A current of 2 A flows through a 2 Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω resistor. The internal resistance of the battery is

Step-by-step solution

Current I = E R + r ⇒ 2 = E 2 + r .... (i) ⇒ 0 .5 = E 9 + r .... (ii) On dividing equation (i) by equation (ii), we get 2 0.5 = 9 + r 2 + r ⇒ 4 = 9 + r 2 + r ⇒ 3 r = 1 ⇒ r = 1 3   Ω