Question

Three resistors of 4 $\Omega$, 6 $\Omega$ and 12 $\Omega$ are connected in parallel and the combination is connected in series with a $1.5\, V$ battery of $1\, \Omega$ internal resistance. The rate of Joule heating in the $4 \, \Omega$ resistor is

• A. 0.55 W
• B. 0.33 W
• C. 0.25 W
• D. 0.86 W

Answer 1

Answer: (C)

Resistors 4 $\Omega$, 6 $\Omega$ and 12 $\Omega$ are connected in parallel, its equivalent resistance (R) is given by
$\frac{1}{R} = \frac{1}{4} +\frac{1}{6}+\frac{1}{12} \Rightarrow R = \frac{12}{6} = 2\Omega$
Again disconnected to $1.5$ Vbattery whose
internal resistance r $= 1 \Omega$.
Equivalent resistance now,
$R'=2\Omega + 1\Omega = 3\Omega$
Current, $I_{total} = \frac{V}{R} = \frac{1.5}{3} = \frac{1}{2} A$
$I_{total} = \frac{1}{2} = 3x + 2x + x = 6x$
$\Rightarrow x = \frac{1}{12}$
$\therefore $ Current through $4\Omega$ resistor $= 3x$
$ = 3\times \frac{1}{12} = \frac{1}{4}A$
Therefore, rate of Joule heating in the $4\Omega$ resistor
$= I^{2} R = \left(\frac{1}{4}\right)^{2} \times 4 = \frac{1}{4} = 0.25 W$