Instruments and Meters

LessonAbout 1 h

Instruments and meters are electrical devices used to measure, monitor, and record physical electrical quantities such as current, voltage, resistance, power, and energy.

1. Ammeter

Ammeter is an electrical instrument connected in series to measure current in amperes.

Low internal resistance minimizes circuit loading effect
Shunt resistor extends ammeter current measuring range
Galvanometer coil carries small full-scale movement current

Ammeter Shunt Resistance and Multiplier

\[ I = I_{\text{sh}} + I_{\text{fs}} \]

\[ R_{\text{sh}} = \dfrac{I_{\text{fs}} R_g}{I_{\text{sh}}} \]

\[ \text{Multiplier} = \dfrac{I}{I_{\text{fs}}} = \dfrac{R_g + R_{\text{sh}}}{R_{\text{sh}}} \]

Where: \(I\) = total line current (A), \(I_{\text{fs}}\) = full-scale meter current (A), \(I_{\text{sh}}\) = shunt current (A), \(R_g\) = galvanometer coil resistance (\(\Omega\)), \(R_{\text{sh}}\) = shunt resistance (\(\Omega\))

EE Board Exam March 1998

Example. A DC ammeter has an internal resistance of \(0.1\ \Omega\). A shunt of \(1.010 \times 10^{-3}\ \Omega\) is connected to the ammeter. What is the multiplier of the setup?

Solution.

\[ \text{Multiplier} = \dfrac{R_g + R_{\text{sh}}}{R_{\text{sh}}} = \dfrac{0.1 + 0.00101}{0.00101} = 100 \]

Answer: \(100\)

2. Voltmeter

Voltmeter is an electrical instrument connected in parallel to measure potential difference in volts.

High internal resistance prevents drawing significant circuit current
Series multiplier extends voltmeter voltage measuring range
Sensitivity defined as reciprocal of full-scale current

Voltmeter Series Multiplier and Sensitivity

\[ E_{\text{fs}} = I_{\text{fs}}(R_g + R_{\text{se}}) \]

\[ R_{\text{se}} = \dfrac{E_{\text{fs}}}{I_{\text{fs}}} - R_g \]

\[ \text{Multiplier} = \dfrac{R_g + R_{\text{se}}}{R_g} \]

\[ \text{Sensitivity} = \dfrac{1}{I_{\text{fs}}} \]

Where: \(E_{\text{fs}}\) = full-scale voltage (V), \(I_{\text{fs}}\) = full-scale meter current (A), \(R_g\) = galvanometer coil resistance (\(\Omega\)), \(R_{\text{se}}\) = series multiplier resistance (\(\Omega\)), \(\text{Sensitivity}\) = voltmeter sensitivity (\(\Omega/\text{V}\))

EE Board Exam October 1992

Example. A certain \(50\ \mu\text{A}\) meter movement has a resistance of \(90\ \Omega\). What value of series multiplier is needed to make the instrument read \(5\ \text{V}\) maximum?

Solution.

\[ R_{\text{se}} = \dfrac{E_{\text{fs}}}{I_{\text{fs}}} - R_g = \dfrac{5}{50 \times 10^{-6}} - 90 = 99910\ \Omega = 99.91\ \text{k}\Omega \]

Answer: \(99.91\ \text{k}\Omega\)

3. Ohmmeter

Ohmmeter is an instrument equipped with an internal battery used to measure unknown resistance in ohms.

Internal DC battery supplies testing voltage to circuit
Zero-adjust resistor compensates for battery voltage drops
Non-linear scale reads zero at full-scale deflection

Ohmmeter Current and Resistance Relations

\[ I_{\text{max}} = \dfrac{E}{R_L + R_v + R_g} \]

\[ R_L = \dfrac{E}{I_{\text{max}}} - (R_g + R_v) \]

\[ I = \dfrac{E}{R_L + R_v + R_g + R_x} \]

Where: \(I_{\text{max}}\) = full-scale current (A), \(I\) = deflection current (A), \(E\) = internal battery voltage (V), \(R_L\) = limiting resistance (\(\Omega\)), \(R_v\) = zero-adjust resistance (\(\Omega\)), \(R_g\) = meter movement resistance (\(\Omega\)), \(R_x\) = unknown resistance (\(\Omega\))

4. Watt-Hour Meter

Watt-hour meter is an integrating instrument that measures total electric energy consumed over a period of time.

