Question

The circuit diagram Fig shows three resistors 2 Ω, 4 Ω and R Ω connected to a battery of e.m.f. 2 V and internal resistance 3 Ω. If main current of 0.25 A flows through the circuit, find:

1. the p.d. across the 4 Ω resistor,
2. the p.d. across the internal resistance of the cell,
3. the p.d. across the R Ω or 2 Ω resistors
4. the value of R.

Answer 1

Current I = 0.25 A = `1/4` A,

(i) Potential difference across the 4Ω resistor = current through the resistor × its resistance

= 0.25 × 4

= 1V

(ii) Potential difference across internal resistance= current through internal resistance × internal resistance

= 0.25 × 3

= 0.75V

(iii) P.d. across the R Ω or 3 Ω resistor= emf - (p.d. across 4Ω resistor + p.d. across internal resistance)

= 2 - (1 + 0.75)

= 0.25V

(iv) We know, e = I `[("r" + 4 Ω) + (1/2 + 1/"R")^-1]`

2 = 0.25 `[7 + (2"R"/"R" + 2)]`

or, `(2"R")/("R" + 2) = 1`

or 2R = R + 2

or R = 2 Ω