Two resistors A and B of resistance 4 Ω and 6 Ω respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate : (i) the power supplied by the battery, (ii) the power dissipated in each resistor.

(i) Given,

Resistance, R_A = 4 Ω

Resistance, R_B = 6 Ω

Let equivalent resistance of the two resistors connected in parallel be R_P

$\dfrac{1}{R$P} = \dfrac{1}{4} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{RP} = \dfrac{3 + 2}{12} \\[0.5em] \dfrac{1}{RP} = \dfrac{5}{12} \\[0.5em] RP = \dfrac{12}{5} \\[0.5em] R_P = 2.4 Ω \\[0.5em]

Hence, equivalent resistance = 2.4 Ω

From relation

$P = \dfrac{V^2}{R} \\[0.5em]$

We get,

$P = \dfrac{6^2}{2.4} \\[0.5em] \Rightarrow P = 15 W$

Hence, power supplied = 15 W

(ii) Power dissipation across each resistor = ?

From relation, P = VI

Current across A (4 Ω) resistor —

I_A = 6 / 4 = 1.5 A

Power dissipation across A = VI_A

Substituting we get,

P = 6 x 1.5 = 9 W

Similarly,

Current across B ( 6 Ω) resistor —

I_B = 6 / 6 = 1 A

Power dissipation across B resistor = VI_B

Substituting the values in the formula above, we get,

P = 6 x 1 = 6 W

Hence, power dissipated across A = 9 W and across B = 6 W