Aluminum disk rotates at speed proportional to power
Voltage and current coils create driving magnetic flux
Meter constant defines disk revolutions per unit energy

Watt-Hour Meter Energy Equation

\[ W = k N \]

\[ W = P t \]

\[ P = \dfrac{k N}{t} \]

Where: \(W\) = recorded energy (\(\text{W}\cdot\text{hr}\) or \(\text{kW}\cdot\text{hr}\)), \(k\) = meter constant (\(\text{W}\cdot\text{hr/revolution}\)), \(N\) = disk revolutions, \(P\) = active power (W or kW), \(t\) = time (hr)

Example. A single-phase three-wire, \(60\ \text{Hz}\), \(220\ \text{V}\), watt-hour meter has a disc constant of \(1.2\). Determine the energy recorded by the meter after \(18,000\) revolutions of the disc.

Solution.

\[ W = k N = 1.2(18000) = 21600\ \text{W}\cdot\text{hr} = 21.6\ \text{kW}\cdot\text{hr} \]

Answer: \(21.6\ \text{kW}\cdot\text{hr}\)

5. Wattmeter Reading Using PT and CT

Instrument transformer metering utilizes potential transformers (PT) and current transformers (CT) to measure high active power safely.

Potential transformer reduces high line voltage for meter
Current transformer reduces high line current for meter
Overall multiplier equals product of CT and PT ratios

PT and CT Metering Multiplier

\[ \text{Multiplier} = \text{CT ratio} \times \text{PT ratio} \]

\[ P_{\text{actual}} = W_{\text{reading}} \times \text{Multiplier} \]

Where: \(\text{Multiplier}\) = overall power multiplying factor, \(\text{CT ratio}\) = current transformer ratio, \(\text{PT ratio}\) = potential transformer ratio, \(P_{\text{actual}}\) = actual load power (W), \(W_{\text{reading}}\) = wattmeter reading (W)

Example. A current transformer with a turns ratio of \(100:5\) and a potential transformer with a turns ratio of \(10:1\) are connected to a single-phase wattmeter measuring load power. If the wattmeter reading is \(240\ \text{W}\), what is the actual power delivered to the load?

Solution.

\[ \text{Multiplier} = \left(\dfrac{100}{5}\right)(10) = 200 \]

\[ P_{\text{actual}} = W_{\text{reading}} \times \text{Multiplier} = 240(200) = 48000\ \text{W} \]

Answer: \(48000\ \text{W}\)

6. Resistance Using Ammeter and Voltmeter

Voltmeter-ammeter method determines unknown resistance by taking simultaneous current and voltage measurements.

High resistance load connects voltmeter directly across load
Low resistance load connects voltmeter across load and ammeter
Ohm’s law ratio gives approximate load resistance

Voltmeter-Ammeter Resistance Formula

\[ R_x = \dfrac{V}{I} \]

Where: \(R_x\) = unknown resistance (\(\Omega\)), \(V\) = voltmeter reading (V), \(I\) = ammeter reading (A)

7. Insulation Resistance of Cables

Cable insulation resistance is the opposition offered by the insulating jacket to leakage currents flowing from conductor to sheath.

Inversely proportional to length of cable
Logarithmic relationship with outer to inner radius ratio
Voltmeter method evaluates insulation resistance using series deflection

Cable Insulation Resistance Formulas

\[ R = \dfrac{\rho}{2\pi \ell} \ln\left(\dfrac{r_2}{r_1}\right) \]

\[ R = R_v \left(\dfrac{E}{E_v} - 1\right) \]

Where: \(R\) = insulation resistance (\(\Omega\)), \(\rho\) = insulation resistivity (\(\Omega\cdot\text{cm}\) or \(\Omega\cdot\text{m}\)), \(\ell\) = length of cable (m or ft), \(r_1\) = conductor radius (mm or in), \(r_2\) = outer insulation radius (mm or in), \(R_v\) = voltmeter resistance (\(\Omega\)), \(E\) = source voltage (V), \(E_v\) = series voltmeter reading (V)

Example. It is desired to measure the resistance of the insulation between a motor winding and the motor frame. A \(300\ \text{V}\), \(50000\ \Omega\) voltmeter is used in making the measurements. The voltmeter when connected across the source reads \(230\ \text{V}\) and when connected in series with the insulation, it reads \(5\ \text{V}\). Find the insulation resistance.

Solution.

\[ R = R_v \left(\dfrac{E}{E_v} - 1\right) = 50000 \left(\dfrac{230}{5} - 1\right) = 2250000\ \Omega = 2.25\ \text{M}\Omega \]

Answer: \(2.25\ \text{M}\Omega\)

8. Wheatstone Bridge

Wheatstone bridge is a four-arm bridge network used for precise measurement of medium resistance values.

