A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery? Can this battery drive starting motor of car?

Question

Accepted Answer

The MAXIMUM current an emf device can deliver occurs when the external resistance is zero (Eq. 11.43), $I_{max}=\dfrac{\varepsilon}{r}=\dfrac{24}{380}\approx0.0632\,\text{A}$ , matching the book's printed answer of 0.063 A. A car's starting (cranking) motor typically needs a current of the order of 100-200 A to turn over a cold engine -- three orders of magnitude larger than the 0.063 A this worn-out battery (with its now-large 380 Ω internal resistance) can supply even in the best case. So this battery, after long use, can no longer drive a car's starting motor; the large internal resistance (which grows as a battery ages/degrades) is exactly what chokes off the available current, even though the emf (24 V, oddly stated but taken as given) may look normal. [!ANSWER] I_max ≈ 0.063 A; No -- a starting motor needs on the order of 100+ A, far beyond what this aged, high-internal-resistance battery can deliver.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set ANNUAL1 markMCQ
$Q. A current of 3 A flows through a 3 Ω resistor when connected across a battery. The same battery supplies a current 0.75 A when connected across a 15 Ω resistor. The internal resistance of the battery is$
(a) 0.25 Ω
(b) 0.5 Ω
(c) 1.0 Ω
(d) 1.5 Ω
›Reveal solutionSolution

Writing the EMF equation for both cases and solving the two equations simultaneously gives r = 1.0 Ω.

Let the battery have EMF $E$ and internal resistance $r$ .

Case 1: $I_1 = 3\ A$ through $R_1 = 3\ \Omega$ :

$E = I_1(R_1 + r) = 3(3 + r) = 9 + 3r \quad \text{...(i)}$

Case 2: $I_2 = 0.75\ A$ through $R_2 = 15\ \Omega$ :

$E = I_2(R_2 + r) = 0.75(15 + r) = 11.25 + 0.75r \quad \text{...(ii)}$

Equating (i) and (ii):

$9 + 3r = 11.25 + 0.75r \implies 2.25r = 2.25 \implies r = 1.0\ \Omega$

✓Final answer
$Internal resistance r = 1.0 Ω (option c).$
CBSE 2016Set ANNUAL1 markMCQ
$Q. Two cells each of emf e but internal resistances r1 and r2 are connected in series through an external resistance R. If the potential difference across the first cell is zero while current flows, the relation of R in terms of r1 and r2 is$
(a) R = r1 + r2
(b) R = r1 − r2
(c) R = (1/2)(r1 + r2)
(d) R = (1/2)(r1 − r2)
›Reveal solutionSolution

Setting the potential difference across the first cell to zero and using the series-circuit current gives R = r₁ − r₂.

Two cells, each of EMF e, with internal resistances r1 and r2, connected in series with external resistance R. Current in the circuit: I = 2e/(R + r1 + r2).

The potential difference across the first cell equals e − I·r1 (terminal voltage of a cell delivering current). Setting this to zero:

e = I·r1 = [2e/(R+r1+r2)]·r1

R + r1 + r2 = 2r1

R = r1 − r2

✓Final answer
$(b) R = r₁ - r₂.$

Q.A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery? Can this battery drive starting motor of car?

Concept understanding — EMF and Internal Resistance

Q.A battery after a long use has an emf 24 V and an internal resistance 380 Ω. Calculate the maximum current drawn from the battery? Can this battery drive starting motor of car?

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