Q.(a) Describe briefly, with the help of a circuit diagram, the method of measuring the internal resistance of a cell.

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if (A) R = r/2 (B) R = r (C) R > r (D) R < r

   ›Reveal solutionSolution

   I = ε/(R+r); smaller R ⇒ larger current, so current is maximum for R < r (ideally R → 0).

   The circuit current is $I = \dfrac{\varepsilon}{R+r}$.

   For a fixed emf ε and fixed internal resistance r, the current increases as the external resistance R decreases. Hence the current through R is maximum when R is as small as possible. Among the given choices, this corresponds t …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.The electromotive force of an accumulator battery is 10 V and internal resistance 0.5Ω. The maximum electric current obtained from the battery will be

   (a) 5 A

   (b) 10 A

   (c) 20 A

   (d) 0.05 A

   ›Reveal solutionSolution

   The maximum current a cell can deliver is its short-circuit current, I = EMF / internal resistance.

   A real battery has EMF (epsilon) and internal resistance r. When connected to an external circuit of resistance R, the current is I = epsilon/(R+r), which is largest when R = 0 (sho …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: The quantity measured across a cell without drawing any current from it, is ..... .

   ›Reveal solutionSolution

   The potential difference measured across a cell's terminals when no current is drawn from it equals the cell's EMF.

   When current $I$ flows, the terminal voltage is $V = \varepsilon - Ir$ (less than the EMF $\varepsilon$ due to the voltage drop across internal resistance $r$). When no current is drawn ($I=0$, open circuit, e.g. measured with a …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The internal resistance of a cell depends on:

   (a) the area of the plates

   (b) the distance between the plates

   (c) the concentration of the electrolyte

   (d) All of the above

   ›Reveal solutionSolution

   A cell's internal resistance behaves like the resistance of the electrolyte column between its electrodes, so it depends on every geometric and chemical factor that affects that column.

   Inside a cell, current flows through the electrolyte between the two electrodes. Treating the electrolyte as a conducting medium, its resistance follows the same rule as any conductor: $r=\rho\dfrac{l}{A}$, where $\rho$ is the electrolyte's resistivity, $l$ the distance between the plates, and $A$ the area of the plates. So (i) a larger plate area $A$ gives a lower resistance (more parallel paths for current), (ii) a larger separation $l$ between plates gives a higher resistance (lon …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.An unknown resistance $R_1$ is connected in series with a resistance of 10 ohms. This combination is connected to one gap of a meter bridge, while a resistance $R_2$ is connected in the other gap, the balance point is obtained at a distance of 50cm. When 10 ohms resistance is removed the balance point shifts to 40cm. The value of $R_1$ is –

   (a) 10 ohms

   (b) 20 ohms

   (c) 40 ohms

   (d) 60 ohms

   ›Reveal solutionSolution

   Two meter-bridge balance equations (with and without the extra 10$\Omega$) solve for $R_1$.

   Meter bridge balance condition: $\dfrac{P}{Q} = \dfrac{l}{100-l}$ (ratio of the two gap resistances equals the ratio of the wire lengths).

   With the 10$\Omega$ in series with $R_1$, balance at 50 cm:

   $\frac{R_1+10}{R_2} = \frac{50}{50} = 1 \implies R_2 = R_1+10$

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The maximum current that can be drawn from a cell is when

   (a) R = 0

   (b) r = 0

   (c) R > r

   (d) r > R

   ›Reveal solutionSolution

   Current from a cell $I = E/(R+r)$ is largest when the total resistance $(R+r)$ is smallest, i.e. when the external resistance R = 0.

   For a cell of emf E and internal resistance r connected to an external resistance R, the current is

   $I = \frac{E}{R+r}$

   …

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7. CBSE 2024Set 55/5/11 markMCQ
   Q.A battery supplies 0.9 A current through a $2\,\Omega$ resistor and 0.3 A current through a $7\,\Omega$ resistor when connected one by one. The internal resistance of the battery is ______. (A) $2\,\Omega$ (B) $1.2\,\Omega$ (C) $1\,\Omega$ (D) $0.5\,\Omega$

   ›Reveal solutionSolution

   A real battery has an internal resistance that causes its terminal voltage to drop when current is drawn. By analyzing the battery's behavior under two different load conditions, we can determine its internal resistance, which is $\boxed{0.5\,\Omega}$.