Null indicator galvanometer detects zero current at balance
Ratio arms specify fixed resistance ratio
Rheostat arm adjusted to achieve bridge balance

Wheatstone Bridge Balance Formula

\[ R_x = R \left(\dfrac{R_1}{R_2}\right) \]

Where: \(R_x\) = unknown resistance (\(\Omega\)), \(R\) = rheostat arm resistance (\(\Omega\)), \(R_1, R_2\) = ratio arm resistances (\(\Omega\))

Example. An unknown resistor \(R_x\) is measured by means of a Wheatstone bridge. When the ratio arms are \(R_1 = 10\ \Omega\) and \(R_2 = 1200\ \Omega\), balance is obtained when the rheostat arm is \(1256\ \Omega\). What is the value of the unknown resistor?

Solution.

\[ R_x = R \left(\dfrac{R_1}{R_2}\right) = 1256 \left(\dfrac{10}{1200}\right) = 10.47\ \Omega \]

Answer: \(10.47\ \Omega\)

9. Slide-Wire Bridge

Slide-wire bridge is a modified Wheatstone bridge employing a calibrated uniform wire as variable ratio arms.

Uniform 100-cm resistance wire serves as ratio arm scale
Sliding contact key shifts galvanometer contact point
Balance position determines unknown resistance ratio

Slide-Wire Bridge Formula

\[ R_x = R \left(\dfrac{100 - s}{s}\right) \]

Where: \(R_x\) = unknown resistance (\(\Omega\)), \(R\) = known standard resistance (\(\Omega\)), \(s\) = distance along 100-cm wire scale (cm)

Example. An unknown resistance \(x\) is measured by a slide-wire bridge \(AC\) on a \(100\ \text{cm}\) uniform wire with \(R = 100\ \Omega\). Balance is obtained when segment \(AD\) equals \(32.5\ \text{cm}\). Find the unknown resistance \(x\).

Solution.

\[ R_x = R \left(\dfrac{100 - s}{s}\right) = 100 \left(\dfrac{100 - 32.5}{32.5}\right) = 48.14\ \Omega \]

Answer: \(48.14\ \Omega\)

10. Varley-Loop Test

Varley-loop test is a bridge method utilizing known ratio arms and a variable resistance arm to locate cable ground faults.

Two-position SPDT switch configures total loop and fault balance
Calculates total loop resistance in initial switch position
Determines fault distance from second switch balance reading

Varley-Loop Test Equations

\[ R = \dfrac{A C}{B} \]

\[ r = \dfrac{R}{2 D} \]

\[ d = \dfrac{2 D}{R}\left(\dfrac{B R - A C}{A + B}\right) \]

Where: \(d\) = fault distance (m or ft), \(D\) = route length of transmission line (m or ft), \(R\) = total loop resistance (\(\Omega\)), \(r\) = cable resistance per unit length (\(\Omega/\text{m}\) or \(\Omega/\text{ft}\)), \(A, B, C\) = bridge arm resistances (\(\Omega\))

11. Murray-Loop Test

Murray-loop test is a bridge circuit method designed to locate cable ground faults in low-resistance loops.

Formed by looped healthy and faulted conductors of equal length
Adjustable ratio arms balanced until galvanometer indicates zero
Fault location computed directly as fraction of total loop length

Murray-Loop Test Equation

\[ d = 2 D \left(\dfrac{A}{A + B}\right) \]

Where: \(d\) = fault distance (m or ft), \(D\) = length of transmission cable line (m or ft), \(A, B\) = adjustable ratio arm resistances (\(\Omega\))

12. Slide-Wire Murray-Loop Test

Slide-wire Murray-loop test incorporates a calibrated slide-wire scale to measure cable fault distance directly by length ratio.

Calibrated uniform wire scale replaces discrete resistor ratio arms
Movable slider contact moved along wire until bridge balances
Direct ratio of slide-wire segments locates fault location

Slide-Wire Murray-Loop Equation

\[ d = 2 D \left(\dfrac{\ell'}{\ell + \ell'}\right) = 2 D \left(\dfrac{100 - s}{100}\right) \]

Where: \(d\) = distance to fault point (m or ft), \(D\) = route length of cable line (m or ft), \(\ell, \ell'\) = slide-wire segment lengths (cm), \(s\) = slider scale reading (cm)

Example. A two-conductor underground cable \(2800\ \text{ft}\) long is used to interconnect a home base station to another station. One conductor is grounded at some point. A Murray loop test was conducted at the home base with a \(100\ \text{cm}\) slide-wire bridge. If a balance on the bridge was obtained at \(76\ \text{cm}\), at what distance is the fault from the home base?

