The circuit in FIGURE $EX28.18$is called a voltage divider. What value of $R$will make $V_{out}=V_{in}/10$?

We need to find the value of $R$whichwill make $V_{out}=V_{in}/10$.

When the resistors are connected without any junctions then they are called to be connected in series. As shown in the given figure, $R$and $100\Omega$are connected in series. For a series of resistors, the equivalent resistance for their combination is given by equation (28.15) in the form

$R_{eq}=R+100\Omega$

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

$I=\frac{V_{in}}{R_{eq}}=\frac{V_{in}}{R+100\Omega }$

The potential difference across the resistor $100\Omega$is $V_{out}$and from Ohm's law it can be expressed as

$V_{out}=I\left(100\Omega \right)=\frac{V_{in}}{R+100\Omega }\left(100\Omega \right)$

When the out voltage is $V_{out}=\frac{V_{in}}{10}$, we can get the value of $R$by

$$\begin{gathered} \frac{Vin}{10}=\frac{V_{in}}{R+100\Omega }\left(100\Omega \right) \\ 10=\frac{R+100\Omega }{100\Omega } \\ R=1000\Omega -100\Omega \\ R=900\Omega \end{gathered}$$