Problem — resistors in series

Problem

Three resistors $R_1 = 100\ \Omega$, $R_2 = 220\ \Omega$, $R_3 = 470\ \Omega$ are connected in series to a $9\ \text{V}$ battery. Calculate the current in the circuit and the voltage across each resistor.

Three resistors in series with the battery: same current in all of them.

Solution

Equivalent resistance — in series the resistances add up: $R_\text{eq} = R_1 + R_2 + R_3 = 100 + 220 + 470 = 790\ \Omega$

Current — it is the same in all resistors (series): $i = \frac{\mathcal{E}}{R_\text{eq}} = \frac{9\ \text{V}}{790\ \Omega} \approx 1{,}14\cdot 10^{-2}\ \text{A} = 11{,}4\ \text{mA}$

Voltages across each — Ohm’s law on each resistor ($V_k = i\,R_k$): $V_1 = i\,R_1 = 0{,}0114 \cdot 100 \approx 1{,}14\ \text{V}$ $V_2 = i\,R_2 = 0{,}0114 \cdot 220 \approx 2{,}51\ \text{V}$ $V_3 = i\,R_3 = 0{,}0114 \cdot 470 \approx 5{,}35\ \text{V}$

Check — the sum of the drops must equal the e.m.f.: $1{,}14 + 2{,}51 + 5{,}35 \approx 9{,}0\ \text{V}$. ✓ The voltage is shared out in proportion to the resistances.

$\ev{i \approx 11{,}4\ \text{mA}; \quad V_1 \approx 1{,}14\ \text{V},\ V_2 \approx 2{,}51\ \text{V},\ V_3 \approx 5{,}35\ \text{V}}$

Links

Topics: Circuiti elettrici Concepts: Resistenze in serie e parallelo