91Ó°ÊÓ

Given :

In the following figure, $R_1>R_2$

Theory used :

Ohm's law, description of the relationship between current, voltage, and resistance. The amount of steady current through a large number of materials is directly proportional to the potential difference, or voltage, across the materials. That is

$∆V_R=I·R$

When current passes through a resistor, it dissipates energy, and the rate at which this energy is dissipated is the power. $P_R=I∆V_R$is the rate at which energy is transmitted from the current to the resistor.

We may get an alternate formula for the dissipated power from Ohm's law : $∆V_R=I·R$.

The current flowing through the resistors and the circuit is the same in series.

As a result, $I$is the same for $R_1\mathrm{and}{R}_2$. $P_R$is proportional to resistance, as said earlier. Because the current is the same and $R_1>R_2$, the power dissipated by $\underline{R}$ is greater than the power dissipated by $\underline{R_2}$.

So,$\boxed{{\mathit{P}}_{\mathbf{1}}\mathbf{>}\mathbf{}{\mathit{P}}_{\mathbf{2}}}$