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 :
- 100
- 1
- 1000
- 10
Answer
(1) 100
Solution
Series: Rs = 10R
Is = E10R
Parallel: Rp = R10
Ip = 10ER
n = IpIs = 10E/RE/10R = 100
Answer: (1)