A resistor${\mathbf{R}}_1$is wired to a battery, then resistor$\mathbf{}{\mathbf{R}}_2$is added in series. Are

(a) the potential difference across${\mathbf{R}}_1$and

(b) the current${\mathbf{i}}_1$through${\mathbf{R}}_1$now more than, less than, or the same as previously?

(c) Is the equivalent resistance${\mathbf{R}}_1$of${\mathbf{R}}_1$and$\mathbf{}{\mathbf{R}}_2$more than, less than, or equal to${\mathbf{R}}_1$?

$R_1$And$R_2$ are connected in series.

When resistances are connected in series then current is same for both and voltage gets divided. And equivalent resistance is more than individual resistance as net resistance is sum of both resistors.

Formulae are as follow:

$R=R_1+R_2$

Where, R is resistance.

Initially, both terminals of resistors$R_1$ are wired to battery and potential difference is measured across the resistor.

When resistor$R_2$is connected in series with$R_1$the potential of the battery would be divided across both the resistors. So potential difference measured across$R_1$would be less.

Hence, potential difference across $R_1$is less

When $R_1$is connected with a potential, the current would be,

$i=\frac{V}{R_1}$

When $R_2$is added in series with,$R_1$ the equivalent resistance would increase as it is equal to sum of both the resistor values. Now potential of the battery is same, so the current through the$R_1$is equal to,

$i=\frac{V}{R_1+R_2}$

As, $R_1+R_2>R_1$the current in the$R_1$would be lesser.

Hence, current $i_1$through $R_1$is less.

When resistances are connected in series then equivalent resistance is found as follows:

$R_{eq}=R_1+R_2$

Since,$R_2>0,$the equivalent resistance is greater than$R_1$

Hence, equivalent resistance of $R_1$and $R_2$is greater than $R_1$

Therefore, using the concept of series connection, compare the potential difference, current and resistance when new resistance in series is added to the circuit.