Question

First, a set of n equal resistor of $10\Omega$ each are connected in series to a battery of emf 20 V and internal resistance $10\Omega$. A current I is observed to flow. Then,the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is

• A. 20
• B. 20.0

Answer 1

Answer: (A), (B)

When resistances are connected in the series combination :-
$R_s=10+10+10+...\text{n times}$
$R_s=10n$
The current is given by :
$I=\frac{E}{R_s+r}=\frac{20}{10n+10}....\left(1\right)$

When resistances are connected in the parallel combination :-
$R_p=\frac{R}{n}=\frac{10}{n}$
The current is given by :
$I_p=20I=\frac{E}{R_p+r}=\frac{20}{\frac{10}{n}+10}$
From $\left(1\right)$ :
$\frac{20\times 20}{10n+10}=\frac{20}{\frac{10}{n}+10}$
$\Rightarrow n=20$