Question

A battery of unknown e.m.f. is connected across resistances as shown in Fig. 1.50. The voltage drop across the 8 Ω resistor is 20 V. What will be the current reading in the ammeter ? What is the e.m.f. of the battery?

Answer 1

Current through 8 Ω resistance = 20/8 = 2.5 A This current is divided into two parts at point A; one part going along path AC and the other along path ABC which has a resistance of 28Ω

I₂ = 2.5 x 11/(11 + 28) = 0.7

Hence, ammeter reads 0.7 A. 

Resistance between A and C = (28 × 11/39) ohm. 

Total circuit resistance = 8 + 11 + (308/39) = 1049/39Ω 

∴ E = 2.5 × 1049/39 = 67.3V