Q.A $44\ \text{mH}$ inductor is connected to $220\ \text{V}$, $50\ \text{Hz}$ ac supply. Determine the rms value of the current in the circuit.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the energy stored in an inductor is $18\,\mathrm{mJ}$ when a current of $3\,\mathrm{A}$ is passed through it, then the magnetic flux linked with the inductor is (A) $36\,\mathrm{mWb}$ (B) $24\,\mathrm{mWb}$ (C) $18\,\mathrm{mWb}$ (D) $12\,\mathrm{mWb}$

   ›Reveal solutionSolution

   The energy stored in an inductor is $U = \frac12 L I^2$, and the magnetic flux linkage is $\Phi = L I$. Combining these gives $\Phi = \frac{2U}{I}$, which yields $12\,\mathrm{mWb}$.

   The key idea is that the energy stored in an inductor is directly related to both its inductance and the current, while the magnetic flux linkage is the product of inductance and current. By eliminating the inductance, we can find the flux directly from the given energy and current.

   1. Recall the two fundamental formulas for an inductor The energy stored in an inductor is

   $U = \frac12 L I^2,$

   and the magnetic flux linked (often called flux linkage) is

   $\Phi = L I.$

   Here, $L$ is the inductance, $I$ is the current, $U$ is the energy, and $\Phi$ is the flux linkage.

   2. We want $\Phi$, but we don’t know $L$ directly However, we can express $L$ from the energy formula:

   $L = \frac{2U}{I^2}.$

   Then substitute this into the flux formula:

   $\Phi = L I = \left( \frac{2U}{I^2} \right) I = \frac{2U}{I}.$

   3. Plug in the given values $U = 18\,\mathrm{mJ} = 18 \times 10^{-3}\,\mathrm{J}$ and $I = 3\,\mathrm{A}$.

   $\Phi = \frac{2 \times 18 \times 10^{-3}}{3} = \frac{36 \times 10^{-3}}{3} = 12 \times 10^{-3}\,\mathrm{Wb} = 12\,\mathrm{mWb}.$

   Tip

   Notice that the units work out: energy in joules (J) divided by current in amperes (A) gives webers (Wb), since $1\,\mathrm{J/A} = 1\,\mathrm{Wb}$.

   Watch out

   A common mistake is to use $U = \frac12 \Phi I$ directly without derivation, but then forget the factor of 2. Always check: $\Phi = 2U/I$, not $U/I$.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.When an inductor and a resistor are connected in series to an ac source, the power factor of the circuit is $\dfrac{2}{\sqrt{13}}$. If the same resistor and a capacitor are connected in series to the same ac source, then the power factor of the circuit is $\dfrac{1}{\sqrt{2}}$. If these inductor, capacitor and resistor are connected in series to the same ac source, then the ratio of the resistance and impedance of the LCR circuit is (A) $1 : \sqrt{7}$ (B) $2 : \sqrt{5}$ (C) $2 : \sqrt{7}$ (D) $1 : \sqrt{5}$

   ›Reveal solutionSolution

   The power factor in an RL or RC circuit gives the ratio of resistance to impedance; combining these yields the resistance and net reactance in the LCR circuit, leading to the ratio $R : Z = 2 : \sqrt{7}$, which corresponds to option (C).

   ---

   Concept & Intuition

   The power factor of an AC circuit is defined as $\cos \phi = \frac{R}{Z}$, where $R$ is the resistance and $Z$ is the impedance. For a series RL circuit, $Z = \sqrt{R^2 + X_L^2}$; for a series RC circuit, $Z = \sqrt{R^2 + X_C^2}$. Given the power factors, we can find the ratios $X_L/R$ and $X_C/R$. Then, for the series LCR circuit, the net reactance is $|X_L - X_C|$, and the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$. The problem asks for $R : Z$, which we can compute directly from these ratios.

   ---

   Step-by-step solution

   1. RL circuit power factor Given $\cos \phi_{RL} = \frac{2}{\sqrt{13}}$. For RL: $\cos \phi = \frac{R}{\sqrt{R^2 + X_L^2}} = \frac{2}{\sqrt{13}}$. Square both sides:

   $\frac{R^2}{R^2 + X_L^2} = \frac{4}{13}$

   Cross-multiply:

   $13R^2 = 4R^2 + 4X_L^2 \quad\Rightarrow\quad 9R^2 = 4X_L^2$

   So $X_L^2 = \frac{9}{4}R^2$ and thus $X_L = \frac{3}{2}R$ (taking positive reactance).

