The heart of every analogue meter
A galvanometer is a coil of fine wire that can turn between the poles of a magnet. A current through the coil makes it turn, and a pointer turns with it, further for a larger current. How the magnet turns the coil belongs to the chapter Magnetic Force on Currents and the Magnetic Dipole. Here we need only two numbers: the coil's resistance \(G\), and the current \(I_g\) that swings the pointer to full scale.
How much it swings for a given current is its current sensitivity, \(\theta/I\), in divisions per milliampere. How much it swings for a given voltage across it is its voltage sensitivity, \(\theta/V = \theta/(IG)\). A galvanometer of \(50\,\Omega\) that deflects \(20\) divisions for \(1\,\text{mA}\) has a current sensitivity of \(20\,\text{div/mA}\). The \(1\,\text{mA}\) needs \(50\,\text{mV}\) across it, so its voltage sensitivity is \(20/50 = 0.4\,\text{div/mV}\).
The Class 12 practical uses the reciprocal, the figure of merit: the current per division. For the galvanometer above it is \(1/20 = 0.05\,\text{mA}\) per division. A good galvanometer has a small figure of merit and a large sensitivity.
Why can the bare coil not measure amperes or volts? Its wire is fine, and \(I_g\) is often a milliampere or less. A current of a few amperes would drive the pointer hard against its stop and burn the coil. Full scale needs only \(I_gG\) across it, \(0.1\,\text{V}\) for \(1\,\text{mA}\) and \(100\,\Omega\). So a few volts across it would wreck it too.
A galvanometer of resistance \(100\,\Omega\) reads full scale, \(30\) divisions, at \(1\,\text{mA}\). Find its current sensitivity, its voltage sensitivity and its figure of merit.
Current sensitivity: \(30/1 = 30\,\text{div/mA}\).
Full scale needs \(1\,\text{mA} \times 100\,\Omega = 100\,\text{mV}\), so the voltage sensitivity is \(30/100 = 0.3\,\text{div/mV}\).
Figure of merit: \(1/30\,\text{mA}\), about \(0.033\,\text{mA}\) per division.
Values used in this lesson: the galvanometer of the workbook's third sheet, \(G = 100\,\Omega\) with \(I_g = 1\,\text{mA}\), and a second one of \(50\,\Omega\) with \(2\,\text{mA}\).
A shunt makes an ammeter
An ammeter must carry the whole current it measures, so it goes in series with the branch. To read \(1\,\text{A}\) with a coil that takes only \(1\,\text{mA}\), give the rest of the current another road. A small resistance, the shunt \(S\), is joined in parallel with the coil. Most of the current takes the shunt, and a fixed small share passes through the coil.
The coil and the shunt are in parallel, so they have the same voltage across them. At full scale the coil carries \(I_g\) and the shunt the rest, \(I - I_g\). So \(I_gG = (I - I_g)S\), and the shunt must be \(S = I_gG/(I - I_g)\).
For \(G = 100\,\Omega\) and \(I_g = 1\,\text{mA}\), a \(1\,\text{A}\) range needs \(S = 0.1/0.999 \approx 0.1\,\Omega\). In the figure above the shunt carries \(999\,\text{mA}\) of every ampere, \(99.9\,\%\), and the coil \(0.1\,\%\). The ammeter's own resistance is \(GS/(G + S) = 0.1\,\Omega\). A \(5\,\text{A}\) range needs \(0.1/4.999 \approx 0.02\,\Omega\): a larger range needs a smaller shunt.
The split is fixed by the ratio of the resistances, so the coil always carries the same fraction \(S/(S + G)\) of the current. Half an ampere through the \(1\,\text{A}\) ammeter sends \(0.5\,\text{mA}\) through the coil, and the pointer stands at half scale. The scale stays linear. The shunt trades sensitivity for range: if full scale is \(20\) divisions, one division now means \(1/20\,\text{A} = 50\,\text{mA}\).
When \(S\) is much smaller than \(G\), the ammeter's resistance \(GS/(G + S) = S/(1 + S/G)\) is almost exactly \(S\). A shunt of \(1\,\Omega\) on a \(100\,\Omega\) coil lets \(1/101\) of the current, about \(1\,\%\), through the coil. A second \(0.1\,\Omega\) joined across a \(0.1\,\Omega\) ammeter takes half of every current, so the meter's full scale becomes \(2\,\text{A}\).
A galvanometer of resistance \(50\,\Omega\) reads full scale at \(2\,\text{mA}\). Find the shunt that makes it an ammeter of range \(2\,\text{A}\), and the resistance of that ammeter.
At full scale the coil carries \(2\,\text{mA}\) and the shunt the remaining \(1.998\,\text{A}\).
\(S = I_gG/(I - I_g)\); the ammeter is \(G\) and \(S\) in parallel.
\(S = 0.002 \times 50/1.998\) \(= 0.1/1.998 \approx 0.05\,\Omega\). In parallel with \(50\,\Omega\) this is still about \(0.05\,\Omega\).
A shunt of about \(0.05\,\Omega\) gives a \(2\,\text{A}\) ammeter of resistance about \(0.05\,\Omega\).
A galvanometer of resistance \(40\,\Omega\) reads full scale at \(5\,\text{mA}\). What shunt makes it an ammeter of range \(2\,\text{A}\), in ohms?
A series resistance makes a voltmeter
A voltmeter must have the voltage it measures across it, so it goes in parallel with the element. To read \(10\,\text{V}\) with a coil that needs only \(0.1\,\text{V}\), a large resistance, the multiplier, is joined in series with the coil. At full scale the current through both is \(I_g\), so \(V = I_g(G + R)\) and \(R = V/I_g - G\).
