Question

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.

Answer 1

Given,
Resistance, R_A = 4Ω
Resistance, R_B = 6Ω
Voltage, V = 6V
(i) As resistance are connected in parallel

Equivalent Resistance = ${\displaystyle \frac{1}{R}=\frac{1}{R_A}+\frac{1}{R_B}}$

                                      ${\displaystyle \frac{1}{R}=\frac{1}{4}+\frac{1}{6}=\frac{1}{24}}$

                                    R  = 2.4 Ω

As  Power ,                 p= ${\displaystyle \frac{V^2}{R}}$

                                    = ${\displaystyle \frac{{\left(6\right)}^2}{2.4}}$ = 15 W

(ii) Power dissipation across each resistor, p = VI 

Current across resistor  ${\displaystyle R_A,I_A}$ = ${\displaystyle \frac{V}{R_A}}$

                                                ${\displaystyle I_A=\frac{6}{4}=1.5A}$

Power dissipation across resistor RA,

P = ${\displaystyle V{I}_A}$ = 6  x 1.5 = 9w 

(ii) Current across resistor  ${\displaystyle R_B,I_B=\frac{V}{R_B}}$

                                         ${\displaystyle I_B=\frac{6}{6}=1A}$

Power dissipation across resistor ${\displaystyle R_B,}$

                                                     P = ${\displaystyle V{I}_B=6\times 1=6}$w