91Ó°ÊÓ

We need to find the magnitude of the potential difference across each resistor.

As we can see the two resistors and the battery are connected in series. Due to the charge conservation, the current is same across this circuit but the voltage is not same. For equivalent resistance. The equivalent resistance $R_{eq}$in the circuit. The strategy is to calculate the combination of the resistors in series, so, their combination $R_{eq}$could be calculated by taking the summation of both resistor

$R_{eq}=R_1+R_2...\left(1\right)$

Putting values for $R_1$and $R_2$into equation (1) to get the equivalent resistance $R_{eq}$.

$$\begin{gathered} R_{eq}=R_1+R_2 \\ =40\mathrm{Î\copyright }+60\mathrm{Î\copyright } \\ =100\mathrm{Î\copyright } \end{gathered}$$

Ohm's law states that the current $I$flows through a resistance due to potential difference$V$between the resistance is

$I=\frac{V}{R_{eq}}$

Calculating the current through the two resistors using the values for $R_{eq}$and $V$of the battery by

$I=\frac{V}{R_{eq}}=\frac{50\mathrm{Î\copyright }}{100\mathrm{Î\copyright }}=0.5A$

This is the current for both resistors. To get the potentials differences across each resistor we use Ohm's law again by multiplying the current by the value of the resistance as next

$∆V_{40\mathrm{Î\copyright }}=(0.5A)\left(40\mathrm{Î\copyright }\right)=24V$

And the highest resistance $60\mathrm{Î\copyright }$has

$∆V_{60\mathrm{Î\copyright }}=(0.5A)\left(60\mathrm{Î\copyright }\right)=30V$