(a) In Fig. 27-18a, with,$\mathbf{\text{R1>R2}}$is the potential difference across more than, less$\mathbf{R}\mathbf{2}$than, or equal to that across$\mathbf{R}\mathbf{1}$?

(b) Is the current through resistor R2 more than, less than, or equal to that through resistor$\mathbf{R}\mathbf{1}$?

$\text{R1>R2}$

If the resistances are connected in parallel, then the voltage across the resistances is same only the current gets divided. And the value of the current depends only on the resistance because the voltage is same in parallel arrangement.

$\mathbf{V}\mathbf{=}\mathbf{IR}$,where, V is voltage, I is current and R is resistance.

As in parallel arrangement V is same so$\mathbf{I}\mathbf{=}\frac{1}{R}$means greater the resistance, less is current and vise- versa.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

$I=\frac{V}{R}$

Where, I is current, V is voltage, R is resistance.

Here, in$fig27-18a$, the resistance$R_1\&R_2$are in parallel arrangement. So, the voltage difference across them is same.

Hence, the potential difference across$R_2$ is same that of across $R_1$.

As the voltage is same in parallel arrangement, the current depends only on the resistance from Ohm’s law.

$I=\frac{V}{R}$

V is constant for parallel arrangement. So,

$I\propto \frac{1}{R}$

in $fig27-18a,$the resistance $R_1>R_2$hence,$I_2>I_1$ because the current is inversely proportional to the resistance.

Hence, the current across$R_2$ is more than that of across.$R_1$

Therefore, compare the current and the potential difference across the resistances using the concept of parallel arrangement of the resistances and the equation of Ohm’s law.