Question

. Two wires of resistances 3 and 2 ohms respectively are connected in parallel. The combination is

then connected in series with a third wire of resistance 4 ohm, when the circuit is completed with

battery main current flow in the circuit will be 0.5 Amp. Calculate the terminal voltage as well as emf.

The internal resistance of the battery is 0.8 ohm.​

Answer 1

1. combination of resistance = 1/3 + 1/2

= 6/5 = 1.2 ohm

2. then the third wire connected in series so,

1.2 + 4 = 6.2 ohm,

V = IR

V= 0.5 × 6.2

= 3.10 v

Explanation:

V= I × R

I = V/R