A moving-coil galvanometer has a rectangular coil of 60 turns, 2.0 cm wide and 2.5 cm high…

A moving-coil galvanometer has a rectangular coil of $60$ turns, $2.0\ \text{cm}$ wide and $2.5\ \text{cm}$ high. The coil turns in the gap between curved pole pieces and a soft iron core, where the field is radial and has a size of $0.25\ \text{T}$ , as in the figure. A spring of torsion constant $5.0 \times 10^{-6}\ \text{N m/rad}$ holds it. The coil has a resistance of $40\ \Omega$ . Its scale has $30$ divisions, each $3.0^\circ$ of turn of the coil.

Top view of a moving-coil galvanometer. Two curved magnet poles, N on the left and S on the right, surround a round soft iron core. Arrows across the gaps point along radii, showing a radial field of 0.25 T. The coil lies across the core, with one side in each gap: the left side carries current out of the page and the right side into it. A note gives the coil as 60 turns, 2.0 cm wide and 2.5 cm high.

What is the current sensitivity of the galvanometer, in divisions per milliampere?
What current gives a full-scale deflection?
What shunt resistance turns it into an ammeter that reads up to $5.0\ \text{A}$ ?
What resistance in series turns it instead into a voltmeter that reads up to $15\ \text{V}$ ?
Two resistors of $10\ \text{k}\Omega$ each are joined in series to a $12\ \text{V}$ battery of negligible internal resistance. The voltmeter of (d) is connected across one of them. What does it read?
A second galvanometer is made in the same way, except that its coil has twice as many turns of the same wire on the same frame, so its resistance is doubled too. By what factor is its voltage sensitivity (divisions per volt) larger than that of the first?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. The radial field makes the deflection grow in step with the current. The shunt and the series resistance follow from the full-scale current. Then treat the voltmeter as one more resistor in the circuit.

Step 1. Sensitivity (a). $NAB = 60 \times 5.0 \times 10^{-4}\ \text{m}^2 \times 0.25\ \text{T} = 7.50 \times 10^{-3}\ \text{N m/A}$ . So $\dfrac{\theta}{I} = \dfrac{NAB}{k} = 1500\ \text{rad/A} = 85.9^\circ$ per mA, which is $28.6$ divisions per mA.

Step 2. Full scale (b). $30$ divisions need $I_g = \dfrac{30}{28.65}\ \text{mA} = 1.047\ \text{mA}$ .

Step 3. Shunt (c). At full scale the coil takes $I_g$ and the shunt the rest, at the same voltage: $S = \dfrac{I_gG}{I - I_g} = \dfrac{1.047 \times 10^{-3}\ \text{A} \times 40\ \Omega}{5.0\ \text{A} - 1.047 \times 10^{-3}\ \text{A}} = 8.38\ \text{m}\Omega$ .

Step 4. Series resistance (d). The whole voltmeter must have $R_V = \dfrac{V}{I_g} = 14.32\ \text{k}\Omega$ , so the series part is $R_V - G = 14.28\ \text{k}\Omega$ .

Step 5. The reading (e). The voltmeter and the $10\ \text{k}\Omega$ resistor in parallel make $5.89\ \text{k}\Omega$ . In series with the other $10\ \text{k}\Omega$ :

\displaystyle V = 12\ \text{V} \times \frac{5.89}{5.89 + 10} = 4.45\ \text{V}

Step 6. Twice the turns (f). The current sensitivity $NAB/k$ doubles. The voltage sensitivity is the current sensitivity divided by the resistance, and the resistance doubles too. So the factor is $1.00$ : no gain at all.

Answer: (a) $28.6$ divisions per mA. (b) $1.047\ \text{mA}$ . (c) $8.38\ \text{m}\Omega$ . (d) $14.28\ \text{k}\Omega$ . (e) $4.45\ \text{V}$ . (f) $1.00$ .

Watch out. Without the voltmeter, the voltage across each resistor is $6.0\ \text{V}$ . The meter reads less because it draws current: its resistance is not large compared with $10\ \text{k}\Omega$ .

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