   2. RC circuit power factor Given $\cos \phi_{RC} = \frac{1}{\sqrt{2}}$. For RC: $\cos \phi = \frac{R}{\sqrt{R^2 + X_C^2}} = \frac{1}{\sqrt{2}}$. Square:

   $\frac{R^2}{R^2 + X_C^2} = \frac{1}{2}$

   Cross-multiply:

   $2R^2 = R^2 + X_C^2 \quad\Rightarrow\quad X_C^2 = R^2$

   So $X_C = R$ (taking positive capacitive reactance).

   3. LCR series circuit The net reactance is $|X_L - X_C| = \left|\frac{3}{2}R - R\right| = \frac{1}{2}R$. The impedance is:

   $Z = \sqrt{R^2 + \left(\frac{R}{2}\right)^2} = \sqrt{R^2 + \frac{R^2}{4}} = \sqrt{\frac{5R^2}{4}} = \frac{R\sqrt{5}}{2}$

   Therefore, the ratio of resistance to impedance is:

   $\frac{R}{Z} = \frac{R}{\frac{R\sqrt{5}}{2}} = \frac{2}{\sqrt{5}}$

   So $R : Z = 2 : \sqrt{5}$.

   4. Check the options Option (B) is $2 : \sqrt{5}$. But wait — the problem asks for the ratio of resistance and impedance of the LCR circuit. That is exactly $R : Z = 2 : \sqrt{5}$. However, let’s double-check: the power factor of the LCR circuit would be $\frac{R}{Z} = \frac{2}{\sqrt{5}}$, which is not directly given, but we computed it. So the correct option is (B).

   Watch out

   A common mistake is to confuse the ratio $R:Z$ with the power factor itself — they are the same thing! Here the power factor of the LCR circuit is $\frac{2}{\sqrt{5}}$, so the ratio is $2:\sqrt{5}$. Option (C) $2:\sqrt{7}$ would arise if one mistakenly used $X_L + X_C$ instead of $|X_L - X_C|$.

   Tip

   Notice that the power factors given for RL and RC circuits directly give the tangent of the phase angle: $\tan \phi_{RL} = X_L/R = 3/2$ and $\tan \phi_{RC} = X_C/R = 1$. Then for LCR, net $\tan \phi = |3/2 - 1| = 1/2$, so $\cos \phi = \frac{1}{\sqrt{1+(1/2)^2}} = \frac{2}{\sqrt{5}}$, confirming the ratio.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The permeability of the material of the core used in a solenoid of length 1.4 m, radius 7 cm having $10^3$ turns and a self-inductance of 2.2 H is (A) $1 \times 10^{-4} \ \mathrm{Hm}^{-1}$ (B) $2 \times 10^{-4} \ \mathrm{Hm}^{-1}$ (C) $3 \times 10^{-4} \ \mathrm{Hm}^{-1}$ (D) $4 \times 10^{-4} \ \mathrm{Hm}^{-1}$

   ›Reveal solutionSolution

   The self-inductance of a solenoid depends on the permeability of its core, its geometry, and the number of turns. Using $L = \mu \frac{N^2 A}{l}$, the permeability works out to $2 \times 10^{-4} \ \mathrm{Hm}^{-1}$, which is option (B).

   The core idea here is that self-inductance $L$ of a solenoid is a direct measure of how much magnetic flux it can produce per unit current — and that depends linearly on the permeability $\mu$ of the material inside. For an air-core solenoid, $\mu = \mu_0$, but here the core is some other material, so we solve for $\mu$ from the given $L$.

   The formula for the self-inductance of a long solenoid (length >> radius, so the field inside is nearly uniform) is:

   $L = \mu \frac{N^2 A}{l}$

   where $N$ is the total number of turns, $A$ is the cross-sectional area, $l$ is the length, and $\mu$ is the absolute permeability of the core material. This comes from $L = N \Phi / I$ and $\Phi = B A = \mu (N I / l) A$.

   We are given $L = 2.2 \ \mathrm{H}$, $l = 1.4 \ \mathrm{m}$, radius $r = 7 \ \mathrm{cm} = 0.07 \ \mathrm{m}$, and $N = 10^3$. We need $\mu$.

   1. Find the cross-sectional area $A$ of the solenoid. Since it's cylindrical:

   $A = \pi r^2 = \pi (0.07)^2 = \pi \times 0.0049 = 0.015394 \ \mathrm{m}^2$

   (Keep it symbolic: $A = \pi \times 49 \times 10^{-4} \ \mathrm{m}^2$.)

