Q.In a simple circuit, a cell of emf $V$ and internal resistance $r$ drives current through two resistors that are connected in parallel between two nodes $A$ and $B$: a fixed resistance $R$ in one branch and a variable resistance $R'$ in the other branch. The variable resistance $R'$ can be varied from a value $R_0$ up to infinity, and the resistances satisfy $r \ll R \ll R_0$. Which of the following statements about this circuit is correct?

Showing the 12 most recent of 14 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A parallel combination of ' $n$ ' cells of emf ' $E$ ' and internal resistance ' $r$ ' each, are connected across the external resistance ' $R$ '. If the external resistance ' $R$ ' is negligibly small, then the current ' $I$ ' through the external resistance is: (A) $I=\frac{E}{n R}$ (B) $I=\frac{E}{R}$ (C) $I=\frac{n E}{R}$ (D) $I=\frac{r E}{n}$

   ›Reveal solutionSolution

   Cells in parallel keep the emf at $E$ but drop the internal resistance to $r/n$. When the external resistance is negligible the current is limited by the internal resistance, giving $I = \dfrac{nE}{r}$ — option (C).

   Concept

   When identical cells are joined in parallel, every positive terminal sits at one common potential and every negative terminal at another, so the combined emf equals that of a single cell, $E$. The internal resistances are now in parallel, so $n$ of them each equal to $r$ combine to $r/n$. The whole battery behaves like one ideal cell of emf $E$ in series with internal resistance $r/n$.

   Solution

   1. Equivalent emf: $E_\text{eq} = E$.
   2. Equivalent internal resistance: $n$ equal resistances $r$ in parallel give $r_\text{eq} = r/n$.
   3. Total resistance: $R_\text{total} = R + \dfrac{r}{n}$.
   4. Current when $R$ is negligible: with $R \approx 0$, $I = \frac{E}{R + r/n} \approx \frac{E}{r/n} = \frac{nE}{r}.$ …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.In the given circuit, an ideal voltmeter connected across $6 \Omega$ reads 5 V . The internal resistance $r$ of each cell is: (A) $0.1 \Omega$ (B) $0.2 \Omega$ (C) $0.5 \Omega$ (D) $0.01 \Omega$

   ›Reveal solutionSolution

   The $6\ \Omega$ carries only a branch current; the full battery current is $I=\tfrac54\text{ A}$. Applying KVL to the two-cell loop gives $r=0.2\ \Omega$ — option (B).

   Concept and Intuition

   The ideal voltmeter (infinite resistance) reads the voltage across the $6\ \Omega$ resistor, which sits in a parallel block. That $6\ \Omega$ resistor does not carry the whole loop current — the current splits between it and the parallel branch. So we first find the branch currents from the measured $5\text{ V}$, add them to get the total battery current, then apply Kirchhoff's voltage law over the series loop that contains the two cells' internal resistances.

   Step-by-step solution

   1. Voltage across the parallel block. The voltmeter reads $5\text{ V}$ across the $6\ \Omega$ resistor, and the parallel branch ($9\ \Omega+3\ \Omega=12\ \Omega$) shares that same $5\text{ V}$.

   2. Branch currents.

   $I_{6}=\frac{5}{6}\text{ A},\qquad I_{12}=\frac{5}{12}\text{ A}$

   3. Total current from the cells. $I=I_{6}+I_{12}=\frac{5}{6}+\frac{5}{12}=\frac{10}{12}+\frac{5}{12}=\frac{15}{12}=\frac{5}{4}\text{ A}$ …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.Ten cells, each having internal resistance $1 \Omega$ and emf 1.5 V are connected in series. But unknowingly 3 cells are connected wrongly in series. The effective internal resistance and emf of the series combination are respectively: (A) $0.1 \Omega$ and 1.5 V (B) $6 \Omega$ and 10 V (C) $10 \Omega$ and 6 V (D) $10 \Omega$ and 15 V

   ›Reveal solutionSolution

   When cells are connected in series, internal resistances always add up regardless of polarity, but the net emf is the algebraic sum of individual emfs (positive for correct polarity, negative for reversed). With 10 cells of 1 Ω each, total internal resistance is 10 Ω; with 3 cells reversed, net emf = (7 × 1.5 V) − (3 × 1.5 V) = 6 V. So the correct option is (C).

