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 $10 \mathrm{I}$. The value of ' $n$ ' is(1) 20(2) 11(3) 10(4) 9

Home » 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 ‘ ‘ resistors are connected in parallel to the same battery. Then the current drawn from battery becomes . The value of ‘ ‘ is(1) 20(2) 11(3) 10(4) 9

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 $10 \mathrm{I}$ . The value of ‘ $n$ ‘ is
(1) 20
(2) 11
(3) 10
(4) 9

Physics

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 $10 \mathrm{I}$ . The value of ‘ $n$ ‘ is
(1) 20
(2) 11
(3) 10
(4) 9

Answer is (3)

$I=\frac{E}{n R+R}$ …(i)

$10 I=\frac{E}{\frac{R}{n}+R}$ …(ii)

Dividing (ii) by (i) we get,

$10=\frac{(n+1) R}{\left(\frac{1}{n}+1\right) R}$

After solving the equation, $n=10$

i

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