Q.An electrical device draws $2\ \text{kW}$ power from AC mains (voltage $223\ \text{V}$ (rms) $= \sqrt{50000}\ \text{V}$). The current differs (lags) in phase by $\phi$ $\left(\tan\phi = -\dfrac{3}{4}\right)$ as compared to voltage. Find

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.In the circuit given below, each of three resistors of $4\ \Omega$ can have a maximum power of 20 W (otherwise, it will melt) (diagram: from terminal A, two $4\ \Omega$ resistors are connected in parallel between A and a middle node; the middle node then connects through a third $4\ \Omega$ resistor in series to terminal B). The maximum power the whole circuit can take is (A) 30 W (B) 40 W (C) 20 W (D) 10 W

   ›Reveal solutionSolution

   This tests identifying which resistor in a series-parallel network reaches its power limit first. Answer: 30 W.

   Concept and Intuition

   All the current supplied to the circuit must pass through the single series resistor, while it splits (here, equally, since the two parallel resistors are identical) between the two parallel ones. So the series resistor carries the most current of any single resistor and will hit its power (hence current) limit before the parallel ones do. The overall power limit of the circuit is set by whichever resistor melts first — the series one.

   Step-by-Step Solution

   1. Parallel combination of the two 4 Ω resistors: $R_p = \dfrac{4\times4}{4+4} = 2\ \Omega$.
   2. Total resistance: $R_{total} = R_p + 4 = 2+4 = 6\ \Omega$.
   3. Let $I$ be the total current (equal to the current through the series resistor). Its power is $P_{series}=I^2(4)$. Setting this to the 20 W limit: $I^2 = 5\ A^2$. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.An alternating supply of 225 V is applied across a circuit with resistance $20\Omega$ and impedance of $45\Omega$. The power dissipated in the circuit is (A) 500 W (B) 1000 W (C) 550 W (D) 2100 W

   ›Reveal solutionSolution

   Average power in an AC circuit is dissipated only in the resistive part; compute rms current from $V/Z$, then use $P=I^2R$.

   Concept and Intuition

   In any series AC circuit with resistance, inductance and/or capacitance, energy is dissipated (as heat) only in the resistor — inductors and capacitors store and return energy over a cycle with zero net dissipation. The rms current through the series circuit is set by the total impedance $Z$ (which already accounts for the reactive elements), and once we have that current, the average power is simply $I_{rms}^2R$.

   Step-by-Step Solution

   1. rms current: $I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{225}{45} = 5\,\text{A}$.
   2. Average power dissipated (only in $R$): $P = I_{rms}^2 R = (5)^2 \times 20 = 25 \times 20 = 500\,\text{W}$.

   Common Mistakes …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.If power dissipated in the $9\ \Omega$ resistor in the circuit shown is 36 W, the potential difference across the $2\ \Omega$ resistor is [FIGURE] (a circuit with two parallel branches, a $9\ \Omega$ resistor carrying current $i_1$ in the top branch and a $6\ \Omega$ resistor in the bottom branch, both branches joined at both ends; this parallel combination is connected in series with a battery of emf V and a $2\ \Omega$ resistor, with total current $i$ flowing from the battery) (A) 2 volt (B) 4 volt (C) 8 volt (D) 10 volt

   ›Reveal solutionSolution

   From $P=i_1^2 R$ get $i_1=2\,\text{A}$ in the $9\,\Omega$; the parallel $6\,\Omega$ carries $3\,\text{A}$, so total $i=5\,\text{A}$ and the $2\,\Omega$ drops $10\,\text{V}$.

