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Byju's Answer

Standard IX

Physics

Equivalent Resistance in Series Circuit

A set of 'n' ...

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 $10$ I. The value of 'n' is?

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Solution

The correct option is B $10$
Electrical resistance = nR
$I = \frac{E}{(n + 1) R}$ ...............(1)
So, $10 I = \frac{E}{\frac{R}{n} + R} = \frac{n E}{R (n + 1)}$ ............(2)
equating 1 and 2
we get n = 10
Option 2 is correct

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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?

The correct option is B 10Electrical resistance = nRI=E(n+1)R ...............(1)So, 10I=ERn+R=nER(n+1) ............(2)equating 1 and 2we get n = 10 Option 2 is correct

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 $10$ I. The value of 'n' is?

The correct option is B $10$
Electrical resistance = nR
$I = \frac{E}{(n + 1) R}$ ...............(1)
So, $10 I = \frac{E}{\frac{R}{n} + R} = \frac{n E}{R (n + 1)}$ ............(2)
equating 1 and 2
we get n = 10
Option 2 is correct