   2. Rearrange the inductance formula for $\mu$:

   $\mu = \frac{L l}{N^2 A}$

   3. Substitute the numbers:

   $\mu = \frac{2.2 \times 1.4}{(10^3)^2 \times \pi (0.07)^2}$

   Compute step by step:

   • Numerator: $2.2 \times 1.4 = 3.08$
   • Denominator: $N^2 = 10^6$, and $A = \pi \times 0.0049 = \pi \times 4.9 \times 10^{-3}$ So denominator = $10^6 \times \pi \times 4.9 \times 10^{-3} = \pi \times 4.9 \times 10^3$

   Thus:

   $\mu = \frac{3.08}{\pi \times 4.9 \times 10^3}$

   4. Simplify numerically: $\frac{3.08}{4.9} \approx 0.62857$, so:

   $\mu \approx \frac{0.62857}{\pi \times 10^3} = \frac{0.62857}{3141.59} \approx 2.00 \times 10^{-4} \ \mathrm{Hm}^{-1}$

   More exactly, using $\pi \approx 3.1416$:

   $\mu = \frac{3.08}{3.1416 \times 4.9 \times 10^3} = \frac{3.08}{15.3938 \times 10^3} = \frac{3.08}{15393.8} = 2.000 \times 10^{-4}$

   Watch out

   A common mistake is to forget that the radius is given in cm — converting to metres is essential. Also, note that the formula uses absolute permeability $\mu$, not relative permeability $\mu_r$. The answer here is in $\mathrm{Hm}^{-1}$, which is the SI unit for $\mu$.

   Tip

   You can avoid messy decimals by keeping everything in powers of 10 from the start: $r = 7 \times 10^{-2} \ \mathrm{m}$, so $A = \pi \times 49 \times 10^{-4} = 4.9\pi \times 10^{-3}$. Then $\mu = \frac{2.2 \times 1.4}{10^6 \times 4.9\pi \times 10^{-3}} = \frac{3.08}{4.9\pi \times 10^3}$, which simplifies cleanly.

   ✓Final answer

   The permeability of the core material is $2 \times 10^{-4} \ \mathrm{Hm}^{-1}$, which corresponds to option (B).

4. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.A resistor of resistance $160~\Omega$, an inductor of inductance $280~\mathrm{mH}$ and a capacitor are connected in series to an ac source of $80~\mathrm{V}$ supply and $50~\mathrm{Hz}$ frequency. If the the circuit is in resonance, then the potential difference across the capacitor is (A) $44~\mathrm{V}$ (B) $40~\mathrm{V}$ (C) $88~\mathrm{V}$ (D) $80~\mathrm{V}$

   ›Reveal solutionSolution

   In a series RLC circuit at resonance, the inductive and capacitive reactances cancel each other out, making the circuit purely resistive. This allows us to calculate the current and subsequently the potential difference across the capacitor. The potential difference across the capacitor is $\boxed{44~\mathrm{V}}$.

   An RLC series circuit consists of a resistor ($R$), an inductor ($L$), and a capacitor ($C$) connected in series to an alternating current (AC) source. Each component offers an opposition to the current flow: the resistor offers resistance ($R$), the inductor offers inductive reactance ($X_L$), and the capacitor offers capacitive reactance ($X_C$).

   The inductive reactance is given by $X_L = \omega L$, where $\omega$ is the angular frequency of the AC source and $L$ is the inductance. The capacitive reactance is given by $X_C = \frac{1}{\omega C}$, where $C$ is the capacitance.

   Important

   Resonance in a series RLC circuit occurs when the inductive reactance exactly equals the capacitive reactance.

   $X_L = X_C$

   At resonance, the net reactance ($X_L - X_C$) becomes zero. This means the total impedance ($Z$) of the circuit, which is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$, simplifies to $Z = \sqrt{R^2 + 0^2} = R$.

   Since the impedance is at its minimum value (equal to the resistance $R$), the current flowing through the circuit at resonance is maximum. The circuit behaves purely resistively, meaning the current and voltage are in phase.

   We can use this understanding to find the potential difference across the capacitor.

   1. Identify the given parameters:

      We are given the following values for the series RLC circuit:

      • Resistance, $R = 160~\Omega$
      • Inductance, $L = 280~\mathrm{mH} = 0.28~\mathrm{H}$
      • Source voltage, $V_{source} = 80~\mathrm{V}$ (This is the RMS voltage, as it's an AC supply value)
      • Frequency, $f = 50~\mathrm{Hz}$
      • The circuit is in resonance.

   2. Calculate the angular frequency ($\omega$):

      The angular frequency is related to the linear frequency by the formula $\omega = 2\pi f$.

   $\omega = 2\pi (50~\mathrm{Hz}) = 100\pi~\mathrm{rad/s}$

   3. Calculate the inductive reactance ($X_L$): Using the formula $X_L = \omega L$:

   $X_L = (100\pi~\mathrm{rad/s}) \times (0.28~\mathrm{H}) = 28\pi~\Omega$

   4. Determine the capacitive reactance ($X_C$) at resonance: Since the circuit is in resonance, the inductive reactance must be equal to the capacitive reactance.

