Question

(i) Design a voltage divider to produce an open-circuit output voltage of 4V from a voltage source of 10V. The current flowing in the circuit must be smaller than 1mA. (ii) If you connect a load of 1kΩ into the output of this voltage divider, what is the voltage drop across the load?

Answer 1

(i) The circuit is shown on the picture below.

Here $U_{in} = 10V$ and $U{out} = 4V$.

The current which flows through this circuit is:

$$I = \dfrac{U_{in}}{R_1 +R_2}$$

Thus, the total resistance should be:

$$R_1 + R_2 > \dfrac{U_{in}}{I} = \dfrac{10V}{1\times 10^{-3}A} = 10\times10^3\Omega = 10k\Omega$$

Let's choose $R_1 = 10k\Omega$. Then, according the voltage divider formula:

$$U_{out} = U_{in}\dfrac{R_2}{R_1 + R_2}$$

obtain the $R_2$:

$$R_2 = \dfrac{R_1U_{out}}{U_{in} - U_{out}} = \dfrac{10\cdot4}{10-4} \approx 6.67 k\Omega$$

(ii) The circuit is shown on the picture below.

Let's conside resistors $R_2$ and $R_{load}$ as the lower leg of the voltage divider. Their total resistance will be (two resistors in parallel):

$$R = \dfrac{R_2R_{load}}{R_2 + R_{load}} \approx0.87k\Omega$$

Again, according the voltage divider formula:

$$U_{out} = U_{in}\dfrac{R}{R_1 + R} = 10\dfrac{0.87}{10 + 0.87} = 0.8V$$

Answer. (i) R1 = 10k, R2 = 0.87k, (ii) 0.8V.