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Electrical Meters

Level 1 - Physics topic page in DC Circuits.

Principle

Electrical meters infer current or voltage by becoming part of the circuit. A good meter has a resistance that makes its disturbance small compared with the resistance being measured.

Notation

\(R_A\)
ammeter resistance
\(\mathrm{\Omega}\)
\(R_V\)
voltmeter resistance
\(\mathrm{\Omega}\)
\(R_g\)
galvanometer coil resistance
\(\mathrm{\Omega}\)
\(I_g\)
full-scale galvanometer current
\(\mathrm{A}\)
\(R_s\)
shunt resistance for an ammeter range
\(\mathrm{\Omega}\)
\(R_m\)
series multiplier resistance for a voltmeter range
\(\mathrm{\Omega}\)

Method

Derivation 1: Ammeter loading

An ammeter must be placed in series with the element whose current is being measured. Its resistance adds to the circuit resistance, so an ideal ammeter has zero resistance.

Measured current with meter inserted
\[I_{\mathrm{meas}}=\frac{V}{R+R_A}\]
Ideal ammeter condition
\[R_A\rightarrow 0\]

Derivation 2: Voltmeter loading

A voltmeter must be placed in parallel with the element whose potential difference is being measured. Its resistance forms an extra parallel path, so an ideal voltmeter has infinite resistance.

Loaded resistance of measured element
\[R_{\mathrm{loaded}}=\left(\frac{1}{R}+\frac{1}{R_V}\right)^{-1}\]
Ideal voltmeter condition
\[R_V\rightarrow\infty\]

Derivation 3: Extending a galvanometer

A galvanometer reaches full scale at a small current. A shunt creates an ammeter range by carrying the extra current. A series multiplier creates a voltmeter range by dropping the extra voltage.

Ammeter shunt condition
\[I_gR_g=(I-I_g)R_s\]
Shunt resistance
\[R_s=\frac{I_gR_g}{I-I_g}\]
Voltmeter range condition
\[V=I_g(R_g+R_m)\]
Multiplier resistance
\[R_m=\frac{V}{I_g}-R_g\]

Rules

Ammeter placement
\[\text{series connection; ideal }R_A=0\]
Voltmeter placement
\[\text{parallel connection; ideal }R_V=\infty\]
Parallel loading
\[R_{\mathrm{loaded}}=R\parallel R_V\]
Ammeter shunt
\[R_s=\frac{I_gR_g}{I-I_g}\]
Voltmeter multiplier
\[R_m=\frac{V}{I_g}-R_g\]

Examples

Question
A
\[1.0\,\mathrm{mA}\]
\[50\,\Omega\]
galvanometer is converted to a
\[1.0\,\mathrm{A}\]
ammeter. Find the shunt.
Answer
\[R_s=\frac{(1.0\times10^{-3})(50)}{1.0-1.0\times10^{-3}}=5.0\times10^{-2}\,\Omega\]

Checks

  • An ammeter in parallel with a low-resistance source is close to a short circuit.
  • A voltmeter with resistance comparable to the measured resistance changes the voltage it is trying to measure.
  • For low-resistance measurements, ammeter resistance can dominate the error.
  • For high-resistance measurements, voltmeter current can dominate the error.
  • Always distinguish the meter reading from the undisturbed circuit value.

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Kirchhoff Analysis

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RC Transients