Verified Solution • 2025–26 Session moderate

QPages 198-203: Q21: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q21 Solution: Consider identical cells, each of emf and internal resistance…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that connecting $n_2$ identical series rows in parallel increases the total internal resistance to $n_1 n_2 r$ or leaves it as $n_1 r$. Remember that parallel branches divide the resistance, so the effective internal resistance is $n_1 r / n_2$, not the sum of all individual resistances.

Consider

N = n_1 n_2

identical cells, each of emf

\varepsilon

and internal resistance r. Suppose

n_1

cells are joined in series to form a line and

n_2

such lines are connected in parallel. The combination drives a current in an external resistance R. a) Find the current in the external resistance. b) Assuming that

n_1

and

n_2

can be continuously varied, find the relation between

n_1

,

n_2

, R and r for which the current in R is maximum.

Step-by-Step Verified Solution

**Answer:** (a)

i = \\frac{n_1 \\varepsilon}{R + n_1 r/n_2}

(b) Condition for maximum:

R = n_1 r/n_2

. **Step-by-Step Explanation:** (a)

n_1

cells in series give emf

n_1\\varepsilon

, internal resistance

n_1 r

.

n_2

such rows in parallel: effective emf

= n_1\\varepsilon

, effective internal resistance

= n_1 r/n_2

.

i = n_1\\varepsilon/(R + n_1 r/n_2)

. (b) For maximum current,

R = n_1 r/n_2

[citation:5].

All Questions in this Chapter

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