Question

Two resistors of 4 ohm and 6 ohm are connected in parallel. The combination is connected across a 6 V battery of negligible resistance. Calculate the power supplied by the battery and the power dissipated in each resistor

Answer 1

Explanation:

(a) V= 6V

Req = 1/R1 + 1 /R2 = 1/4 + 1/6 = 5/12 = 12/5 ohm = 2.4 ohm

According to ohm's law ,

V = IR

I = V/R

I = 6/2.4

I = 2.5 A

(b) 1) V = 6 V , R= 6 ohm , I = ?

I = V/R

I = 6/6

I = 1 A

2) V = 6V , R = 4 ohm , I = ?

I = V/R

I = 6 /4

I = 1.5 A

Answer 2

Given data,

$R_{1}=4\Omega$

$R_{2}=6\Omega$

$V=6V$

$\because$$R_{1}$ and $R_{2}$ are connected in parallel

Equivalent resistance,

$\frac{1}{R} } =\frac{1}{4} +\frac{1}{6} =\frac{3+2}{12} =\frac{5}{12}$

$\therefore R=\frac{12}{5} =2.4\Omega$

(i) To find the power supplied by the battery,

We know that,

$P=\frac{V^{2} }{R} =\frac{6\times 6}{2.4} =\frac{45}{3}=15W$

Hence, power supplied by battery is $15W$

(ii) To find the power dissipated in each resistor,

We know that,

$P=VI$

1. current across resistor $R_{4\Omega}$,

$I_{4\Omega} =\frac{V}{R_{4\Omega} } =\frac{6}{4}=1.5A$

The power dissipated across the resistor $R_{4\Omega}$,

$P=VI_{4\Omega}=6\times 1.5=9W$

2. current across resistor $R_{6\Omega}$,

$I_{6\Omega} =\frac{V}{R_{6\Omega} } =\frac{6}{6}=1A$

The power dissipated across the resistor $R_{6\Omega}$,

$P=VI_{6\Omega}=6\times 1=6W$

Hence, the power dissipated across $R_{4\Omega}$ and $R_{6\Omega}$ is $9W$ and $6W$ respectively.