Problem — ranking batteries in series

Problem

Ranking: batteries in series. Four combinations of ideal batteries power the same lamp of $R = 10\ \Omega$: (A) one battery of $1{,}5\ \text{V}$. (B) two batteries of $0{,}75\ \text{V}$ in series. (C) two batteries of $1{,}5\ \text{V}$ in series. (D) three batteries of $1{,}5\ \text{V}$ in series. Rank from smallest to largest the power dissipated in the lamp.

Solution

Reasoning. Batteries in series add their EMFs. The power dissipated in the lamp depends only on the total voltage applied: $P = \frac{V_\text{tot}^2}{R}$ So it’s enough to compare the total voltages (then squared).

Total voltages:

• A: $V = 1{,}5\ \text{V}$
• B: $V = 2 \cdot 0{,}75 = 1{,}5\ \text{V}$
• C: $V = 2 \cdot 1{,}5 = 3{,}0\ \text{V}$
• D: $V = 3 \cdot 1{,}5 = 4{,}5\ \text{V}$

Powers (with $R = 10\ \Omega$): $P_A = P_B = \frac{1{,}5^2}{10} = 0{,}225\ \text{W}, \quad P_C = \frac{3{,}0^2}{10} = 0{,}90\ \text{W}, \quad P_D = \frac{4{,}5^2}{10} = 2{,}025\ \text{W}$

A and B have the same total voltage, so the same power; C and D grow with the square.

$\ev{P_A = P_B < P_C < P_D}$

Links

Topics: Circuiti elettrici Concepts: Forza elettromotrice Skills: Ragionamento di ranking