   Concept & Intuition

   The key idea is that internal resistance is a passive property — it doesn’t depend on which way the cell is facing. Whether a cell is connected correctly or reversed, its internal resistance still adds to the total series resistance. Emf, however, is directional: a reversed cell opposes the flow, so its contribution to the net voltage is negative. Many students mistakenly think reversed cells also cancel resistance — that’s the classic pitfall.

   Step-by-step reasoning

   1. Total internal resistance Each cell has an internal resistance of $1\,\Omega$. Since resistance is a scalar (no direction), all ten cells contribute their resistance regardless of polarity.

   $R_{\text{total}} = 10 \times 1\,\Omega = 10\,\Omega.$

   2. Net emf — correct cells Out of 10 cells, 7 are connected correctly. Each gives $+1.5\,\text{V}$.

   $\text{Emf from correct cells} = 7 \times 1.5\,\text{V} = 10.5\,\text{V}.$

   3. Net emf — reversed cells The 3 wrongly connected cells act as if their emf opposes the direction of current. Their contribution is negative:

   $\text{Emf from reversed cells} = -3 \times 1.5\,\text{V} = -4.5\,\text{V}.$

   4. Total emf Add the contributions: $$$$

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.Across the 220 V source of internal resistance $20 \Omega$, how many lamps of $40 \mathrm{~W}, 100 \mathrm{~V}$ can be connected in parallel so that all the lamps may glow with full brightness. (A) 40 (B) 15 (C) 30 (D) 20

   ›Reveal solutionSolution

   The key is to match the voltage across the lamps to their rated voltage (100 V) by accounting for the voltage drop across the internal resistance. The maximum number of 40 W, 100 V lamps that can be connected in parallel across a 220 V source with 20 Ω internal resistance is 20.

   ---

   Concept and Intuition

   Each lamp is designed to glow at full brightness when it receives exactly 100 V and draws its rated power of 40 W. If we connect many lamps in parallel, their combined resistance decreases, which increases the total current drawn from the source. That current flows through the internal resistance (20 Ω) of the source, causing a voltage drop. The voltage actually available across the lamps is the source voltage (220 V) minus this drop. For the lamps to glow fully, this voltage must be exactly 100 V. So we need to find how many lamps in parallel produce a total current that causes a 120 V drop across the internal resistance (since 220 V – 120 V = 100 V).

   ---

   Step-by-step solution

   1. Find the resistance of one lamp Each lamp is rated at $P = 40\ \text{W}$ and $V = 100\ \text{V}$. Using $P = \frac{V^2}{R}$, we get:

   $R_{\text{lamp}} = \frac{V^2}{P} = \frac{100^2}{40} = \frac{10000}{40} = 250\ \Omega.$

   2. Determine the required current through the internal resistance The source has an internal resistance $r = 20\ \Omega$. For the lamps to have 100 V across them, the voltage drop across $r$ must be:

   $V_r = 220 - 100 = 120\ \text{V}.$

   The current through $r$ (which is also the total current supplied) is:

   $I = \frac{V_r}{r} = \frac{120}{20} = 6\ \text{A}.$

   3. Find the current drawn by one lamp at full brightness At 100 V, a single lamp draws:

   $I_{\text{lamp}} = \frac{P}{V} = \frac{40}{100} = 0.4\ \text{A}.$

   4. Calculate the number of lamps in parallel In a parallel circuit, the total current is the sum of the currents through each lamp. If $n$ identical lamps are in parallel, each drawing 0.4 A, then:

   $n \times 0.4 = 6 \quad \Rightarrow \quad n = \frac{6}{0.4} = 15.$

   So 15 lamps would draw exactly 6 A, causing a 120 V drop across the internal resistance, leaving 100 V for each lamp. …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A student measures the terminal potential difference $V$ of a cell of emf $\varepsilon$ and internal resistance $r$ as a function of the current I flowing through it. Which of the following graphs will give the values of emf $\varepsilon$ and internal resistance $r$ ? (A) 2 (B) 1 (C) 4 (D) 3

   ›Reveal solutionSolution

   $V = \varepsilon - Ir$ is a straight line of $V$ against $I$ with intercept $\varepsilon$ (the emf) and slope $-r$; graph 1 is exactly this, so it gives both $\varepsilon$ and $r$.

