Q.Power $P$ is to be delivered to a device via transmission cables having resistance $R_C$. If $V$ is the voltage across $R$ and $I$ the current through it, find the power wasted and how can it be reduced.

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The value of 'R' in the given circuit is [circuit: a source of 25 V driving a current of 0.5 A connects in series to a resistor R; this then connects to three resistors of $10\ \Omega$, $10\ \Omega$, and $20\ \Omega$ arranged in parallel, which return to the source] (A) $46\ \Omega$ (B) $10\ \Omega$ (C) $28\ \Omega$ (D) $50\ \Omega$

   ›Reveal solutionSolution

   This tests series-parallel circuit reduction; solving for R in series with a 3-resistor parallel bank gives $R = 46\ \Omega$.

   Concept and Intuition

   When a resistor R is in series with a parallel combination, the total voltage across the whole loop equals the current times the total effective resistance (R plus the parallel bank's equivalent resistance). So the first job is to collapse the parallel network into one equivalent resistor, then apply Ohm's law to the whole series loop.

   Step-by-Step Solution

   1. Combine the parallel resistors: $\dfrac{1}{R_p} = \dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{20} = \dfrac{2}{20}+\dfrac{2}{20}+\dfrac{1}{20} = \dfrac{5}{20} = \dfrac{1}{4}$, so $R_p = 4\ \Omega$.
   2. The source (25 V) drives a current of 0.5 A through the series combination of $R$ and $R_p$: $25 = 0.5\times(R+R_p)$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Two batteries of emf's 4V and 8V with internal resistances 1Ω and 2Ω are connected in a circuit with a resistance of 9Ω as shown in the figure. The current and potential difference between the points P and Q are (Figure: point P connects through a resistor $r_1=1Ω$ to the positive terminal of a 4V battery; the negative terminal of the 4V battery connects to the positive terminal of an 8V battery; the negative terminal of the 8V battery connects through a resistor $r_2=2Ω$ to point Q — this is the top branch from P to Q. A separate bottom branch directly connects P to Q through a 9Ω resistor.) (A) $\dfrac{1}{3}A$ and 3V (B) $\dfrac{1}{6}A$ and 4V (C) $\dfrac{1}{9}A$ and 9V (D) $\dfrac{1}{2}A$ and 12V

   ›Reveal solutionSolution

   This is a single loop (two cells with internal resistance opposing each other, external 9Ω resistor closes the loop). Net EMF = 8−4 = 4 V, total resistance = 1+2+9 = 12Ω, giving $I=\dfrac13$A and $V_{PQ}=3$V.

   Concept and Intuition

   When two cells and a resistor form a single loop (only two nodes, P and Q, joined by exactly two branches), the same current flows through every element — there is no branching to apply the current-divider rule to. The two EMFs combine algebraically depending on how their polarities are arranged around the loop: if they drive current in the same rotational sense they add, if in opposite senses they subtract. Since P and Q are bridged directly only by the 9Ω resistor, the potential difference across P and Q must always equal (loop current) × (9Ω) — a structural fact independent of how the EMFs combine.

   Step-by-Step Solution

   1. The circuit is one single loop: cell (4V, 1Ω) and cell (8V, 2Ω) in the top branch, resistor 9Ω forming the bottom branch, both bridging the same nodes P and Q.
   2. The two cells oppose each other around the loop, so the net EMF driving the current is $8-4=4$ V (the larger cell wins).
   3. Total resistance around the loop (all elements are in series in a single loop): $R = r_1+r_2+9 = 1+2+9 = 12\ \Omega$.
   4. Current: $I = \dfrac{4}{12} = \dfrac13\text{ A}$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A potentiometer wire is 10m long and has a resistance of $18\ \Omega$. It is connected to a battery of emf 5V and internal resistance $2\ \Omega$. Calculate the potential gradient along the wire. (A) $0.65\ Vm^{-1}$ (B) $0.45\ Vm^{-1}$ (C) $0.35\ Vm^{-1}$ (D) $0.25\ Vm^{-1}$

   ›Reveal solutionSolution

   Find the current through the potentiometer wire using the full circuit (wire + internal resistance), then divide the voltage across the wire by its length to get the potential gradient — the answer is $0.45\ Vm^{-1}$.

