Problem — voltage divider

Problem

Two resistors $R_1 = 1\ \text{k}\Omega$ and $R_2 = 4\ \text{k}\Omega$ are in series with an input voltage $V_\text{in} = 10\ \text{V}$. (a) Calculate the voltage $V_2$ across $R_2$. (b) Calculate the power dissipated in each resistor. (c) What happens if an ideal voltmeter is connected across $R_2$ to measure $V_2$?

Voltage divider: $V_2$ is taken across $R_2$.

Solution

(a) Voltage $V_2$ — the two resistors in series are traversed by the same current, so the voltage splits in proportion to the resistances (divider formula): $V_2 = V_\text{in}\,\frac{R_2}{R_1 + R_2} = 10\ \text{V} \cdot \frac{4\ \text{k}\Omega}{1 + 4\ \text{k}\Omega} = 10 \cdot \frac{4}{5} = 8\ \text{V}$

(b) Power in each resistor — the current in the circuit is $i = \frac{V_\text{in}}{R_1 + R_2} = \frac{10\ \text{V}}{5\ \text{k}\Omega} = 2\ \text{mA}$ Joule dissipation ($P = R\,i^2$, with $i = 2\cdot 10^{-3}\ \text{A}$): $P_1 = R_1\,i^2 = 1000 \cdot (2\cdot 10^{-3})^2 = 1000 \cdot 4\cdot 10^{-6} = 4\ \text{mW}$ $P_2 = R_2\,i^2 = 4000 \cdot 4\cdot 10^{-6} = 16\ \text{mW}$ (check: $P_1 + P_2 = 20\ \text{mW} = V_\text{in}\,i = 10\cdot 2\cdot 10^{-3}$ ✓)

(c) Ideal voltmeter — an ideal voltmeter has infinite resistance: connected across $R_2$ it draws no current and does not alter the circuit. The current stays $2\ \text{mA}$, the split does not change and the instrument reads exactly $V_2 = 8\ \text{V}$. (A real voltmeter, with finite resistance, would be placed in parallel with $R_2$, lowering its equivalent resistance and slightly reducing the measured voltage: this is the loading effect of the instrument.)

$\ev{V_2 = 8\ \text{V}; \quad P_1 = 4\ \text{mW}, \quad P_2 = 16\ \text{mW}}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Resistenze in serie e parallelo · Potenza