Irodov 3.175: The Weaker Source Goes Silent – Amit & Deepali Sinha Foundation

Problem Irodov 3.175

Two sources with the same emf are connected in series; their internal resistances are $R_1$ and $R_2$ , with $R_2 > R_1$ . Find the external resistance $R$ for which the voltage across the terminals of one of the sources becomes zero, and say which source it is.

The two sources in series, each an emf E with its internal resistance (R_1 between the terminals A and B, R_2 between B and C), closed through the external resistance R. — The two sources in series, each an emf $\mathcal{E}$ with its internal resistance ( $R_1$ between the terminals $A$ and $B$ , $R_2$ between $B$ and $C$ ), closed through the external resistance $R$ .

Given

equal emfs $\mathcal{E}$ in series
internal resistances $R_1$ , $R_2$
$R_2 > R_1$

Find

R

for zero terminal voltage, and which source

Solution

Step 1The current in the loop

The two emfs act the same way round the single loop, so they add, and all three resistances are in series:

I = \frac{2\mathcal{E}}{R + R_1 + R_2} \tag{1}

Step 2The terminal voltage of each source

Between its terminals ( $A$ and $B$ for source 1, $B$ and $C$ for source 2) a source shows its emf minus the drop across its own internal resistance. For source $k$ ( $k = 1, 2$ ):

U_k = \mathcal{E} - IR_k

It vanishes when $IR_k = \mathcal{E}$ . With (1):

\frac{2\mathcal{E}R_k}{R + R_1 + R_2} = \mathcal{E}

Multiplying out and solving for the external resistance:

R = 2R_k - R_1 - R_2 \tag{2}

Step 3Which source can reach zero

For source 1, (2) asks for $R = R_1 - R_2$ , which is negative since $R_2 > R_1$ : no external resistance does it. For source 2 it gives a positive value:

R = R_2 - R_1

(3)

At this resistance the voltage between the terminals of the source with the larger internal resistance, $R_2$ , is zero.

Check

With $R = R_2 - R_1$ , (1) gives

I = \frac{2\mathcal{E}}{2R_2} = \frac{\mathcal{E}}{R_2},

so the internal drop of source 2 is exactly $\mathcal{E}$ . Source 1 then shows $\mathcal{E}(1 - R_1/R_2)$ , positive, and this must be the voltage across $R$ ; indeed $IR$ is $\mathcal{E}(R_2 - R_1)/R_2$ , the same. With equal internal resistances the formula asks for $R = 0$ , a short circuit, where both terminal voltages vanish together.

Remark

At this load the second source contributes nothing to the external circuit: its terminals are at one potential, so they could be joined by a wire without changing any current. Its emf is spent entirely on driving the current through its own internal resistance, and it acts, as far as the rest of the circuit can tell, like a plain piece of wire.