Problem — three lamps A, B, C

Problem

Three lamps. An ideal battery $\mathcal{E} = 6{,}0\ \text{V}$ powers three identical lamps of resistance $R = 12\ \Omega$ as in the figure: lamps A and B in parallel, the group A$\parallel$B in parallel with C (so the three lamps are all in parallel across the battery).

Diagram: lamps A, B, C are three branches in parallel between the same two nodes, powered by the battery $\mathcal{E}$.

(a) Calculate the equivalent resistance. (b) If lamp B burns out (open circuit), does the brightness of A increase, decrease or stay the same? And that of C? (c) If instead C burns out, what happens to A and B? Answer first by reasoning, then check with calculations.

Solution

(a) Equivalent resistance. The three identical lamps are in parallel: $\frac{1}{R_\text{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \;\Rightarrow\; R_\text{eq} = \frac{R}{3} = \frac{12}{3} = 4{,}0\ \Omega$ Total current delivered: $i_\text{tot} = \mathcal{E}/R_\text{eq} = 6{,}0/4{,}0 = 1{,}5\ \text{A}$.

(b) and (c) Reasoning. The battery is ideal: it maintains $\mathcal{E} = 6{,}0\ \text{V}$ across the parallel combination whatever happens to the other branches. Each lamp always sees the same $6{,}0\ \text{V}$, independently of the others. So:

• If B burns out: A and C stay the same (they still see $6{,}0\ \text{V}$); only the total current delivered decreases.
• If C burns out: A and B stay the same, for the same reason.

In a parallel combination on an ideal source, switching off one branch does not affect the others.

Check with calculations. Each lamp, subjected to $6{,}0\ \text{V}$: $i_A = i_B = i_C = \frac{\mathcal{E}}{R} = \frac{6{,}0}{12} = 0{,}5\ \text{A}$ The current in each branch depends only on $\mathcal{E}$ and the $R$ of that branch, not on the others: removing one, the others keep $0{,}5\ \text{A}$ and hence the same brightness. The total current instead drops from $3 \cdot 0{,}5 = 1{,}5\ \text{A}$ to $2 \cdot 0{,}5 = 1{,}0\ \text{A}$.

$\ev{R_\text{eq} = 4{,}0\ \Omega \quad i_\text{tot} = 1{,}5\ \text{A} \quad i_A=i_B=i_C=0{,}5\ \text{A}\;;\quad \text{the other branches remain unchanged}}$

Links

Topics: Circuiti elettrici Concepts: Resistenze in serie e parallelo