Question 652: Current Electricity

Question

A set of n equal resistors, of value R each, are
connected in series to a battery of emf E and
internal resistance R. The current drawn is I. Now,
the n resistors are connected in parallel to the
same battery. Then the current drawn from battery
becomes 10I. The value of n is

Accepted Answer

n series grouping equivalent resistance
R
series
= nR
In parallel grouping equivalent resistance
R
parallel
= $${R \over n}$$
Current drawn from a battery when n
resistors are connected in series is
I = $${E \over {nR + R}}$$ ......(i)
Current drawn from same battery when n resistors
are connected in parallel is
10I = $${E \over {{R \over n} + R}}$$ ......(ii)
On dividing eqn. (ii) by (i),
10 = $${{\left( {n + 1} \right)R} \over {\left( {{1 \over n} + 1} \right)R}}$$
Solving we get, n = 10

Answer

10

Answer

11

Answer

20

Answer

9

Current Electricity

A set of n equal resistors, of value R each, are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10I. The value of n is

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