For \(G = 100\,\Omega\) and \(I_g = 1\,\text{mA}\), a \(10\,\text{V}\) range needs \(10/0.001 - 100 = 9900\,\Omega\). The voltmeter's resistance is \(10\,000\,\Omega\). Take the voltmeter chip in the figure above: at \(5\,\text{V}\) the coil carries \(0.5\,\text{mA}\), and the \(9900\,\Omega\) drops \(4.95\,\text{V}\), \(99\,\%\) of the voltage. The same galvanometer with \(9900\,\Omega\) across a \(5\,\text{V}\) cell therefore stands at half scale.
A galvanometer of \(50\,\Omega\) and \(2\,\text{mA}\) needs \(5/0.002 - 50 = 2450\,\Omega\) for a \(5\,\text{V}\) range. For \(20\,\text{V}\) it needs \(9950\,\Omega\), so \(7500\,\Omega\) more must be added in series. A meter with two ranges uses two resistances in a chain. With the \(100\,\Omega\), \(1\,\text{mA}\) coil, \(R_1 = 9900\,\Omega\) gives \(10\,\text{V}\). Adding \(R_2 = 100/0.001 - 100 - 9900\) \(= 90\,000\,\Omega\) gives \(100\,\text{V}\).
A galvanometer of resistance \(40\,\Omega\) reads full scale at \(5\,\text{mA}\). What resistance in series makes it a voltmeter of range \(15\,\text{V}\), in ohms?
Which statement about ideal meters is correct?
Every meter changes the circuit
A meter becomes part of the circuit it measures. An ammeter in series adds its own resistance. A \(12\,\text{V}\) cell of negligible internal resistance drives \(12/5 = 2.4\,\text{A}\) through \(5\,\Omega\). Insert an ammeter of \(1\,\Omega\), and the current falls to \(12/6 = 2\,\text{A}\). The meter reads \(2\,\text{A}\), \(16.7\,\%\) below the current that flowed without it.
A voltmeter in parallel draws current. Two \(1000\,\Omega\) resistors in series on \(12\,\text{V}\) each have \(6\,\text{V}\) across them. Put a \(1000\,\Omega\) voltmeter across one. The pair becomes \(500\,\Omega\), the current rises to \(12/1500 = 8\,\text{mA}\), and the meter reads \(8\,\text{mA} \times 500\,\Omega = 4\,\text{V}\). It has changed the voltage it came to measure.
In Fig 6.1 a meter rated \(1000\,\Omega\) per volt, used on its \(10\,\text{V}\) range, has \(10\,\text{k}\Omega\). Across the lower resistor it makes \(909\,\Omega\) and reads \(5.7\,\text{V}\), much nearer the truth. So an ammeter should have as little resistance as possible, and a voltmeter as much as possible. An ideal meter has zero resistance as an ammeter and infinite resistance as a voltmeter.
Two \(5000\,\Omega\) resistors are joined in series across a \(10\,\text{V}\) supply of negligible internal resistance. A voltmeter of resistance \(5000\,\Omega\) is joined across one of them. What does it read, in volts?
An ammeter must never be joined straight across a cell. Its resistance is a fraction of an ohm, so the current is limited only by the cell's internal resistance. A \(1.5\,\text{V}\) cell of \(0.5\,\Omega\) would push \(1.5/0.6 = 2.5\,\text{A}\) through a \(0.1\,\Omega\) ammeter of \(1\,\text{A}\) range, far past its full scale. The shunt and the coil overheat and may burn.
A student joins an ammeter directly across the terminals of a cell. Why is this dangerous for the meter?
An ammeter has a very small resistance, a fraction of an ohm. Joined across a cell, it forms a circuit whose only other resistance is the cell's internal resistance. The current is then many amperes, far above the meter's full-scale value, so the shunt and the coil overheat and may burn.
An analogue multimeter holds one galvanometer and a rotary switch. On a current range the switch joins one of several shunts across the coil; the smaller the shunt, the larger the range. On a voltage range it puts one of several series resistances in the chain. A meter rated \(20\,\text{k}\Omega\) per volt, a common rating, has a coil of \(50\,\mu\text{A}\) full scale. With a coil of \(100\,\Omega\) its \(10\,\text{V}\) range needs about \(200\,\text{k}\Omega\) in series, and a \(1\,\text{A}\) range needs a shunt of only about \(5\,\text{m}\Omega\).
Early galvanometers used a magnetised needle turned by a current in a coil around it, and nearby iron or the Earth's field upset them. In 1882 the French physician and physicist Jacques-Arsène d'Arsonval, working with Marcel Deprez, turned the design round. A light coil hung between the poles of a fixed permanent magnet and turned itself. The moving-coil galvanometer was steady and sensitive, and every analogue ammeter and voltmeter since has been built round it.
- Joining the shunt in series with the coil.The shunt must share the coil's voltage and take most of the current, so it goes in parallel.
- Forgetting to subtract \(G\) when finding the series resistance.\(V/I_g\) is the whole voltmeter's resistance. The added resistance is \(V/I_g - G\).
- Thinking that a real meter reads the value that was there before it was joined.An ammeter lowers the current by adding resistance; a voltmeter lowers the voltage by drawing current.
- Using an ammeter to check whether a cell is alive.Across a cell an ammeter nearly shorts it and burns. Use a voltmeter, which draws almost nothing.