   A battery is not an ideal voltage source; it possesses an inherent internal resistance, denoted by $r$. This internal resistance is effectively in series with the battery's electromotive force (EMF), $E$. When a current $I$ is drawn from the battery through an external resistor $R$, a voltage drop occurs across this internal resistance, equal to $Ir$.

   The voltage available across the terminals of the battery, known as the terminal voltage $V$, is therefore less than the EMF $E$. It is given by:

   $V = E - Ir$

   This terminal voltage is also the voltage across the external resistor $R$, so $V = IR$.

   Equating these two expressions for $V$, we get:

   $IR = E - Ir$

   Rearranging this equation to solve for the EMF $E$:

   $E = I(R + r)$

   This equation is fundamental for analyzing circuits with real batteries. The EMF $E$ and the internal resistance $r$ are constant properties of the battery. We can use the two given scenarios to form a system of equations and solve for $r$.

   1. Formulate equations for each scenario.

      We are given two distinct situations where the battery is connected to a different external resistor, resulting in a different current. We will apply the formula $E = I(R + r)$ to each case.

      • Scenario 1: The battery supplies a current $I_1 = 0.9\,\text{A}$ through an external resistor $R_1 = 2\,\Omega$. Using the formula $E = I(R + r)$:

   $E = 0.9\,\text{A} \times (2\,\Omega + r)$

   $E = 1.8 + 0.9r \quad \text{(Equation 1)}$

   *   **Scenario 2:** The battery supplies a current $I_2 = 0.3\,\text{A}$ through an external resistor $R_2 = 7\,\Omega$.
       Using the formula $E = I(R + r)$:
   

   $E = 0.3\,\text{A} \times (7\,\Omega + r)$

   $E = 2.1 + 0.3r \quad \text{(Equation 2)}$

   2. Solve the system of equations for $r$. …

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8. CBSE 2024Set 55/1/11 markMCQ
   Q.Consider the circuit shown in the figure. The potential difference between points A and B is : (A) $6\ \text{V}$ (B) $8\ \text{V}$ (C) $9\ \text{V}$ (D) $12\ \text{V}$

   ›Reveal solutionSolution

   Two cells with internal resistances in parallel drive current through each other; the terminal voltage $V_{AB}$ is found by treating each branch as a source with EMF and internal resistance, then using the parallel-source formula. The answer is $\boxed{8 \text{ V}}$.

   ---

   Why this approach works: cells with internal resistance in parallel

   When two cells are connected in parallel between the same two points A and B, they don't simply "add up" like batteries in series. Each cell has an EMF (electromotive force) and an internal resistance, and they can drive current through each other. The stronger cell (higher EMF per unit resistance) will actually charge the weaker one.

   The key insight: treat each branch as a source characterized by its EMF $\mathcal{E}$ and internal resistance $r$. The terminal voltage $V_{AB}$ across the parallel combination is the voltage that appears at the terminals when both sources are connected together. This is given by the weighted average of the EMFs, where the weights are the conductances (reciprocals of internal resistances).

   $V_{AB} = \frac{\frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$

   This formula comes from applying Kirchhoff's laws: the current from each source adjusts so that both branches have the same terminal voltage.

   ---

   Step-by-step solution

   1. Identify the parameters of each branch from the circuit

   

   Reading the circuit:

   • Upper branch: EMF $\mathcal{E}_1 = 12 \text{ V}$, internal resistance $r_1 = 1 \, \Omega$
   • Lower branch: EMF $\mathcal{E}_2 = 6 \text{ V}$, internal resistance $r_2 = 0.5 \, \Omega$

   2. Calculate the "weighted EMFs" (EMF divided by internal resistance)

   These represent the short-circuit current each source can deliver:

   $\frac{\mathcal{E}_1}{r_1} = \frac{12}{1} = 12 \text{ A}$

   $\frac{\mathcal{E}_2}{r_2} = \frac{6}{0.5} = 12 \text{ A}$

   Interestingly, both branches have the same short-circuit current capability.

   3. Calculate the sum of conductances …

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9. CBSE 2023Set 55/1/11 markMCQ
   Q.The potential difference across a cell in an open circuit is $8\ \text{V}$. It falls to $4\ \text{V}$ when a current of $4\ \text{A}$ is drawn from it. The internal resistance of the cell is :

   (a) $4\ \Omega$

   (b) $3\ \Omega$

   (c) $2\ \Omega$

   (d) $1\ \Omega$

   ›Reveal solutionSolution

   The key idea is that the open-circuit voltage is the cell’s EMF ($\mathcal{E} = 8\ \text{V}$), and the drop to $4\ \text{V}$ when $4\ \text{A}$ flows is due entirely to the voltage drop across the internal resistance $r$. Using $V = \mathcal{E} - Ir$, we get $r = 1\ \Omega$, so the correct option is (d).