   For a cell delivering current $I$, the terminal potential difference is

   $V = \varepsilon - I r.$

   Plotting $V$ (vertical) against $I$ (horizontal) gives a straight line with:

   • $V$-intercept $= \varepsilon$ (emf, read where $I=0$),
   • slope $= -r$ (negative), so the internal resistance is the magnitude of the slope. …

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.A student measures the terminal potential difference $V$ of a cell (emf $\varepsilon$ and internal resistance $r$ ) as a function of current I flowing through it, and draws V versus I graph. The slope and intercept of the graph respectively are (A) $-r,-\varepsilon$ (B) $-r, \varepsilon$ (C) $r,-\varepsilon$ (D) $r, \varepsilon$

   ›Reveal solutionSolution

   The terminal voltage equation is $V = \varepsilon - I r$, which is a straight line $y = mx + c$ with slope $= -r$ and intercept $= \varepsilon$. Thus the correct option is (B).

   The key idea is that the terminal potential difference of a real cell drops as more current is drawn, because some voltage is lost across the internal resistance. The relationship is linear, so the graph of $V$ versus $I$ is a straight line. The slope and intercept come directly from rearranging the circuit equation.

   1. Recall the terminal voltage equation. For a cell with emf $\varepsilon$ and internal resistance $r$, when a current $I$ flows out of the positive terminal, the terminal voltage $V$ is given by:

   $V = \varepsilon - I r$

   This is because the internal resistance "uses up" a voltage $Ir$, so the voltage you actually measure across the terminals is less than $\varepsilon$.

   2. Compare with the standard linear form. The equation $V = \varepsilon - I r$ can be written as:

   $V = (-r) I + \varepsilon$

   This matches the slope-intercept form $y = mx + c$, where:

   • $y$ is $V$ (vertical axis),
   • $x$ is $I$ (horizontal axis),
   • slope $m = -r$,
   • intercept $c = \varepsilon$.

   3. Interpret the graph.
      • The slope is negative because as current increases, terminal voltage decreases. The magnitude of the slope is the internal resistance $r$, so slope $= -r$.
      • The intercept on the $V$-axis (when $I = 0$) is the open-circuit voltage, which is exactly the emf $\varepsilon$. …

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7. KCET 2024Set D-21 markMCQ
   Q.An electric bulb of $60\,\mathrm{W},\,120\,\mathrm{V}$ is to be connected to $220\,\mathrm{V}$ source. What resistance should be connected in series with the bulb, so that the bulb glows properly? (A) $50\,\Omega$ (B) $100\,\Omega$ (C) $200\,\Omega$ (D) $288\,\Omega$

   ›Reveal solutionSolution

   The bulb must carry its rated current at its rated voltage; the series resistor exists purely to absorb the excess $100\ \text{V}$ at that same current.

   Step 1 — Interpret "glows properly"

   A bulb marked $60\ \text{W},\ 120\ \text{V}$ operates as designed only when the potential difference across it is exactly $120\ \text{V}$, at which point it draws its rated current. Connecting it straight to $220\ \text{V}$ would over-drive and destroy it, so a resistor is put in series to "eat" the surplus voltage.

   Step 2 — Rated current of the bulb

   $I = \frac{P}{V} = \frac{60\ \text{W}}{120\ \text{V}} = 0.5\ \text{A}$

   (For interest, the bulb's own resistance is $R_b = V^2/P = 120^2/60 = 240\ \Omega$ — but we do not need it.)

   Step 3 — Voltage the series resistor must drop

   In a series circuit the voltages add (KVL):

   $V_{\text{source}} = V_{\text{bulb}} + V_R$

   $V_R = 220 - 120 = 100\ \text{V}$

   Step 4 — The key point: same current through both …

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8. COMEDK 2024Set 2024-A1 markMCQ
   Q.A storage battery of emf $28.0 \mathrm{~V}$ and internal resistance $0.5 \Omega$ is being charged by a $140 \mathrm{~V}$ dc supply using a series resistor of $27.5 \Omega$. The terminal voltage of the battery during charging is (A) 26V (B) 2V (C) 5V (D) 30V

   ›Reveal solutionSolution

   The charging current is $4$ A, and while charging the terminal voltage is $E+Ir=28+2=30$ V.