   $X_C = X_L = 28\pi~\Omega$

   5. Calculate the total impedance ($Z$) of the circuit at resonance: At resonance, the impedance of a series RLC circuit is purely resistive.

   $Z = R = 160~\Omega$

   6. Calculate the current ($I$) flowing through the circuit: Using Ohm's law for AC circuits, $I = \frac{V_{source}}{Z}$.

   $I = \frac{80~\mathrm{V}}{160~\Omega} = 0.5~\mathrm{A}$

   This is the RMS current flowing through all components in the series circuit.
   

   7. Calculate the potential difference across the capacitor ($V_C$):

   The potential difference across the capacitor is given by $V_C = I \cdot X_C$.

   $V_C = (0.5~\mathrm{A}) \times (28\pi~\Omega) = 14\pi~\mathrm{V}$

   To get a numerical value, we use $\pi \approx 3.14159$:
   

   $V_C = 14 \times 3.14159 \approx 43.982~\mathrm{V}$

   Rounding this to the nearest whole number, we get $44~\mathrm{V}$.
   

   Watch out

   It is a common mistake to assume that at resonance, the potential difference across the inductor and capacitor are zero. While their net voltage is zero (they are $180^\circ$ out of phase and cancel each other), the individual potential differences across $L$ and $C$ can be significant and often much larger than the source voltage.

   Comparing this result with the given options, $44~\mathrm{V}$ matches option (A).

   ✓Final answer

   The potential difference across the capacitor is $\boxed{44~\mathrm{V}}$.

5. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The potential difference V across the filament of the bulb shown in the given Wheatstone bridge varies as $V = i(2i + 1)$, where ‘i’ is the current in ampere through the filament of the bulb. The emf of the battery ($V_a$) so that the bridge becomes balanced is [FIGURE] (A) 10 V (B) 15 V (C) 20 V (D) 25 V

   ›Reveal solutionSolution

   Balance demands $R_{\text{bulb}}=6\,\Omega$. Since $R_{\text{bulb}}=V/i=2i+1$, the current must be $2.5$ A; that branch ($4+6=10\,\Omega$) sits across the battery, so $V_a=25$ V — option (D).

   The concept first

   This question is clever because it fuses two ideas.

   (1) Bridge balance. The battery is across the left and right vertices; the two paths from left vertex to right vertex are

   $\text{path 1: } 4\,\Omega \to \text{bulb}, \qquad \text{path 2: } 8\,\Omega \to 12\,\Omega .$

   The bridge is balanced when the midpoints of the two paths (the top and bottom vertices) sit at the same potential — then the diagonal carries no current. Two potential dividers reach the same fraction only if their ratios agree:

   $\frac{4}{4+R_{\text{bulb}}}=\frac{8}{8+12}.$

   Equivalently $\dfrac{4}{R_{\text{bulb}}}=\dfrac{8}{12}$ — the familiar $\dfrac{P}{Q}=\dfrac{R}{S}$.

   (2) A non-ohmic element. A bulb filament is not ohmic: it gets hotter as the current rises, so its resistance rises with current. That is exactly what $V=i(2i+1)$ encodes. Its effective resistance at a given operating point is still defined by

   $R=\frac{V}{i}=2i+1 ,$

   which depends on $i$. So demanding a particular resistance is the same as demanding a particular current — and that is the bridge to the second half of the problem.

   Step-by-step

   1. Impose balance.

   $\frac{4}{R_{\text{bulb}}}=\frac{8}{12}=\frac{2}{3}\qquad\Longrightarrow\qquad R_{\text{bulb}}=\frac{4\times3}{2}=6\,\Omega .$

   (Check with the potential-divider form: $\frac{4}{4+6}=0.4$ and $\frac{8}{8+12}=0.4$ ✓ — both midpoints sit at $40\%$ of the way, so they are at equal potential.)

   2. Convert the resistance requirement into a current.

   $R_{\text{bulb}}=\frac{V}{i}=\frac{i(2i+1)}{i}=2i+1 = 6 \qquad\Longrightarrow\qquad 2i=5 \qquad\Longrightarrow\qquad i=2.5\ \text{A}.$

   So the bulb (and hence the whole $4\,\Omega$–bulb branch, since they are in series) must carry $2.5$ A.

