A certain number of identical resistors are connected in series across an ideal battery of constant emf, and the total power dissipated by the combination is P . When the same resistors are connected in parallel across the same battery, the total power dissipated becomes 400P . The number of resistors is:

Step-by-step solution

Let there be N identical resistors, each of resistance R , connected across a battery of emf V . When connected in series, the equivalent resistance is R_s = NR . The power dissipated in series is P_s = V^2 R_s = V^2 NR = P . When connected in parallel, the equivalent resistance is R_p = R N . The power dissipated in parallel is P_p = V^2 R_p = V^2 R N = NV^2 R = 400P . Taking the ratio of the powers: P_p P_s = NV^2 R V^2 NR = N^2 Given that P_p = 400P and P_s = P , we have: N^2 = 400 N = 20 Thus, there are 20 resi