   Current in the $9\,\Omega$ branch. Power dissipated is $36\,\text{W}$:

   $P=i_1^2\,R\;\Rightarrow\;36=i_1^2(9)\;\Rightarrow\;i_1^2=4\;\Rightarrow\;i_1=2\,\text{A}.$

   Voltage across the parallel section. The voltage across the $9\,\Omega$ resistor equals that across the parallel $6\,\Omega$ resistor:

   $V_{\parallel}=i_1\times 9=2\times 9=18\,\text{V}.$

   Current in the $6\,\Omega$ branch:

   $i_2=\frac{18}{6}=3\,\text{A}.$ …

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4. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A resistor of resistance 30 Ω and a capacitor of reactance 40 Ω are connected in series to an ac supply. If the rms current through the resistor is 2 mA, then the wattless current is (A) zero (B) 2 mA (C) 1.2 mA (D) 1.6 mA

   ›Reveal solutionSolution

   The wattless (idle) current component is the part of the total current that is out of phase with the voltage and does no net work — found using $I\sin\phi$ for a series RC circuit.

   Concept and Intuition

   In an AC circuit with both resistance and reactance, only the in-phase component of current ($I\cos\phi$) contributes to real power dissipation. The out-of-phase (wattless/reactive) component is $I\sin\phi$, where $\phi$ is the phase angle between voltage and current, determined by the ratio of reactance to impedance.

   Step-by-Step Solution

   1. Impedance of the series RC circuit: $Z=\sqrt{R^2+X_C^2}=\sqrt{30^2+40^2}=\sqrt{900+1600}=\sqrt{2500}=50\,\Omega$.
   2. $\sin\phi = \dfrac{X_C}{Z}=\dfrac{40}{50}=0.8$.
   3. Wattless current $=I\sin\phi = 2\,mA\times0.8=1.6\,mA$.

   Common Mistakes …

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5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The power dissipated by a uniform wire of resistance $100\ \Omega$ when a potential difference of 120 V is applied across its ends is (A) 122 W (B) 144 W (C) 160 W (D) 200 W

   ›Reveal solutionSolution

   Tests the power–resistance relation $P=V^2/R$ for a fixed resistor; answer is 144 W.

   Concept and Intuition

   For a resistor with a fixed potential difference $V$ across it, the power dissipated is entirely due to Joule heating, given by $P = VI = I^2R = \dfrac{V^2}{R}$. Since we are given $V$ and $R$ directly, the most direct form to use is $P = V^2/R$.

   Step-by-Step Solution

   1. Given: $R = 100\ \Omega$, $V = 120\ \text{V}$.
   2. Apply $P = \dfrac{V^2}{R} = \dfrac{120^2}{100} = \dfrac{14400}{100}$. …

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6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A wire of resistance 'R' is bent in the form of a circular loop. Two points on the circle separated by a quarter circumference are connected to a battery of emf 'E' and negligible internal resistance. The heat generated in the wire per second is (A) $\dfrac{E^2}{4R}$ (B) $\dfrac{16E^2}{3R}$ (C) $\dfrac{E^2}{R}$ (D) $\dfrac{2E^2}{3R}$

   ›Reveal solutionSolution

   The wire loop splits into two resistive arcs in parallel between the battery terminals; combine them and use $P=E^2/R_{eq}$.

   Concept and Intuition

   A uniform wire bent into a circle has resistance distributed uniformly along its length, so resistance of any arc is proportional to its arc length. When two points on the loop are tapped by a battery, current can flow to the other terminal via either arc — the two arcs are electrically in parallel, not in series, because both start and end at the same two nodes.

   Step-by-Step Solution

   1. Total loop resistance is $R$, uniformly distributed, so resistance is proportional to arc length.
   2. A quarter-circumference arc has resistance $R_1 = \dfrac{1}{4}R$; the remaining three-quarters arc has resistance $R_2 = \dfrac{3}{4}R$.
   3. These two arcs connect the same pair of terminals, so they are in parallel: $R_{eq}=\frac{R_1R_2}{R_1+R_2}=\frac{\left(\tfrac{R}{4}\right)\left(\tfrac{3R}{4}\right)}{\tfrac{R}{4}+\tfrac{3R}{4}}=\frac{\tfrac{3R^2}{16}}{R}=\frac{3R}{16}$ …

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7. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.If an electric bulb is rated at 50 W for a 220 V ac supply, then the resistance of the bulb and the peak voltage of the ac source are respectively (A) $968\ \Omega, 220\sqrt{2}\ \mathrm{V}$ (B) $484\ \Omega, 220\ \mathrm{V}$ (C) $968\ \Omega, 220\ \mathrm{V}$ (D) $484\ \Omega, 220\sqrt{2}\ \mathrm{V}$

   ›Reveal solutionSolution

   The bulb's resistance follows from $P=V_{rms}^2/R$, and the peak voltage is $\sqrt2$ times the rms (rated) voltage — giving $968\,\Omega$ and $220\sqrt2$ V.