   Every real cell behaves like a perfect EMF source $\mathcal{E}$ in series with a small internal resistance $r$. When no current flows (open circuit), the terminal voltage equals $\mathcal{E}$ — there’s no drop across $r$. But the moment you draw current, $r$ steals some voltage: $V_{\text{terminal}} = \mathcal{E} - I r$. That’s the whole physics in one line.

   Here, the open-circuit reading gives $\mathcal{E} = 8\ \text{V}$. When $4\ \text{A}$ is drawn, the terminal voltage crashes to $4\ \text{V}$. That $4\ \text{V}$ loss is $I r$. So:

   1. Write the terminal voltage equation:

   $V = \mathcal{E} - I r$

   2. Plug in the numbers:

   $4 = 8 - (4) r$

   3. Solve for $r$: $4r = 8 - 4 = 4 \quad\Rightarrow\quad r = 1\ \Omega$ …

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10. CBSE 2023Set 55/3/11 markMCQ
    Q.Two statements are given – one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b),

    (c) and

    (d) below.

    (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

    (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

    (c) Assertion (A) is true, but Reason (R) is false.

    (d) Assertion (A) is false and Reason (R) is also false. Assertion (A) : The internal resistance of a cell is constant. Reason (R) : Ionic concentration of the electrolyte remains same during use of a cell.

    ›Reveal solutionSolution

    Both statements are false: a cell's internal resistance increases as the electrolyte concentration changes and ions are consumed during discharge, making the correct answer (d).

    Why internal resistance matters

    The internal resistance of a cell arises from the opposition to ion flow within the electrolyte and at the electrode-electrolyte interfaces. If the cell were an ideal voltage source, it would have zero internal resistance. Real cells, however, have finite $r$ that affects the terminal voltage under load: $V = \mathcal{E} - Ir$, where $\mathcal{E}$ is the emf and $I$ is the current drawn.

    The question asks whether this internal resistance stays constant and whether the electrolyte concentration remains unchanged during use. Both claims touch on what happens inside a cell as it discharges.

    Examining the Assertion

    Assertion (A): The internal resistance of a cell is constant.

    This is false. Internal resistance depends on several factors:

    1. Electrolyte concentration: As a cell discharges, chemical reactions consume the active materials. In a typical electrochemical cell, ions are converted or depleted, changing the ionic concentration of the electrolyte. Lower ion concentration means fewer charge carriers, which increases resistivity and thus internal resistance.

    2. Temperature: Internal resistance decreases with rising temperature (ions move more freely) and increases when the cell cools.

    3. Age and usage: Over time, electrode surfaces may become coated with reaction products (polarization), further increasing resistance.

    4. State of charge: A nearly exhausted cell has significantly higher internal resistance than a fresh one.

    In practice, $r$ increases noticeably as a cell is used, which is why old batteries deliver lower terminal voltages under the same load.

    Examining the Reason

    Reason (R): Ionic concentration of the electrolyte remains same during use of a cell.

    This is also false. During discharge, electrochemical reactions at the electrodes consume reactants and produce products, directly altering the electrolyte composition.

    For example, in a lead-acid cell:

    • At the anode: $\text{Pb} + \text{SO}_4^{2-} \to \text{PbSO}_4 + 2e^-$
    • At the cathode: $\text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \to \text{PbSO}_4 + 2\text{H}_2\text{O}$ …

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11. CBSE 2023Set ANNUAL1 mark
    Q.State the condition for maximum current to be drawn from a cell.

    ›Reveal solutionSolution

    Maximum current is drawn when the external resistance matches the internal resistance.

    …

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12. CBSE 2022Set ANNUAL1 mark
    Q.What do you understand by internal resistance of a Cell?

    ›Reveal solutionSolution

    Internal resistance $r$ is the opposition to current flow offered by the cell's own electrolyte/electrodes.

    Every real cell has some resistance to the flow of charge within itself, arising from the electrolyte and the electrode material -- this is called its internal resistance, denoted $r$. Because of $r$, when a cell of emf $\varepsilon$ drives a current $I$ through an external resistance $R$, the terminal potential difference is $V = \varepsilon - Ir$, which is less than the emf. Internal resistance depends on the nature of the electrolyte, the distance between the electrodes, the area of the electrodes, and the temperat …

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