   3. Voltage across that branch. Because the galvanometer branch carries no current at balance, the $4\,\Omega$ and the bulb form a simple series chain directly across the battery:

   $R_{\text{branch}}=4+6=10\,\Omega .$

   4. Find the emf.

   $V_a=i\,R_{\text{branch}}=2.5\times10=25\ \text{V}.$

   5. Cross-check with the other branch. The $8\,\Omega$–$12\,\Omega$ branch has $20\,\Omega$ across the same $25$ V, carrying $i_2=25/20=1.25$ A. Potential drop to the bottom vertex: $1.25\times8=10$ V. On the other branch, drop to the top vertex: $2.5\times4=10$ V. Equal — the diagonal carries no current, exactly as balance requires. ✓

   6. Bonus check on the bulb. At $i=2.5$ A, $V_{\text{bulb}}=i(2i+1)=2.5\times6=15$ V, and $4+ \text{bulb} \Rightarrow 10+15=25$ V $=V_a$ ✓.

   ✓Final answer

   The emf of the battery must be $25\ \mathrm{V}$, so the correct option is (D).

   ANSWER: D

6. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.The self-induced emf of a coil is 36 V. If the current in the coil is changed from 12 A to 24 A in one second, then the change in the energy stored in the coil is (A) 648 J (B) 462 J (C) 486 J (D) 572 J

   ›Reveal solutionSolution

   The key idea is that the self-induced emf gives the inductance, and the change in stored energy is $\frac{1}{2}L(I_f^2 - I_i^2)$. The result is 648 J, so option (A) is correct.

   Concept and Intuition

   The self-induced emf in a coil opposes the change in current (Lenz’s law). Its magnitude is $|\mathcal{E}| = L \frac{\Delta I}{\Delta t}$, where $L$ is the inductance. Once we find $L$, the energy stored in the magnetic field of an inductor is $U = \frac{1}{2} L I^2$. The change in energy when the current changes from $I_i$ to $I_f$ is simply $\Delta U = \frac{1}{2} L (I_f^2 - I_i^2)$. No integration or calculus is needed here because the current change is linear and the inductance is constant.

   Step-by-step solution

   1. Find the inductance $L$ from the given emf. The self-induced emf is given as 36 V. The current changes from 12 A to 24 A in 1 second, so $\Delta I = 24 - 12 = 12 \, \text{A}$ and $\Delta t = 1 \, \text{s}$. Using $|\mathcal{E}| = L \frac{\Delta I}{\Delta t}$:

   $36 = L \cdot \frac{12}{1} \quad \Rightarrow \quad L = \frac{36}{12} = 3 \, \text{H}.$

   2. Calculate the initial and final stored energy. Initial energy:

   $U_i = \frac{1}{2} L I_i^2 = \frac{1}{2} \cdot 3 \cdot (12)^2 = \frac{1}{2} \cdot 3 \cdot 144 = 216 \, \text{J}.$

   Final energy:

   $U_f = \frac{1}{2} L I_f^2 = \frac{1}{2} \cdot 3 \cdot (24)^2 = \frac{1}{2} \cdot 3 \cdot 576 = 864 \, \text{J}.$

   3. Find the change in stored energy.

   $\Delta U = U_f - U_i = 864 - 216 = 648 \, \text{J}.$

   Tip

   A shortcut: $\Delta U = \frac{1}{2} L (I_f^2 - I_i^2) = \frac{1}{2} \cdot 3 \cdot (576 - 144) = \frac{1}{2} \cdot 3 \cdot 432 = 648 \, \text{J}$. This avoids computing each energy separately.

   Watch out

   A common mistake is to use $\Delta U = \frac{1}{2} L (\Delta I)^2$ instead of the difference of squares. Here $\frac{1}{2} \cdot 3 \cdot (12)^2 = 216 \, \text{J}$, which is not the correct change. Always use $I_f^2 - I_i^2$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

7. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The process of the loss of strength of a signal while propagating through a medium is (A) damping (B) attenuation (C) amplification (D) modulation

   ›Reveal solutionSolution

   The loss of signal strength during propagation is called attenuation; the correct answer is (B).

   The key concept here is signal degradation in transmission. When a signal travels through any medium (like a cable, optical fiber, or air), it inevitably loses energy due to resistance, scattering, or absorption. This reduction in amplitude or power is a fundamental phenomenon in physics and engineering.

   Let’s clarify each option:

   1. Damping – This usually refers to the reduction of oscillation amplitude in mechanical or electrical systems over time (e.g., a swinging pendulum slowing down). While related to energy loss, it’s not the standard term for signal strength loss in a propagation medium.

   2. Attenuation – This is the precise, widely used term for the gradual loss of signal strength (power or amplitude) as it travels through a medium. It’s measured in decibels (dB) per unit distance. For example, in fiber optics, attenuation is caused by absorption and scattering of light.

   3. Amplification – This is the opposite of loss; it increases signal strength. Clearly not the answer.

   4. Modulation – This is the process of varying a carrier wave’s properties (amplitude, frequency, or phase) to encode information. It has nothing to do with signal strength loss.

   Watch out

   A common pitfall is confusing "damping" with "attenuation." Damping is more specific to oscillatory systems (like a spring or an RLC circuit), while attenuation is the general term for signal power loss in any transmission medium.

