Question

Consider the circuit shown in the figure to find the power delivered to 6 Ohm resistance (in W). Given that Vs= 30

Answer 1

66.7 Watts

Explanation:

Let $R_{1}=1.0$ ohms, $R_{2}=3.0$ohms and $R_{3}=6.0\:$ohms. Since $R_{2}$ and $R_{3}$ are in parallel, their combined resistance $R_{23}$ is given by

$\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

or

$R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms$

The total current flowing through the circuit I is given

$I=\dfrac{V_{s}}{R_{Total}}$

where

$R_{Total}=R_{1}+R_{23}= 3.0\:ohms$

Therefore, the total current through the circuit is

$I=\dfrac{30\:V}{3.0\:ohms}=10\:A$

In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor $V_{1}$:

$V_{1}=(10\:A)(1.0\:ohms)=10\:V$

This means that voltage drop across the 6-ohm resistor $V_{3}$ is 20 V. The power dissipated P by the 6-ohm resitor is given by

$P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W$