Problem — series chain with different bulbs

Problem

Three bulbs are connected in series to a battery $V = 42\ \text{V}$: $L_1$ ($V_\text{nom} = 6\ \text{V}$, $R_1 = 60\ \Omega$), $L_2$ ($V_\text{nom} = 12\ \text{V}$, $R_2 = 30\ \Omega$), $L_3$ ($V_\text{nom} = 24\ \text{V}$, $R_3 = 120\ \Omega$). Rank the bulbs by the actual voltage across them and identify which one burns out first.

Solution

Current in the chain — in series the current is the same; the total resistance is $R_\text{tot} = R_1 + R_2 + R_3 = 60 + 30 + 120 = 210\ \Omega$ $i = \frac{V}{R_\text{tot}} = \frac{42\ \text{V}}{210\ \Omega} = 0{,}20\ \text{A}$

Actual voltage across each ($V_k = i\,R_k$): $V_1 = 0{,}20 \cdot 60 = 12\ \text{V}, \quad V_2 = 0{,}20 \cdot 30 = 6\ \text{V}, \quad V_3 = 0{,}20 \cdot 120 = 24\ \text{V}$

Ranking by actual voltage: $V_3 (24\ \text{V}) > V_1 (12\ \text{V}) > V_2 (6\ \text{V})$.

Which one burns out first — what matters is not the absolute voltage, but the ratio between actual and rated voltage (the relative overload): $L_1:\ \frac{V_1}{V_{1,\text{nom}}} = \frac{12}{6} = 2 \qquad L_2:\ \frac{6}{12} = 0{,}5 \qquad L_3:\ \frac{24}{24} = 1$

$L_1$ operates at double its rated voltage. In terms of power relative to the rated value, $P \propto V^2$ at fixed $R$ gives $L_1$ a factor $2^2 = 4$: it dissipates four times the power it is designed for. $L_1$ is the most stressed and burns out first; $L_3$ operates exactly at its rated value, $L_2$ is actually underpowered.

$\ev{V_3 > V_1 > V_2;\quad L_1 \text{ burns out first (at } 2\times V_\text{nom})}$

Links

Topics: Circuiti elettrici Concepts: Resistenze in serie e parallelo Skills: Ragionamento di ranking