91Ó°ÊÓ

We need to find the value of resistor R.

The two points and there are resistors connected at both ends of the points, we called that the resistors are connected in parallel. This is obvious when the resistors are aligned side by side. The potential difference across the resistors is the same in parallel connection and it is the same for the circuit.

The current in the circuit by $I=2A$and the potential difference is $â–³V=8V$. So, use Ohm's law to get the equivalent resistance in the circuit where Ohm's law states that the current $I$flows through a resistance due to the potential difference $â–³V$between the resistance.

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

Plug the values for $â–³V$and $I$into equation (1) to get $R_{eq}$by

$R_{eq}=\frac{V}{I}=\frac{8V}{2A}=4\mathrm{Î\copyright }$

The equation shows the equivalent resistance for the parallel connection in the form

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_N}$

Three resistors is parallel where $R_1=10,R_2=15and{R}_3=R$in equation (2) to get $R$by

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+ \frac{1}{R} \\ \frac{1}{R}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2} \\ R={\left(\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}\right)}^{-1} \end{gathered}$$

and put the values for $R_{eq},R_{2}and{R}_1$into equation(3) to get $R$by

$R={\left(\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}\right)}^{-1}$

$={\left(\frac{1}{4\mathrm{Î\copyright }}-\frac{1}{10\mathrm{Î\copyright }}-\frac{1}{15\mathrm{Î\copyright }}\right)}^{-1}$

localid="1648638572257" $R=12\mathrm{Î\copyright }$