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NVAEasyJEE 2023Cells, EMF & Internal Resistance

JEE Physics 2023 Question with Solution

Two identical cells, when connected either in parallel or in series, give the same current in an external resistance of 5 Ω5 \, \Omega. The internal resistance of each cell will be _____ Ω\Omega.

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: Two identical cells of emf ϵ\epsilon and internal resistance rr each are connected with an external resistance of 5 Ω5 \, \Omega.

Find: The value of internal resistance rr when the current is the same in series and parallel combinations.

Circuit diagrams comparing two identical cells in series and in parallel with internal resistances marked r, equivalent resistance r by 2, and an external resistance of 5 ohm.

For one arrangement, the current is

i=2ϵ5+2ri = \frac{2\epsilon}{5 + 2r}

... (1)

For the other arrangement, the current is

i=ϵr2+5i = \frac{\epsilon}{\frac{r}{2} + 5}

... (2)

Equating (1) and (2):

2ϵ5+2r=ϵr2+5\frac{2\epsilon}{5 + 2r} = \frac{\epsilon}{\frac{r}{2} + 5}

Cross-multiplying,

2ϵ(r2+5)=ϵ(5+2r)2\epsilon \left( \frac{r}{2} + 5 \right) = \epsilon (5 + 2r)

Simplifying,

r+10=5+2rr + 10 = 5 + 2r r=5r = 5

Therefore, the internal resistance of each cell is 5 Ω5 \, \Omega.

Using equivalent cell combinations

Given: Each cell has emf ϵ\epsilon and internal resistance rr. The external resistance is 5 Ω5 \, \Omega.

Find: Internal resistance rr.

When the two identical cells are combined in one way, the effective emf and internal resistance become such that the current is

i=2ϵ5+2ri = \frac{2\epsilon}{5 + 2r}

For the other way, the current becomes

i=ϵr2+5i = \frac{\epsilon}{\frac{r}{2} + 5}

Since the currents are equal,

2ϵ5+2r=ϵr2+5\frac{2\epsilon}{5 + 2r} = \frac{\epsilon}{\frac{r}{2} + 5}

Cancelling ϵ\epsilon from both sides,

25+2r=1r2+5\frac{2}{5 + 2r} = \frac{1}{\frac{r}{2} + 5}

Now cross-multiply:

2(r2+5)=5+2r2\left( \frac{r}{2} + 5 \right) = 5 + 2r r+10=5+2rr + 10 = 5 + 2r r=5r = 5

Hence, the required internal resistance is 5 Ω5 \, \Omega.

Common mistakes

  • Using the series and parallel formulas in the wrong order. This gives incorrect effective emf or internal resistance. First identify the equivalent cell combination correctly, then write the current expression.

  • Forgetting that in the parallel combination of identical cells, the effective internal resistance becomes r2\frac{r}{2}. Taking it as 2r2r is incorrect. Use the equivalent internal resistance carefully.

  • Equating voltages instead of currents. The question states that the same current flows through the external resistance. Therefore, set the two current expressions equal, not the total resistances or emfs directly.

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