   Concept and Intuition

   An AC bulb's power rating (50 W at 220 V) refers to the rms voltage, since power dissipation in a resistor depends on rms values: $P=\dfrac{V_{rms}^2}{R}$. Separately, any sinusoidal AC voltage has a peak value related to its rms value by $V_0=V_{rms}\sqrt2$, purely from the definition of rms for a sine wave — this is independent of the bulb's rating.

   Step-by-Step Solution

   1. Resistance: $R=\dfrac{V_{rms}^2}{P}=\dfrac{(220)^2}{50}=\dfrac{48400}{50}=968\ \Omega$.
   2. Peak (maximum) voltage of the 220 V (rms) AC supply: $V_0=\sqrt2\times V_{rms}=220\sqrt2\ \mathrm{V}$. …

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8. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.A lamp is rated at 240V, 60W. When in use the resistance of the filament of the lamp is 20 times that of cold filament. The resistance of the lamp when not in use is (A) 54 Ω (B) 60 Ω (C) 50 Ω (D) 48 Ω

   ›Reveal solutionSolution

   Rated power and voltage give the filament's hot resistance; dividing by 20 gives the cold resistance.

   Concept and Intuition

   A lamp's "rated" values (240 V, 60 W) describe its operating (hot) state, since that's when it's glowing and consuming that power. The cold filament (before switch-on) has much lower resistance because resistivity of the metal filament increases sharply with temperature. The problem gives us the ratio between the two, so we first get the hot resistance from $P=V^2/R$, then scale down.

   Step-by-Step Solution

   1. Hot resistance: $R_{hot} = \dfrac{V^2}{P} = \dfrac{240\times240}{60} = \dfrac{57600}{60} = 960\ \Omega$.
   2. Given $R_{hot} = 20\,R_{cold}$.
   3. $R_{cold} = \dfrac{R_{hot}}{20} = \dfrac{960}{20} = 48\ \Omega$. …

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9. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
   Q.A resistor of $50\,\Omega$, an inductor and a capacitor are connected in series to an ac source of peak voltage $200\sqrt{2} \, \mathrm{V}$. When the capacitor alone is removed from the circuit, the current lags the voltage by $37^\circ$ and when the inductor alone is removed from the circuit, the current leads the voltage by $37^\circ$. The power dissipated in the LCR circuit is (A) 400 W (B) 800 W (C) 200 W (D) 100 W

   ›Reveal solutionSolution

   Equal-magnitude lag and lead angles show $X_L = X_C$, meaning the LCR circuit is at resonance; power dissipated is simply $V_{rms}^2/R = 800$ W.

   Concept and Intuition

   Removing the capacitor leaves an RL circuit (current lags voltage); removing the inductor leaves an RC circuit (current leads voltage). If both phase angles have the same magnitude ($37^\circ$), it means $X_L$ and $X_C$ individually produce the same magnitude of phase shift relative to $R$ — i.e. $X_L = X_C$. That is exactly the resonance condition for the full LCR circuit, where the net reactance cancels and the circuit behaves purely resistively.

