A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38Ω as shown in the figure. Find the value of current in the circuit.
Solution:
Net emf of the combination of cell E = 200 - 10 = 190 V Net resistance of the circuit R = 38Ω So, current in the circuit I = E/R ⇒ I = 190/38 = 5 A