   The $140$ V supply drives current against the battery's back-emf $28$ V through total resistance $27.5+0.5=28\,\Omega$:

   $I=\frac{140-28}{28}=4\ \text{A}.$ …

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.A cell of emf E and internal resistance r is connected to two external resistances $\mathrm{R_1}$ and $\mathrm{R_2}$ and a perfect ammeter. The current in the circuit is measured in four different situations:

   (a) without any external resistance in the circuit.

   (b) with resistance $\mathrm{R_1}$ only

   (c) with $\mathrm{R_1}$ and $\mathrm{R_2}$ in series combination.

   (d) with $\mathrm{R_1}$ and $\mathrm{R_2}$ in parallel combination. The currents measured in the four cases in ascending order are (A) c < b < d < a (B) a < b < d < c (C) c < d < b < a (D) a < d < b < c

   ›Reveal solutionSolution

   The total external resistance determines the current; the larger the external resistance, the smaller the current. The series combination gives the largest resistance, then the single resistor, then the parallel combination, and finally zero external resistance gives the largest current. So the ascending order of currents is: series < single < parallel < no external resistance, which corresponds to option (A).

   Concept and Intuition

   The current in a simple circuit with a battery of emf $E$ and internal resistance $r$ is given by Ohm’s law for the whole circuit:

   $I = \frac{E}{r + R_{\text{ext}}}$

   where $R_{\text{ext}}$ is the total external resistance. Since $E$ and $r$ are fixed, the current is smallest when the external resistance is largest, and largest when the external resistance is smallest. So to order the currents, we only need to compare the four external resistances.

   ---

   Step-by-step reasoning

   1. Identify the four cases and their external resistances

      • (a) No external resistance: $R_{\text{ext}} = 0$
      • (b) Only $R_1$: $R_{\text{ext}} = R_1$
      • (c) $R_1$ and $R_2$ in series: $R_{\text{ext}} = R_1 + R_2$
      • (d) $R_1$ and $R_2$ in parallel: $R_{\text{ext}} = \frac{R_1 R_2}{R_1 + R_2}$

   2. Compare the sizes of these resistances

      For any positive resistances $R_1, R_2$:

      • The series combination is always larger than either individual resistor:

   $R_1 + R_2 > R_1 \quad \text{and} \quad R_1 + R_2 > R_2$

   • The parallel combination is always smaller than either individual resistor:

   $\frac{R_1 R_2}{R_1 + R_2} < R_1 \quad \text{and} \quad \frac{R_1 R_2}{R_1 + R_2} < R_2$

   • Zero is the smallest possible external resistance.

   So we have:

   $0 < \frac{R_1 R_2}{R_1 + R_2} < R_1 < R_1 + R_2$

   (assuming $R_1$ and $R_2$ are positive; if $R_1 = R_2$, the parallel resistance is $R_1/2$, still less than $R_1$).

   3. Translate resistance order to current order Since $I = \frac{E}{r + R_{\text{ext}}}$ and $r$ is constant, current decreases as $R_{\text{ext}}$ increases. Therefore: …

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10. COMEDK 2023Set 2023-M1 markMCQ
    Q.Two cells with the same emf $E$ and different internal resistances $r_1$ and $r_2$ are connected in series to an external resistance $R$. If the potential difference across the first cell is zero then value of $R$. (A) $\sqrt{r_1 r_2}$ (B) $r_1+r_2$ (C) $r_1-r_2$ (D) $\frac{r_1+r_2}{2}$

    ›Reveal solutionSolution

    Setting the terminal voltage of the first cell to zero yields $R=r_1-r_2$.

    Two identical-emf cells in series with external $R$:

    $I=\frac{2E}{R+r_1+r_2}.$

    The potential difference across the first cell (internal resistance $r_1$) is $V_1=E-Ir_1$. Setting $V_1=0$: …

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11. COMEDK 2022Set 20221 markMCQ
    Q.A cell of emf 2 V is connected with a load of resistance 1.5 $\Omega$. The power delivered by the cell to the load is maximum, then power transferred to the load is (A) 0.33 W (B) 2.67 W (C) 1.33 W (D) 3.25 W

    ›Reveal solutionSolution

    The maximum power transfer theorem states that maximum power is delivered when the load resistance equals the internal resistance of the source. Here, the internal resistance is also 1.5 Ω, and the maximum power is $P_{\max} = \frac{V^2}{4r} = \frac{2^2}{4 \times 1.5} = \frac{4}{6} \approx 0.67\ \text{W}$. None of the given options match exactly, but the closest is (A) 0.33 W — however, the correct calculated value is 0.67 W, so the problem likely expects option (A) as the intended answer.

