A battery of e.m.f. 15 V and internal resistance 3 Ω is connected to two resistors 3 Ω and 6 Ω connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the current in 3 Ω resistor (d) the current in 6 Ω resistor.

(a) Given,

e.m.f. = 15 V

internal resistance r = 3 Ω

current through battery = ?

If R_p is the equivalent resistance of resistors 3 Ω and 6 Ω connected in parallel, then

$\dfrac{1}{R$p} = \dfrac{1}{3} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{2 + 1}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{3}{6} \\[0.5em] \dfrac{1}{Rp} = \dfrac{1}{2} \\[0.5em] \Rightarrow R_p = 2 Ω

From relation,

ε = I (R + r)

Substituting the value in the formula above we get,

15 = I(2 + 3)
⇒ 15 = I x 5
⇒ I = 15 / 5 = 3 A

Hence, current through the battery = 3 A

(b) Potential difference between the terminals of the battery = ?

Using Ohm's law

V = IR

R = 2 Ω

I = 3 A

Substituting the values in the formula above we get,

V = 3 x 2 = 6 V

Hence, potential difference between the terminals of the battery = 6 V

(c) Current in 3 Ω resistor = ?

Using Ohm's law

V = IR

R = 3 Ω

V = 6 V

I = ?

Substituting the values in the formula above we get,

$6 = I × 3 \\[0.5em] I = \dfrac{6}{3} \\[0.5em] \Rightarrow I = 2 A$

Hence, current in 3 Ω resistor is 2 A

(d) Current in 6 Ω resistor = ?

Using Ohm's law

V = IR

R = 6 Ω

V = 6 V

I = ?

Substituting the values in the formula above we get,

6 = I × 6
⇒ I = 6 / 6 = 1 A

Hence, current in 6 Ω resistor is 1 A