   Tip

   Think of attenuation like a flashlight beam fading as it goes farther into fog — the light is still there, but it’s weaker. That’s exactly what happens to radio waves, electrical signals, or light in a cable.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
   Q.In LCR series circuit, the current amplitude becomes $\dfrac{1}{\sqrt{2}}$ times its maximum value at frequencies $212~\text{rad s}^{-1}$ and $232~\text{rad s}^{-1}$. If the value of R in the circuit is $5~\Omega$, then the value of L is (A) $20~\text{mH}$ (B) $250~\text{mH}$ (C) $10~\text{mH}$ (D) $5~\text{mH}$

   ›Reveal solutionSolution

   The key idea is that the two given frequencies are the half-power points of a series RLC circuit, where the current drops to $1/\sqrt{2}$ of its peak. The bandwidth $\Delta \omega = \omega_2 - \omega_1$ equals $R/L$, giving $L = 5 / 20 = 0.25\ \text{H} = 250\ \text{mH}$. The correct option is (B).

   ---

   Concept & Intuition

   In a series LCR circuit, the current amplitude is maximum at the resonant frequency $\omega_0$, where the impedance is purely resistive ($Z = R$). At frequencies slightly above or below resonance, the impedance increases because the net reactance $|\omega L - 1/(\omega C)|$ becomes nonzero, so the current drops. The frequencies at which the current falls to $1/\sqrt{2}$ of its maximum value are called the half-power points. The difference between these two frequencies is the bandwidth $\Delta \omega$, and for a series RLC circuit, a classic result is:

   $\Delta \omega = \frac{R}{L}$

   This is derived from the condition that at the half-power points, the magnitude of the reactance equals the resistance: $|\omega L - 1/(\omega C)| = R$. For a narrow bandwidth (high $Q$), the two solutions are approximately $\omega_0 \pm R/(2L)$, so their difference is $R/L$. Even without the approximation, the exact difference between the two frequencies where $|Z| = \sqrt{2}R$ is indeed $R/L$ — a beautiful, exact result.

   Thus, given the two frequencies and $R$, we can directly find $L$.

   ---

   Step-by-step solution

   1. Identify the given data The two frequencies (in rad/s) are:

   $\omega_1 = 212,\quad \omega_2 = 232$

   Resistance $R = 5\ \Omega$.

   These are the frequencies where current amplitude is $1/\sqrt{2}$ times its maximum — i.e., the half-power points.

   2. Compute the bandwidth The bandwidth is simply the difference:

   $\Delta \omega = \omega_2 - \omega_1 = 232 - 212 = 20\ \text{rad/s}$

   3. Apply the series RLC bandwidth formula For a series RLC circuit, the exact bandwidth between the two half-power frequencies is:

   $\Delta \omega = \frac{R}{L}$

   This is independent of the capacitance $C$ and holds exactly, not just approximately.

   4. Solve for $L$ Rearranging:

   $L = \frac{R}{\Delta \omega} = \frac{5}{20} = 0.25\ \text{H}$

   Convert to millihenries:

   $L = 0.25 \times 1000 = 250\ \text{mH}$

   5. Match with the options The options are: (A) 20 mH, (B) 250 mH, (C) 10 mH, (D) 5 mH. Clearly, $250\ \text{mH}$ corresponds to option (B).

   Tip

   A common pitfall is to mistakenly use the formula $\Delta \omega = R/(2L)$ (which is the half-bandwidth from resonance to one half-power point). Remember: the full bandwidth between the two half-power frequencies is $R/L$.

   Watch out

   Do not confuse angular frequency (rad/s) with ordinary frequency (Hz). Here the units are already rad/s, so no conversion is needed. If frequencies were given in Hz, you would first convert: $\omega = 2\pi f$.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.An alternating emf given by the equation $E = 200 \sin (50 \pi t)$ (where E is in volts and t is in seconds) is applied across a series combination of an inductor and a resistor having inductive reactance $40~\Omega$ and resistance $30~\Omega$ respectively. At time $t = 1~s$, the power dissipated by the resistor is close to $(\cos 53^\circ = 0.6)$ (A) $480~\text{W}$ (B) $240~\text{W}$ (C) $173~\text{W}$ (D) $307~\text{W}$

   ›Reveal solutionSolution

   $i(t) = 4\sin(50\pi t - 53^\circ)$; at $t = 1$ s this is $-3.2$ A, so the instantaneous power in the resistor is $i^2R \approx 307$ W.

   Concept. In a series LR circuit the current lags the applied emf by $\varphi$ with $\tan\varphi = X_L/R$; the instantaneous power dissipated in the resistor is $i^2R$ evaluated at that instant.

