Potential Divider Circuits

A potential divider circuit can be seen below. This circuit is made of a power source Vin and two fixed resistors. It also has a power output across $R_1$.

The potential divider circuit works by splitting the input voltages between the two resistors in a ratio of their resistances. For example, if both resistors were the same value for resistance, each would get half of the share of the input voltage. 

An equation can be used to calculate the output voltage in circuits like the one above. This is not on your data sheet so worth noting and remembering:

$V_{text{out}}=dfrac{R_1}{R_t}times V_{text{in}}$

The equation can be adjusted so that the top value for resistance is the resistor of which the output voltage is over.

Example: Calculate the voltage output across the $200   Omega$ resistor.

[2 marks]

$begin{aligned} bold{V}_{textbf{out}} &= bold{dfrac{R_1}{R_2} times V}_{text{in}} &= dfrac{200}{200+700} times 6 &= bold{1.33} textbf{ V} end{aligned}$

We can do a quick logical check of this. The smaller resistor will always receive the smallest share of the input voltage. So our answer looks good as $R_1$ is much smaller than $R_2$ so will only receive a small share of the input voltage.

For the output of a potential divider circuit to be variable, variable resistors, LDRs and thermistors may be included in the circuit. This is particularly useful if you want the circuit to control an outcome such as a light switching on or a fan switching on. 

The circuit above includes a thermistor. Remember, a thermistor is a variable resistor whose resistance depends upon the temperature. As the temperature increases, the resistance decreases. 

Example: In this potential divider circuit, what would happen to $R_1$, $R_2$ and $V_{text{out}}$ as the temperature of the thermistor increases?

[3 marks]

Firstly, as the temperature increases, $bold{R_2}$ decreases. 

As $R_1$ is a fixed resistor, its resistance remains the same. 

Finally, as $R_2$ has decreased, $R_1$ is now a larger part of the ratio of resistances and will therefore, receive a larger proportion of the input voltage. So $bold{V}_{textbf{out}}$ increases.