   Concept and Intuition

   The potential gradient of a potentiometer wire is simply the voltage drop per unit length along the wire, $k = V_{wire}/L$. To find $V_{wire}$, treat the whole potentiometer circuit (battery of EMF $\varepsilon$, internal resistance $r$, wire resistance $R$) as a simple series loop and use Ohm's law to get the current, then apply that current to just the wire's resistance to get the voltage actually developed across it (the internal resistance "eats" some of the EMF and never appears along the wire).

   Step-by-Step Solution

   1. Total circuit resistance: $R + r = 18 + 2 = 20\ \Omega$.
   2. Current drawn from the battery: $I = \dfrac{\varepsilon}{R+r} = \dfrac{5}{20} = 0.25\ \text{A}$. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.When the current through a battery is 5A, the potential difference across its terminals is 9V and when the current through that battery is 3A, the potential difference across its terminals is 12V. If the same battery is connected to an external resistor of resistance $4\Omega$, then the current through the $4\Omega$ resistor is, (A) 2A (B) 3A (C) 4A (D) 5A

   ›Reveal solutionSolution

   This tests the terminal-voltage equation of a real battery, $V=\varepsilon - Ir$. Two (I, V) readings let us solve for the unknown $\varepsilon$ and internal resistance $r$; the answer is 3 A.

   Concept and Intuition

   A real battery's terminal voltage drops below its EMF because current flowing through its own internal resistance causes a voltage drop: $V = \varepsilon - Ir$. Given two different external loads (hence two different currents), we get two linear equations in the two unknowns $\varepsilon$ and $r$.

   Step-by-Step Solution

   1. First condition: $I_1 = 5\text{A}, V_1 = 9\text{V} \Rightarrow \varepsilon = 9 + 5r$.
   2. Second condition: $I_2 = 3\text{A}, V_2 = 12\text{V} \Rightarrow \varepsilon = 12 + 3r$.
   3. Equating: $9 + 5r = 12 + 3r \Rightarrow 2r = 3 \Rightarrow r = 1.5\ \Omega$.
   4. Then $\varepsilon = 9 + 5(1.5) = 16.5$ V (check: $12+3(1.5)=16.5$ ✓). …

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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.In a potentiometer experiment, the potentiometer wire of length 4 m and resistance 20 $\Omega$ is connected in series with an external resistor of resistance 979 $\Omega$ and a cell of internal resistance 1 $\Omega$. If the emf of the cell is 1.2 V, then the length of the wire between two points where the potential difference is 12 mV is (A) 2 m (B) 1.5 m (C) 2.5 m (D) 3 m

   ›Reveal solutionSolution

   A potentiometer problem: find the primary-circuit current, convert to a potential gradient (volts per metre along the wire), then use that gradient to find the length corresponding to a given potential difference. Answer: 2 m.

   Concept and Intuition

   In a potentiometer, a steady current flows through the uniform wire from a driver cell circuit. Because the wire has uniform cross-section and resistivity, the potential drop is proportional to length — this is the 'potential gradient' $k$ (volts/metre). Any measured potential difference along a segment of length $\ell$ then satisfies $V = k\ell$.

   Step-by-Step Solution

   1. Total resistance in the primary loop: wire ($20\ \Omega$) + external resistor ($979\ \Omega$) + cell's internal resistance ($1\ \Omega$) $= 1000\ \Omega$.
   2. Current through the wire: $I = \dfrac{\varepsilon}{R_{total}} = \dfrac{1.2\text{V}}{1000\ \Omega} = 1.2\times10^{-3}$ A.
   3. Resistance per unit length of the potentiometer wire: $\dfrac{20\ \Omega}{4\text{ m}} = 5\ \Omega/\text{m}$. …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The balancing length of a potentiometer is at 120 cm. On shunting the cell with a resistance of 4 ohm, the balancing point shifts to a length of 60 cm. The internal resistance of the cell is (A) 7 ohm (B) 12 ohm (C) 3 ohm (D) 4 ohm

   ›Reveal solutionSolution

   Comparing the potentiometer balance length for the cell's EMF (120 cm) against its terminal voltage under a 4Ω shunt (60 cm) gives the cell's internal resistance via $r=R(l_1/l_2-1)=4\ \Omega$.

