Question

In the above question, Find individual power consumed.

Given $R_1=1k\Omega$ and $R_2=0.5k\Omega$

Answer 1

The equivalent resistance in the given circuit is

$R_{eq}=R_1+R_2$ ($\because$ series combination)

$\Rightarrow R_{eq}=1+0.5=1.5k\Omega =1500\Omega$

Now, the current through the $R_1$ and $R_2$ will be same as that in the equivalent resistance $R_{eq}.$

The voltage across it is $V=220V$

Applying Ohm's law for $R_{eq}$,

$V=I{R}_{eq}$

$\Rightarrow 220=I\times 1500$

$\Rightarrow I=\frac{220}{1500}=0.146A$

Now we apply Ohm's law individually on $R_1$ and $R_2$ to find $V_1$ and $V_2$ respectively,

$V_1=I{R}_1=0.14666\times 1000=146.66V$

$V_2=I{R}_2=0.14666\times 500=73.33V$

Now we can calculate the individual power by either of the formula

$P=\frac{V^2}{R}$ or $P=I^{2}R$

Let's use the first relation.

• Power consumed by resistance $R_1$ :

$P_1=\frac{(V_1{)}^2}{R_1}=\frac{(146.66{)}^2}{1000}=21.509W$

• Power consumed by resistance $R_2$ :

$P_2=\frac{(V_2{)}^2}{R_2}=\frac{(73.33{)}^2}{500}=10.755W$