Question

For assembling the voltage divider shown next, two of each of the following resistors are available: 250 $\Omega$,

500 $\Omega$, and 1.00 k$\Omega$.

(a) Describe a suitable combination of the resistors that would give the indicated voltages.

(b) What would be the IR drop across R3? Note: R3 could be more than one resistor.

(c) What current would be drawn from the source?

(d) What power is dissipated by the circuit?

Short answer

The resistors required to give the desired voltage are$R_1=500\Omega ,R_2=2k\Omega and{R}_3=2.5k\Omega$

.

Step-by-step solution

Given Information

Given the following diagram indicating the voltages across the resistors:

Part (a)

The resistores $R_{1,}{R}_{2}and{R}_3$are connected across the battery with volatge 10V in series.

The equivalent resistance is$R=R_1+R_2+R_3$.

The voltage drop across $R_1$is given by:

$$\begin{gathered} V_1=\frac{R_1}{R}V \\ 1=\frac{R_1}{R}10 \\ R=10R_1 \end{gathered}$$

Similarly:

$$\begin{gathered} V_2=\frac{R_2}{R}V \\ 4=\frac{R_2}{R}10 \\ 10R_2=4R \end{gathered}$$

it implies that:

$\frac{R_1}{R_2}=\frac{1}{4}$

here $R_1=500\Omega$so $R_2=2000\Omega$

Now,

$$\begin{gathered} \frac{V_2}{V_3}=\frac{R_2}{R_3} \\ \frac{V_2}{V-(V_1+V_2)}=\frac{R_2}{R_3} \\ \frac{4}{5}=\frac{2000}{R_3} \\ {R}_3=2500\Omega \end{gathered}$$

Part (b)

The IR drop across R₃ means the voltage drop across R₃.

Since the resistors are in series combination with the battery the voltage drop across R₃ is:

V₃ = V - (V₁+ V₂) = 5V

OR

$$\begin{gathered} V_3=\frac{R_3}{R}V \\ {V}_3=\frac{2500}{500+2000+2500}\times 10V \\ {V}_3=5V \end{gathered}$$

Part (c)

The total current I from the source will be:

$$\begin{gathered} I=\frac{V}{R} \\ I=\frac{10}{500+2000+2500}A \\ I=2mA \end{gathered}$$

Part (d)

The power dissipated by the circuit is:

P = I²(R₁ + R₂ + R₃)

$$\begin{gathered} P=4\times {10}^{-6}\left(5000\right)W \\ P=20\times {10}^{-3}W \end{gathered}$$