   Concept and Intuition

   A potentiometer measures potential difference by balance length (length $\propto$ voltage, for a fixed potential gradient along the wire). With the cell in open circuit, the potentiometer balances its full EMF at length $l_1$. When a known resistance $R$ is connected across the cell (shunting it), current flows and the cell's terminal voltage drops below its EMF due to the voltage lost across its own internal resistance $r$; the new balance length $l_2$ measures this lower terminal voltage. Comparing the two balance lengths (which are proportional to EMF and terminal voltage respectively) lets us extract $r$.

   Step-by-Step Solution

   1. EMF $\varepsilon \propto l_1 = 120$ cm (open circuit balance).
   2. Terminal voltage $V \propto l_2 = 60$ cm (balance with shunt $R=4\ \Omega$ connected). …

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7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.When an external resistor of resistance $18\ \Omega$ is connected to a cell, the current drawn from the cell is I. if another $18\ \Omega$ resistor is connected parallel to the first resistor, the current drawn from the cell increases by 90%, then the internal resistance of the cell is (A) $2\ \Omega$ (B) $0.5\ \Omega$ (C) $1.5\ \Omega$ (D) $1\ \Omega$

   ›Reveal solutionSolution

   Use $I=\varepsilon/(R_{ext}+r)$ before and after halving the external resistance (two equal resistors in parallel), and solve for $r$ from the stated 90% current increase. Answer: $1\ \Omega$.

   Concept and Intuition

   A cell of EMF $\varepsilon$ and internal resistance $r$ driving an external resistor $R$ delivers current $I = \varepsilon/(R+r)$. Adding an identical resistor in parallel halves the external resistance (two equal $18\,\Omega$ resistors in parallel give $9\,\Omega$), which increases current — but by how much depends on how large $r$ is relative to $R$. The stated 90% increase pins down $r$.

   Step-by-Step Solution

   1. Before: $I = \dfrac{\varepsilon}{18+r}$.
   2. After adding a parallel $18\,\Omega$: external resistance $= \dfrac{18\times18}{18+18} = 9\,\Omega$, and new current $I' = \dfrac{\varepsilon}{9+r}$. …

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8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.A 48 V battery is supplying a current 12 A when connected to an external resistor. If the efficiency of the battery at this current is 75 %, then the internal resistance of the battery is (A) 3 $\Omega$ (B) 1.5 $\Omega$ (C) 2.5 $\Omega$ (D) 1 $\Omega$

   ›Reveal solutionSolution

   Battery "efficiency" is the fraction of its EMF that actually appears across the external circuit; working out the internal voltage drop from the given 75% efficiency gives an internal resistance of 1 Ω.

   Concept and Intuition

   A real battery has internal resistance $r$, so when it drives a current $I$, part of its EMF is "used up" driving current through its own internal resistance ($Ir$), and only the rest appears as useful terminal voltage $V$. Efficiency is defined as $\eta = \dfrac{\text{useful output power}}{\text{total power generated}} = \dfrac{VI}{\text{EMF}\times I} = \dfrac{V}{\text{EMF}}$.

   Step-by-Step Solution

   1. EMF $=48$ V, efficiency $\eta=75\%=0.75$.
   2. Terminal voltage: $V = \eta\times\text{EMF} = 0.75\times48=36$ V. …

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9. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.Two cells of each emf $1.5\ \mathrm{V}$ and internal resistance $1\ \Omega$ are first connected in series to an external resistance, R and then the cells are connected in parallel to the same external resistance, R. If the ratio of the potential differences across the resistor R in the two cases is $4:3$, then the value of R is (A) $5.5\ \Omega$ (B) $4.5\ \Omega$ (C) $3.5\ \Omega$ (D) $2.5\ \Omega$

   ›Reveal solutionSolution

   Comparing the terminal voltage across $R$ for two identical cells in series versus in parallel gives an equation in $R$; solving the given $4:3$ ratio yields $R = 2.5\,\Omega$.