   Step-by-Step Solution

   1. RL circuit (C removed): $\tan 37^\circ = X_L/R \Rightarrow X_L = R\tan 37^\circ = 50 \times 0.75 = 37.5\,\Omega$.
   2. RC circuit (L removed): $\tan 37^\circ = X_C/R \Rightarrow X_C = 50 \times 0.75 = 37.5\,\Omega$. …

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10. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.In the circuit given below, if the bulb is to glow with maximum intensity, the value of 'R' is (neglect internal resistance of the cell) [FIGURE] (a circuit diagram with a 6V battery at the top; below it the circuit splits into a branch with a 3Ω resistor in series with a resistor R, and this joins down to a bulb marked 'W' rated 1.5 V, 0.45 W, which connects back to complete the loop with the battery) (A) 1.25 Ω (B) 4.5 Ω (C) 6 Ω (D) 8.5 Ω

    ›Reveal solutionSolution

    This tests using a bulb's rated voltage/power to find its rated current, then applying KVL and KCL to a series–parallel circuit to find the parallel resistor that makes the bulb operate exactly at its rated (maximum-intensity) point.

    Concept and Intuition

    A bulb glows with "maximum intensity" when it operates at its rated voltage and power — any brighter isn't achievable without exceeding its design rating. Here R and the bulb are in parallel (sharing the same voltage), and this combination is in series with the fixed 3Ω resistor and the ideal 6V battery. For the bulb to sit exactly at 1.5 V, the total current drawn from the battery is fixed by KVL, and R must draw whatever current is "left over" after the bulb takes its rated share.

    Step-by-Step Solution

    1. Bulb rating: 1.5 V, 0.45 W ⇒ rated current $I_{bulb}=\dfrac{P}{V}=\dfrac{0.45}{1.5}=0.3$ A.
    2. For maximum (rated) intensity, the voltage across the parallel section (bulb and R) must equal 1.5 V.
    3. Apply KVL around the loop: EMF = drop across 3Ω + drop across parallel section: $6=3I_{total}+1.5$.
    4. Solving: $3I_{total}=4.5\Rightarrow I_{total}=1.5$ A. …

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11. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.An inductor and a resistor of 25 Ω are connected in series to an ac source of voltage 100 sin (100 πt) volt. If the impedance of the circuit is 50 Ω, the average power dissipated per cycle in the circuit is (A) 10 W (B) 25 W (C) 50 W (D) 100 W

    ›Reveal solutionSolution

    In a series LR AC circuit, average power is dissipated only in the resistor: $P = I_{rms}^2 R = 50$ W here.

    Concept and Intuition

    An ideal inductor stores and releases energy over a cycle without net dissipation — its current and voltage are 90° out of phase, so the average power delivered to it over a full cycle is zero. All the average power in an LR series circuit is therefore dissipated in the resistor, and it can be computed either from $I_{rms}^2 R$ or from $V_{rms}I_{rms}\cos\phi$ (where $\cos\phi = R/Z$ is the power factor) — both give the same result.

    Step-by-Step Solution

    1. Peak voltage from $V = 100\sin(100\pi t)$: $V_0 = 100$ V, so $V_{rms} = V_0/\sqrt2 = 100/\sqrt2$ V.
    2. Given impedance $Z = 50\ \Omega$, find $I_{rms} = V_{rms}/Z = \dfrac{100/\sqrt2}{50} = \dfrac{2}{\sqrt2} = \sqrt2$ A.
    3. Average power is dissipated only in the resistor: $P = I_{rms}^2 R = (\sqrt2)^2 \times 25 = 2\times25 = 50$ W. …

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12. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.10 W is to be delivered to a device via a wire having resistance 2 Ω. If 20 V is the voltage across the device, the power wasted in the process is (A) 3 W (B) 2 W (C) 0.5 W (D) 1.5 W

    ›Reveal solutionSolution

    Finding the current from the device's power and voltage, then applying $I^2R$ to the wire, gives 0.5 W wasted.

    Concept and Intuition

    The same current flows through both the device and the connecting wire (they're in series). Knowing the power delivered to and the voltage across the device lets us find that current, which we then use to find the ohmic loss in the wire.

    Step-by-Step Solution

    1. Current supplied to device: $I=\dfrac{P}{V}=\dfrac{10\ \text{W}}{20\ \text{V}}=0.5$ A.
    2. This same current flows through the 2 Ω wire. …

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