Question 36: 10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is :
(1) 1
(2) 1000
(3) 10
(4) 100
Answer: Option (4)
Explanation:
When 10 resistors each of resistance R are connected in series,
the equivalent resistance is R_s = 10R.
The current drawn from the battery in series combination is
I_s = \frac{E}{R_s} = \frac{E}{10R}.
When the same 10 resistors are connected in parallel,
the equivalent resistance is R_p = \frac{R}{10}.
The current drawn from the battery in parallel combination is
I_p = \frac{E}{R_p} = \frac{E}{R/10} = \frac{10E}{R}.
The factor by which the current increases is n = \frac{I_p}{I_s}.
Substituting the values, n = \frac{\frac{10E}{R}}{\frac{E}{10R}}.
On simplification, n = 100.
Hence, the correct answer is Option (4).