Current Electricity — Question 28

(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 x $\Big(\dfrac{2\text{R}}{2 + \text{R}}\Big)$

⇒ $\dfrac{2\text{R}}{2 + \text{R}}$ = 1

⇒ 2R = 2 + R

⇒ 2R - R = 2

⇒ R = 2 Ω

Hence, value of R = 2 Ω