   Step 1 — impedance and phase.

   $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = 50\,\Omega, \qquad \tan\varphi = \frac{40}{30} = \frac{4}{3} \Rightarrow \varphi = 53^\circ.$

   Step 2 — current expression. Peak current $I_0 = \dfrac{E_0}{Z} = \dfrac{200}{50} = 4$ A, and the current lags:

   $i(t) = 4\sin(50\pi t - 53^\circ).$

   Step 3 — evaluate at $t = 1$ s. $50\pi\,\text{rad} = 25\times 2\pi$, i.e. exactly 25 full cycles, so $\sin(50\pi - 53^\circ) = -\sin 53^\circ = -0.8$:

   $i = 4 \times (-0.8) = -3.2\ \text{A}.$

   Step 4 — instantaneous power in R.

   $P = i^2 R = (3.2)^2 \times 30 = 10.24 \times 30 = 307.2\ \text{W} \approx 307\ \text{W}.$

   ✓Final answer

   The power dissipated by the resistor at $t = 1$ s is close to $307$ W — option (D).

   ANSWER: D

10. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
    Q.A capacitor of capacitance $100\ \mu \mathrm{F}$ and a coil of resistance $20\ \Omega$ and inductance $12.5\ \mathrm{mH}$ are connected in series with a $220\ \mathrm{V},\ \dfrac{200}{\pi}\ \mathrm{Hz}$ AC source. The maximum value of instantaneous current in the circuit is (A) 20 A (B) 10 A (C) 11 A (D) 15 A

    ›Reveal solutionSolution

    The circuit is a series RLC driven by an AC source. The maximum instantaneous current is the peak current $I_0 = V_0 / Z$, where $Z$ is the impedance. The answer is 11 A.

    The key idea here is that in a series AC circuit, the current is not simply $V/R$ because the inductor and capacitor each oppose the flow with a frequency-dependent reactance. The total opposition is the impedance $Z$, which combines resistance $R$, inductive reactance $X_L$, and capacitive reactance $X_C$. The maximum (peak) current occurs when the instantaneous voltage is at its peak, and Ohm's law for AC gives $I_0 = V_0 / Z$.

    Let’s work through the numbers step by step.

    1. Find the angular frequency $\omega$.

       The source frequency is $f = \frac{200}{\pi}\ \mathrm{Hz}$.

       Angular frequency $\omega = 2\pi f = 2\pi \cdot \frac{200}{\pi} = 400\ \mathrm{rad/s}$.

    2. Compute the inductive reactance $X_L$.

       $X_L = \omega L = 400 \times 12.5 \times 10^{-3} = 5\ \Omega$.

    3. Compute the capacitive reactance $X_C$.

       $X_C = \frac{1}{\omega C} = \frac{1}{400 \times 100 \times 10^{-6}} = \frac{1}{0.04} = 25\ \Omega$.

    4. Find the net reactance $X$.

       Since $X_L$ and $X_C$ oppose each other, $X = X_L - X_C = 5 - 25 = -20\ \Omega$.

       The negative sign just means the circuit is capacitive overall; impedance uses the magnitude.

    5. Calculate the impedance $Z$.

       $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + (-20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2}\ \Omega$.

    6. Determine the peak voltage $V_0$.

       The given $220\ \mathrm{V}$ is the RMS voltage. For a sinusoidal source, $V_0 = V_{\text{rms}} \times \sqrt{2} = 220\sqrt{2}\ \mathrm{V}$.

    7. Compute the peak current $I_0$.

       $I_0 = \frac{V_0}{Z} = \frac{220\sqrt{2}}{20\sqrt{2}} = \frac{220}{20} = 11\ \mathrm{A}$.

    Watch out

    A common mistake is to forget that the given voltage is RMS, not peak. Using $220\ \mathrm{V}$ directly as $V_0$ would give $I_0 = 220 / (20\sqrt{2}) \approx 7.78\ \mathrm{A}$, which is not among the options. Always check whether the problem states RMS or peak.

    Tip

    Notice that the $\sqrt{2}$ factors cancel neatly here because both $V_0$ and $Z$ contain $\sqrt{2}$ terms. This is a handy shortcut: if the impedance has a $\sqrt{2}$ factor, the peak current often simplifies to a clean number.

    ✓Final answer

    The maximum value of instantaneous current is $\boxed{11\ \mathrm{A}}$, which corresponds to option (C).

11. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
    Q.A cell phone charger consists of a stepdown transformer to convert AC voltage of 120 V to AC voltage of 5.0 V. If the secondary coil contains 30 turns and charger supplies 720 mA, calculate the current in the primary coil. (A) 12 mA (B) 30 mA (C) 85 mA (D) 160 mA

    ›Reveal solutionSolution

    For an ideal stepdown transformer, the ratio of currents is the inverse of the turns ratio. Using $I_p / I_s = N_s / N_p$, we find the primary current is 30 mA, which corresponds to option (B).

