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

(a) the p.d. across the 4 Ω resistor

(b) the p.d. across the internal resistance of the cell,

(c) the p.d. across the R Ω or 2 Ω resistor, and

(d) the value of R.

(a) Given,

resistor = 4 Ω

I = 0.25 A

the p.d. across the 4 Ω resistor (V) = ?

Using Ohm's law

V = IR

Substituting the values in the formula above we get,

V = 0.25 x 4 = 1 V

Hence, the p.d. across the 4 Ω resistor (V) = 1 V

(b) Given,

internal resistance = 3 Ω

I = 0.25 A

the p.d. across the internal resistance of the cell = ?

Using Ohm's law

V = IR

Substituting the values in the formula above we get,

V = 0.25 x 3 = 0.75 V

Hence, the p.d. across the internal resistance (V) = 0.75 V

(c) Potential difference across R Ω or 2 Ω

V = V_net - V_{across 4 Ω} - V_{across 3 Ω}

Hence, we get,

V = 2 - 1 - 0.75 = 0.25 V

(d) The p.d. across resistor of R Ω = 0.25 V

Let the equivalent resistance of the resistors of 2 Ω and R Ω connected in parallel be R'_p

$\dfrac{1}{R'$p} = \dfrac{1}{R} + \dfrac{1}{2} \\[0.5em] \Rightarrow \dfrac{1}{R'p} = \dfrac{2 + R}{2R} \\[0.5em] \Rightarrow R'_p = \dfrac{2R}{2 + R} \\[0.5em]

Using Ohm's law

V = IR

0.25 = 0.25 x R'_p

Substituting the value of R'_p from above:

$0.25 = 0.25 \times \Big(\dfrac{2R}{2 + R}\Big) \\[0.5em] \Rightarrow \dfrac{2R}{2 + R} = 1 \\[0.5em] \Rightarrow 2R = 2 + R \\[0.5em] \Rightarrow 2R - R = 2 \\[0.5em] \Rightarrow R = 2 \\[0.5em]$

Hence, value of R = 2 Ω