Irodov 3.173: Two Readings Fix the Voltmeter – Amit & Deepali Sinha Foundation

Problem Irodov 3.173

An ammeter and a voltmeter are connected in series across a battery of emf $\mathcal{E} = 6.0$ V. A certain resistance is then connected in parallel with the voltmeter; as a result the voltmeter reading falls by a factor $\eta = 2.0$ , while the ammeter reading rises by the same factor. Find the voltmeter reading after the resistance has been connected.

The battery E with the ammeter and the voltmeter in series, and the resistance R that is then connected across the voltmeter. — The battery $\mathcal{E}$ with the ammeter and the voltmeter in series, and the resistance $R$ that is then connected across the voltmeter.

Given

$\mathcal{E} = 6.0$ V
voltmeter reading falls $\eta$ -fold
ammeter reading rises $\eta$ -fold
$\eta = 2.0$

Find

the voltmeter reading

U'

after the connection

Solution

Step 1The loop before and after

Let $R_0$ be the total resistance of the loop outside the voltmeter: the ammeter’s together with the battery’s internal resistance, whatever they are. Before the connection the current is $I$ and the voltmeter reads $U$ ; after it, the ammeter reads $\eta I$ and the voltmeter $U' = U/\eta$ . The emf is the sum of the drops round the loop in both states, and the voltmeter reading is the drop across the part of the loop it spans (the voltmeter alone before, the voltmeter and $R$ in parallel after):

\begin{gathered} \mathcal{E} = IR_0 + U \\ \mathcal{E} = \eta IR_0 + \frac{U}{\eta} \end{gathered} \tag{1}

Step 2Eliminating the unknown loop resistance

Solve the first line of (1) for the drop $IR_0$ and substitute it into the second:

\begin{aligned} IR_0 &= \mathcal{E} - U\\[2pt] \mathcal{E} &= \eta(\mathcal{E} - U) + \frac{U}{\eta}\\[2pt] (\eta - 1)\,\mathcal{E} &= U\left(\eta - \frac{1}{\eta}\right) = U\,\frac{(\eta - 1)(\eta + 1)}{\eta} \end{aligned}

Dividing by $\eta - 1$ , which is not zero:

U = \frac{\eta\,\mathcal{E}}{\eta + 1} \tag{2}

Step 3The reading after the connection

After the connection: the resistance R in parallel with the voltmeter. — After the connection: the resistance $R$ in parallel with the voltmeter.

The new reading is $U' = U/\eta$ ; dividing (2) by $\eta$ :

U' = \frac{\mathcal{E}}{\eta + 1} = 2.0~\text{V}

(3)

Check

The shunt the data imply must be a real resistor. By (2) and the first line of (1), the voltmeter’s resistance is $U/I$ ; after the connection it passes $U'/R_V$ of the total $\eta I$ , and $R$ takes the rest at the voltage $U'$ :

\begin{aligned} R_V &= \frac{U}{I} = \eta R_0\\[2pt] R &= \frac{U'}{\eta I - U'/R_V} = \frac{U/\eta}{I(\eta - 1/\eta)} = \frac{R_V}{\eta^2 - 1} \end{aligned}

This is positive only for $\eta > 1$ , as it must be for a resistor added in parallel, which can only lower the reading; and as $\eta \to 1$ it tends to infinity (nothing connected), where (3) and (2) both give $\mathcal{E}/2$ , the two readings becoming equal.

Remark

The answer needs neither meter’s resistance, yet the data fix both ratios: here $R = R_V/3$ and $R_V = 2R_0$ . A typical voltmeter, with a resistance far above the ammeter’s, could not behave this way: the data describe a meter only twice as resistive as the rest of the loop.