    The key idea here is the transformer power balance: in an ideal transformer (no losses), the power delivered to the primary equals the power delivered by the secondary. Since power is voltage times current, this gives $V_p I_p = V_s I_s$. Rearranging, the current ratio is the inverse of the voltage ratio, which is also the inverse of the turns ratio. So we don’t need to know the primary voltage directly — we can use the turns ratio instead.

    1. Identify the known quantities

       • Secondary voltage $V_s = 5.0\ \text{V}$
       • Primary voltage $V_p = 120\ \text{V}$
       • Secondary turns $N_s = 30$
       • Secondary current $I_s = 720\ \text{mA} = 0.720\ \text{A}$ We need the primary current $I_p$.

    2. Use the turns ratio to find primary turns

       For a transformer, $\frac{V_p}{V_s} = \frac{N_p}{N_s}$.

       So $N_p = N_s \cdot \frac{V_p}{V_s} = 30 \cdot \frac{120}{5.0} = 30 \cdot 24 = 720$ turns.

    3. Apply the current ratio

       For an ideal transformer, $\frac{I_p}{I_s} = \frac{N_s}{N_p}$.

       Therefore, $I_p = I_s \cdot \frac{N_s}{N_p} = 0.720 \cdot \frac{30}{720}$.

       Simplify: $\frac{30}{720} = \frac{1}{24}$, so $I_p = 0.720 \cdot \frac{1}{24} = 0.030\ \text{A} = 30\ \text{mA}$.

    Tip

    You can skip finding $N_p$ entirely: since $\frac{V_p}{V_s} = \frac{N_p}{N_s}$, the current ratio becomes $\frac{I_p}{I_s} = \frac{V_s}{V_p}$. So directly $I_p = 0.720 \cdot \frac{5.0}{120} = 0.720 \cdot \frac{1}{24} = 30\ \text{mA}$. This is faster and avoids extra steps.

    Watch out

    A common mistake is to invert the ratio incorrectly — students sometimes multiply by $\frac{N_p}{N_s}$ instead of $\frac{N_s}{N_p}$. Remember: higher voltage side has lower current, so the primary (120 V) should have a much smaller current than the secondary (5 V). 30 mA is indeed much smaller than 720 mA, which checks out.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
    Q.A light bulb is rated at 110 W for a 220 V supply. The resistance of the bulb is (A) $440~\Omega$ (B) $220~\Omega$ (C) $55~\Omega$ (D) $110~\Omega$

    ›Reveal solutionSolution

    Using the power rating and voltage, we apply $P = V^2 / R$ to find the resistance. The bulb’s resistance is $440~\Omega$, so the correct option is (A).

    The key concept here is the relationship between electrical power, voltage, and resistance for a device operating under steady conditions. A light bulb’s rating (110 W at 220 V) tells us the power it consumes when connected to a 220 V supply. Since the bulb is a resistive load (it heats up and glows), we can use the formula $P = V^2 / R$, which comes from combining Ohm’s law ($V = IR$) with the power formula $P = VI$. This avoids needing the current, which isn’t given.

    Why this works: The bulb’s resistance is assumed constant (though in reality it changes with temperature, the problem treats it as fixed for calculation). The formula $P = V^2 / R$ directly links the three quantities, so rearranging gives $R = V^2 / P$.

    Now, step by step:

    1. Identify the given values:

       Power $P = 110$ W, voltage $V = 220$ V.

    2. Recall the power formula for a resistor:

       $P = \frac{V^2}{R}$. This is derived from $P = VI$ and $V = IR$, eliminating $I$.

    3. Rearrange to solve for resistance $R$:

       Multiply both sides by $R$: $P R = V^2$.

       Then divide by $P$: $R = \frac{V^2}{P}$.

    4. Substitute the numbers:

       $R = \frac{(220)^2}{110} = \frac{48400}{110}$.

    5. Simplify:

       $48400 \div 110 = 440$. So $R = 440~\Omega$.

    Tip

    A common shortcut: $220^2 = 48400$, and dividing by 110 is the same as multiplying by $1/110$. Notice $48400/110 = 4840/11 = 440$ — you can cancel a factor of 10 first.

    Watch out

    A classic mistake is to use $P = I^2 R$ or $P = VI$ without finding current first, leading to errors. Always pick the formula that uses the two given quantities directly — here $P$ and $V$ — to avoid extra steps.

    Thus, the bulb’s resistance is $440~\Omega$, which matches option (A).

    ✓Final answer

    The correct option is (A).

    ANSWER: A