   Concept and Intuition

   Two identical cells (emf $\varepsilon$, internal resistance $r$) in series behave like a single cell of emf $2\varepsilon$ and internal resistance $2r$. The same two cells in parallel behave like a single cell of the same emf $\varepsilon$ but internal resistance $r/2$. The voltage delivered to an external resistor $R$ in each configuration is found from a simple series-circuit current calculation.

   Step-by-Step Solution

   1. Series case: total emf $=2\varepsilon = 3\,\mathrm{V}$, total internal resistance $=2r=2\,\Omega$. Current $I_s = \dfrac{2\varepsilon}{R+2r} = \dfrac{3}{R+2}$. Voltage across $R$: $V_s = I_s R = \dfrac{3R}{R+2}$.
   2. Parallel case: emf $=\varepsilon=1.5\,\mathrm{V}$, internal resistance $=\dfrac{r}{2}=0.5\,\Omega$. Current $I_p = \dfrac{\varepsilon}{R+r/2} = \dfrac{1.5}{R+0.5}$. Voltage across $R$: $V_p = I_p R = \dfrac{1.5R}{R+0.5}$. …

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10. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.If a generator of emf 440 V and internal resistance $400\Omega$ is connected to an external resistance of $4000\Omega$, then the potential difference across the external resistance is (A) 220 V (B) 440 V (C) 200 V (D) 400 V

    ›Reveal solutionSolution

    A simple series circuit calculation: find the current using total resistance, then the terminal voltage across the external resistor — 400 V.

    Concept and Intuition

    A real source has internal resistance, so the terminal voltage across an external load is always less than the EMF, dropping by $Ir$ across the internal resistance.

    Step-by-Step Solution

    1. Total resistance in the circuit: $R+r = 4000+400=4400\,\Omega$.
    2. Current: $I=\dfrac{\varepsilon}{R+r}=\dfrac{440}{4400}=0.1$ A.
    3. Potential difference across the external resistance: $V=IR=0.1\times4000=400$ V.

    Common Mistakes …

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11. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.A dc supply of 160 V is used to charge a battery of emf 10 V and internal resistance $1\ \Omega$ by connecting a series resistance of $24\ \Omega$. The terminal voltage of the battery during charging is (A) 8 V (B) 12 V (C) 16 V (D) 4 V

    ›Reveal solutionSolution

    This tests the terminal-voltage relation for a battery being charged: current is forced backward through it, so the terminal voltage exceeds the emf by $Ir$ — here 16 V.

    Concept and Intuition

    When a battery is being charged, an external, higher-voltage source drives current into the battery against its own emf. Because the current direction inside the battery is reversed relative to normal discharge, the terminal voltage is now the emf plus the internal-resistance drop, $V=\varepsilon+Ir$ (unlike discharge, where $V=\varepsilon-Ir$). The current itself is found by treating the supply, series resistor, and battery (opposing emf plus internal resistance) as one loop.

    Step-by-Step Solution

    1. Loop equation: supply voltage drives current through the series resistor and against the battery's emf and internal resistance: $V_{supply} = IR_{series} + \varepsilon + Ir$. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.The readings of the voltmeter and ammeter in the circuit shown in the diagram are respectively [FIGURE] (a circuit: a 6 V battery with internal series resistance $1\,\Omega$ in series with a $0.6\,\Omega$ resistor and a 12 V battery, forming the top branch; a voltmeter V is connected across a $4\,\Omega$ resistor on the left branch (with + on the right, - on the left of the voltmeter), and a $0.4\,\Omega$ resistor is on the right branch; an ammeter A is in the bottom branch connecting the left and right branches, with + on the left and - on the right of the ammeter) (A) 5 V, 3 A (B) 7 V, 3 A (C) 5 V, 1 A (D) 7 V, 1 A

    ›Reveal solutionSolution

    Loop current $=1\,\text{A}$ and the voltmeter reads $7\,\text{V}$ — option (D).

    The two cells are connected so that they oppose each other, so the net EMF driving the loop is

    $\varepsilon=12-6=6\,\text{V}.$

    Total resistance around the loop:

    $R=\underbrace{1}_{\text{internal of }6\,\text{V}}+0.6+0.4+4=6\,\Omega.$

    Current (ammeter reading):

    $I=\frac{\varepsilon}{R}=\frac{6}{6}=1\,\text{A}.$ …

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