Question

A resistance of $3ohms$ is connected in series with another resistance of $5ohms$. A potential difference of $6$ volts is applied across the combination. Calculate the current through the circuit and potential difference across the $3ohm$ resistance.

Answer 1

Given: A resistance of $3\Omega$ is connected in series with another resistance of $5\Omega$. A potential difference of $6V$ is applied across the combination.

To find the current through the circuit and potential difference across the $3\Omega$ resistance.

Solution:

as the resistance are connected in series

so $R=R_1+R_2=3+5=8\Omega$

now according to Ohm's law,

$R=IR$

$⟹I=\frac{V}{R}⟹I=\frac{6}{8}⟹I=0.75A$

So, $0.75A$ of current flow through circuit.

Now , Potential Difference across the $3\Omega$ resistance is

$V_1=I{R}_1⟹V_1=0.75\times 3⟹V_1=2.25V$

Now the corresponding current will be,

$I_1=\frac{V_1}{R_1}⟹I_1=\frac{2.25}{3}⟹I_1=0.75A=I$

and we know that the same current flow through each of the resistance in series.