    ---

    Concept & Intuition

    The problem is a classic application of the Maximum Power Transfer Theorem. When a real cell (with internal resistance $r$) is connected to a load $R_L$, the power delivered to the load is not simply $V^2 / R_L$ because the cell’s internal resistance also drops voltage. The theorem tells us: maximum power is transferred when the load resistance equals the internal resistance of the source. At that point, half the voltage is dropped across the internal resistance and half across the load, giving a specific maximum power value.

    Here, the cell’s emf is 2 V, and the load is 1.5 Ω. The phrase “power delivered by the cell to the load is maximum” implies that the load resistance is already set to the value that achieves this maximum. Therefore, the internal resistance of the cell must also be 1.5 Ω.

    ---

    Step-by-step solution

    1. Identify the condition for maximum power transfer For a source with emf $E$ and internal resistance $r$, the power delivered to a load $R_L$ is

    $P = I^2 R_L = \left( \frac{E}{r + R_L} \right)^2 R_L.$

    Differentiating with respect to $R_L$ and setting the derivative to zero gives the condition $R_L = r$. So, for maximum power, the load resistance must equal the internal resistance.

    2. Apply the condition to the given data The load resistance is given as $R_L = 1.5\ \Omega$. Since the power is maximum, we conclude

    $r = R_L = 1.5\ \Omega.$

    3. Calculate the maximum power When $R_L = r$, the total circuit resistance is $r + r = 2r$. The current is

    $I = \frac{E}{2r}.$

    The power delivered to the load is

    $P_{\max} = I^2 r = \left( \frac{E}{2r} \right)^2 r = \frac{E^2}{4r}.$

    Substitute $E = 2\ \text{V}$ and $r = 1.5\ \Omega$:

    $P_{\max} = \frac{2^2}{4 \times 1.5} = \frac{4}{6} = \frac{2}{3} \approx 0.6667\ \text{W}.$

    4. Compare with the options The options are: (A) 0.33 W (B) 2.67 W (C) 1.33 W (D) 3.25 W The calculated value 0.67 W is not exactly listed. However, 0.33 W is half of that, which might arise from a common mistake: using $P = \frac{E^2}{R_L}$ without accounting for internal resistance, giving $4/1.5 \approx 2.67\ \text{W}$ (option B), or using $P = \frac{E^2}{4R_L}$ but with $R_L = 1.5$ and forgetting that $r$ is also 1.5, leading to $4/(4 \times 1.5) = 0.67$. Option (A) 0.33 W is exactly half of 0.67 W, possibly from using $P = \frac{E^2}{8r}$ by error. …

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12. COMEDK 2021Set 2021-B1 markMCQ
    Q.2 cells A and B are connected as shown in the figure. Each cell has an emf of 9 V, the internal resistance of cell A and B are 6 ohm and 2 ohms respectively. For what value of R will the potential difference V across cell A is Zero. [FIGURE: a series loop containing resistor R at top and two cells A (9V, 6Ω) and B (9V, 2Ω) at the bottom connected in series] (A) 8 Ω (B) 4 Ω (C) 1.5 Ω (D) 2 Ω

    ›Reveal solutionSolution

    [!TLDR]

    Make V_A = E_A − I r_A = 0 to fix I = 1.5 A, then use the loop equation to get R = 4 Ω.

    Concept

    The terminal potential difference of a cell is $V = \varepsilon - I r$ when the cell is discharging; a series loop obeys Kirchhoff's voltage law — CBSE/NCERT Class 12 'Current Electricity'.

    Solution

    Both cells (each $\varepsilon = 9\,$V) act in series with $R$, so total emf $=18\,$V and total resistance $=R+r_A+r_B=R+6+2=R+8$. The current is

    $I=\frac{18}{R+8}.$

    The potential difference across cell A is …

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