Question

In the following figure, a voltmeter of resistance $R_{V} = 300 Ω$ and an ammeter of resistance $R_{A} = 3.00 Ω$ are being used to measure a resistance $R$ in a circuit that also contains a resistance $R_{0} = 100 Ω$ and an ideal battery of emf $ε = 28.5 V$ . Resistance $R$ is given by $R = V / i$ , where $V$ is the voltmeter reading and $i$ is the current in resistance $R$ . However, the ammeter reading is not $i$ but rather $i^{'},$ which is $i$ plus the current through the voltmeter. Thus, the ratio of the two meter readings is not $R$ but only an apparent resistance $R^{'} = V / i^{'}$ . If $R = 85.0 Ω$ , what are (a) the ammeter reading, (b) the voltmeter reading, and (c) $R^{'}$ ? (d) If $R_{V}$ is increased, does the difference between $R^{'}$ and $R$ increase, decrease, or remain the same?

Accepted Answer

The current through resistance $R$ is $i$ , the current through the ammeter is $i^{'}$ . Then the current through voltmeter is $i^{'} - i$ .

The resistor $R$ , voltmeter $V$ are connected in parallel.

The resistance offered by voltmeter is $R_{v}$ .

The net resistance $ε = 28.5 V$

$R' = \frac{R R_{v}}{R + R_{v}}$

The resistance offered by ammeter, $R_{A}$

In the given circuit the equivalent resistance, $R_{eq} = R' + R_{A} + R_{o}$

$R_{eq} = \frac{R R_{v}}{R + R_{v}} + R_{A} + R_{o}$

(a) The ammeter reading

$i^{'} = \frac{E}{R_{eq}} = \frac{E}{\frac{R R_{v}}{R R_{v}} + R_{A} + R_{ο}}$

Since, $E = 28.5 V, R = 85, R_{A} = 3, R_{ο} = 100$

$i^{'} = \frac{28.5 V}{\frac{(85) (300)}{(85 + 300)} + 3 + 100}$

$i^{'} = 1.68 A$

(b) The voltmeter reading is:

$V = E - i^{'} (R_{A} + R_{O})$

$V = 28.5 - (1.68 \times 10^{- 1}) (103)$

$= 28.5 - 17.3$

$= 11.2 V$

(c) The apparent resistance

$R^{'} = \frac{V}{i^{'}}$

$= \frac{11.2 V}{1.68 A}$

$= 6.6$ ohm

(d) If the resistance $R_{v}$ across voltmeter is increased the difference between $R$ and $R^{'}$ decreases. In fact $R^{'} \to R$ as $R_{r} \to \infty$ .

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