In the above question, Find individual power consumed.

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

The equivalent resistance in the given circuit is

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

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

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 $V=220 \ V$

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

$V=IR_{eq}$

$\Rightarrow 220= I \times 1500$

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

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

$V_1=IR_1 =0.14666 \times 1000=146.66 \ V$

$V_2=IR_2 =0.14666 \times 500=73.33 \ V$

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

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

Let's use the first relation.

• Power consumed by resistance $R_1$ :

${P}_{1}=\dfrac{(V_1)^2}{R_1}= \dfrac{(146.66)^2}{1000}=21.509 \ W$

• Power consumed by resistance $R_2$ :

${P}_{2}=\dfrac{(V_2)^2}{R_2}= \dfrac{(73.33)^2}{500}=10.755 \ W$