What is the value of resistor $R$in FIGURE $EX28.16$?

We need to find that the value of resistor $R$.

When the resistors are connected without junction then they are called to be connected in series. As shown in the given figure, the $3$resistors are connected in series and as there are no junctions, the current in three resistors is the same $I=100mA=0.1\hspace{0.17em}A$.

For series resistors, the equivalent resistance for their combination is given by equation (28.15) in the form

$R_{eq}=R_1+R_2+R_3...\left(1\right)$

As we mentioned above, the current is the same for the resistors and is also the same for the circuit. So, using Ohm's law to get the current in the circuit

$I=\frac{∆V}{R_{eq}}...\left(2\right)$

Putting the value of $R_{eq}$into equation (2) and solve it for $R$

localid="1648895883564" $$\begin{gathered} I=\frac{∆V}{R_1+R_2+R} \\ {R}_1+R_2+R=\frac{∆V}{I} \\ R=\frac{∆V}{I}-(R_1+R_2)...\left(3\right) \end{gathered}$$

Now, putting the values for $I,∆V,R_1$and $R_2$into equation (3) to get $R$

$$\begin{gathered} R=\frac{∆V}{I}-\left(R_1+R_2\right) \\ =\frac{5.0V}{0.1A}-\left(10\Omega +15\Omega \right) \\ =25\Omega \end{gathered}$$