Question

Three resistors of 4 $\Omega$, 6 $\Omega$ and 12 $\Omega$ are connected in parallel and the combination is connected in series with a $1.5V$ 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,
$R^{\prime }=2\Omega +1\Omega =3\Omega$
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
$=I^{2}R={\left(\frac{1}{4}\right)}^2\times 4=\frac{1}{4}=0.25W$