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:

Step-by-step solution

When the resistors are connected in series with the battery, the equivalent resistance R s can be written as R s = R + R + . . . . 10   times = 10 R       . . . 1 Hence, the current i s is given by i s = E R s = E 10 R       . . . 2 When the resistors are connected in parallel, the equivalent resistance R p is given by R p = 1 1 R + 1 R + . . . .   10   times =   R 10       . . . 3 Hence, the current i p is given by i p = E R p = E R 10 = 10 E R &#160