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 = ${\displaystyle \frac{1}{4}}$ 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 ${\displaystyle [(\text{r}+4\Omega )+{(\frac{1}{2}+\frac{1}{\text{R}})}^{-1}]}$

2 = 0.25 ${\displaystyle [7+(2\frac{\text{R}}{\text{R}}+2)]}$

or, ${\displaystyle \frac{2\text{R}}{\text{R}+2}=1}$

or 2R = R + 2

or R = 2 Ω