ$ phase difference between

   voltage and current, but that doesn't affect the rms magnitude relation).

   Step-by-Step Solution

   1. Compute angular frequency: $\omega = 2\pi f = 2\pi(50) = 100\pi\ \mathrm{rad/s}$.
   2. Inductive reactance: $X_L = \omega L = 100\pi \times 0.070 = 7\pi\ \Omega \approx 21.99\ \Omega$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.The impedance of an LR circuit with $L = \dfrac{60}{\pi}$ mH, $R = 8\ \Omega$ and frequency 50 Hz is (A) $1.3\ \Omega$ (B) $14.3\ \Omega$ (C) $20\ \Omega$ (D) $10\ \Omega$

   ›Reveal solutionSolution

   Compute the inductive reactance $X_L=\omega L$ first (the $\pi$'s conveniently cancel), then combine with $R$ in quadrature — a clean 6-8-10 right triangle.

   Concept and Intuition

   In an AC circuit with a resistor and inductor in series, the voltage across $R$ and the voltage across $L$ are $90°$ out of phase (resistor voltage in phase with current, inductor voltage leading by $90°$). So their contributions to the total impedance add as a vector sum (Pythagorean combination), not a simple algebraic sum: $Z = \sqrt{R^2+X_L^2}$.

   Step-by-Step Solution

   1. Angular frequency: $\omega = 2\pi f = 2\pi(50) = 100\pi\ \text{rad/s}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
    Q.A circuit containing inductance of $\dfrac{1}{6\pi}$ H and a resistance of 15 Ω in series. If an AC voltage of 100 V and 60 Hz is applied to above circuit, then the current in the circuit and phase difference between voltage and current respectively are (A) 4 A and $\tan^{-1}\left(\dfrac{4}{5}\right)$ (B) 5.3 A and $\tan^{-1}\left(\dfrac{3}{4}\right)$ (C) 4 A and $\tan^{-1}\left(\dfrac{4}{3}\right)$ (D) 5.3 A and $\tan^{-1}\left(\dfrac{4}{3}\right)$

    ›Reveal solutionSolution

    Find the inductive reactance, combine with resistance to get impedance, then use $I=V/Z$ and $\tan\phi=X_L/R$.

    Concept and Intuition

    In a series RL circuit driven by AC, current lags voltage by a phase angle $\phi$ determined by the ratio of reactance to resistance. The impedance $Z=\sqrt{R^2+X_L^2}$ sets the current magnitude via Ohm's law for AC circuits.

    Step-by-Step Solution

    1. $X_L = 2\pi f L = 2\pi\times 60\times\dfrac{1}{6\pi} = \dfrac{2\times60}{6} = 20\,\Omega$.
    2. $Z = \sqrt{R^2+X_L^2} = \sqrt{15^2+20^2} = \sqrt{225+400}=\sqrt{625}=25\,\Omega$.
    3. $I = \dfrac{V}{Z} = \dfrac{100}{25} = 4$ A. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.The reactance of an inductor at $50\ Hz$ is $10\ \Omega$. The reactance of it at $200\ Hz$ is: (A) $10\ \Omega$ (B) $40\ \Omega$ (C) $2.5\ \Omega$ (D) $20\ \Omega$

    ›Reveal solutionSolution

    Tests that inductive reactance is directly proportional to frequency. Since frequency increases 4×, reactance increases 4× to 40 Ω.

    Concept and Intuition

    An inductor opposes changes in current; the faster the current alternates (higher $f$), the more strongly it opposes, so $X_L = 2\pi f L$ grows linearly with frequency (unlike capacitive reactance, which falls with frequency).

    Step-by-Step Solution

    1. At $f_1 = 50\ Hz$: $X_{L1} = 2\pi f_1 L = 10\ \Omega$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.The current and emf through an inductance differ by phase by: (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{2}$ (C) $\frac{\pi}{3}$ (D) $\pi$

    ›Reveal solutionSolution

    This is a standard AC circuits fact: current through a pure inductor lags the EMF/voltage across it by exactly $\pi/2$ radians (90°).

    Concept and Intuition

    In a pure inductor, the voltage across it is $V = L\dfrac{dI}{dt}$. If the current varies sinusoidally as $I = I_0\sin(\omega t)$, then $V = L I_0 \omega \cos(\omega t) = LI_0\omega \sin(\omega t + \pi/2)$ — the voltage leads the current (equivalently, current lags voltage) by exactly a quarter cycle, $\pi/2$. This is because an inductor opposes CHANGES in current, so the current's peak (zero rate of change) lines up with the moment the voltage crosses zero, and vice versa.

    Step-by-Step Solution

    1. For a pure inductor, $V_L = L\dfrac{dI}{dt}$.
    2. Let $I(t) = I_0\sin(\omega t)$; differentiate: $\dfrac{dI}{dt} = I_0\omega\cos(\omega t)$.
    3. So $V_L = L I_0 \omega \cos(\omega t) = L I_0 \omega \sin(\omega t + \pi/2)$. …

    🔒 Start your 14-day free trial to see the full solution →