Question

When a "dry-cell" flashlight battery with an internal resistance of 0.24 Ω is connected to a 1.65-Ω light bulb, the bulb shines dimly. However, when a lead-acid "wet-cell" battery with an internal resistance of 0.068 Ω is connected, the bulb is noticeably brighter. Both batteries have the same emf. Find the ratio Pwet/Pdry of the power delivered to the bulb by the wet-cell battery to that delivered by the dry-cell battery.

Answer 1

Pwet = E^2/(R+rwet)......(1); where E is the emf of the battery, R is resistance of the bulb and rwet is the internal resistance of the wet cell battery.

Pdry = E^2/(R+rdry)......(2); where E is the emf of the battery, R is resistance of the bulb and rdry is the internal resistance of the dry cell battery.

Dividing (1) by (2), we get, Pwet/Pdry = (R + rdry)/(R+rwet)....(3)

It is given that R = 1.65 ohm, rdry is 0.24 ohm and rwet is 0.068 ohm.

Substituting the above values in (3), we get Pwet/Pdry = 1.1 Ans.