Solution.

\[ d = 2 D \left(\dfrac{100 - s}{100}\right) = 2(2800)\left(\dfrac{100 - 76}{100}\right) = 1344\ \text{ft} \]

Answer: \(1344\ \text{ft}\)

13. Formula Summary

Formula Name	Formula	Variables
Ammeter Shunt Resistance and Multiplier	\(I = I_{\text{sh}} + I_{\text{fs}}\) \(R_{\text{sh}} = \dfrac{I_{\text{fs}} R_g}{I_{\text{sh}}}\) \(\text{Multiplier} = \dfrac{I}{I_{\text{fs}}} = \dfrac{R_g + R_{\text{sh}}}{R_{\text{sh}}}\)	\(I\) = total current (A), \(I_{\text{fs}}\) = full-scale current (A), \(I_{\text{sh}}\) = shunt current (A), \(R_g\) = meter resistance (\(\Omega\)), \(R_{\text{sh}}\) = shunt resistance (\(\Omega\))
Voltmeter Series Multiplier and Sensitivity	\(E_{\text{fs}} = I_{\text{fs}}(R_g + R_{\text{se}})\) \(R_{\text{se}} = \dfrac{E_{\text{fs}}}{I_{\text{fs}}} - R_g\) \(\text{Multiplier} = \dfrac{R_g + R_{\text{se}}}{R_g}\) \(\text{Sensitivity} = \dfrac{1}{I_{\text{fs}}}\)	\(E_{\text{fs}}\) = full-scale voltage (V), \(I_{\text{fs}}\) = full-scale current (A), \(R_g\) = meter resistance (\(\Omega\)), \(R_{\text{se}}\) = series resistance (\(\Omega\)), \(\text{Sensitivity}\) = sensitivity (\(\Omega/\text{V}\))
Ohmmeter Current and Resistance Relations	\(I_{\text{max}} = \dfrac{E}{R_L + R_v + R_g}\) \(R_L = \dfrac{E}{I_{\text{max}}} - (R_g + R_v)\) \(I = \dfrac{E}{R_L + R_v + R_g + R_x}\)	\(I_{\text{max}}\) = full-scale current (A), \(I\) = deflection current (A), \(E\) = voltage (V), \(R_L\) = limiting resistance (\(\Omega\)), \(R_v\) = zero-adjust resistance (\(\Omega\)), \(R_g\) = meter resistance (\(\Omega\)), \(R_x\) = unknown resistance (\(\Omega\))
Watt-Hour Meter Energy Equation	\(W = k N\) \(W = P t\) \(P = \dfrac{k N}{t}\)	\(W\) = energy (\(\text{W}\cdot\text{hr}\) or \(\text{kW}\cdot\text{hr}\)), \(k\) = meter constant (\(\text{W}\cdot\text{hr/rev}\)), \(N\) = disk revolutions, \(P\) = power (W or kW), \(t\) = time (hr)
PT and CT Metering Multiplier	\(\text{Multiplier} = \text{CT ratio} \times \text{PT ratio}\) \(P_{\text{actual}} = W_{\text{reading}} \times \text{Multiplier}\)	\(\text{Multiplier}\) = overall factor, \(\text{CT ratio}\) = current ratio, \(\text{PT ratio}\) = voltage ratio, \(P_{\text{actual}}\) = actual power (W), \(W_{\text{reading}}\) = reading (W)
Voltmeter-Ammeter Resistance Formula	\(R_x = \dfrac{V}{I}\)	\(R_x\) = resistance (\(\Omega\)), \(V\) = voltage (V), \(I\) = current (A)
Cable Insulation Resistance Formulas	\(R = \dfrac{\rho}{2\pi \ell} \ln\left(\dfrac{r_2}{r_1}\right)\) \(R = R_v \left(\dfrac{E}{E_v} - 1\right)\)	\(R\) = insulation resistance (\(\Omega\)), \(\rho\) = resistivity (\(\Omega\cdot\text{m}\)), \(\ell\) = length (m), \(r_1, r_2\) = radii (m), \(R_v\) = voltmeter resistance (\(\Omega\)), \(E\) = source voltage (V), \(E_v\) = voltmeter reading (V)
Wheatstone Bridge Balance Formula	\(R_x = R \left(\dfrac{R_1}{R_2}\right)\)	\(R_x\) = unknown resistance (\(\Omega\)), \(R\) = rheostat resistance (\(\Omega\)), \(R_1, R_2\) = ratio arm resistances (\(\Omega\))
Slide-Wire Bridge Formula	\(R_x = R \left(\dfrac{100 - s}{s}\right)\)	\(R_x\) = unknown resistance (\(\Omega\)), \(R\) = standard resistance (\(\Omega\)), \(s\) = scale reading (cm)
Varley-Loop Test Equations	\(R = \dfrac{A C}{B}\) \(r = \dfrac{R}{2 D}\) \(d = \dfrac{2 D}{R}\left(\dfrac{B R - A C}{A + B}\right)\)	\(d\) = fault distance (m), \(D\) = route length (m), \(R\) = loop resistance (\(\Omega\)), \(r\) = resistance per unit length (\(\Omega/\text{m}\)), \(A, B, C\) = bridge arms (\(\Omega\))
Murray-Loop Test Equation	\(d = 2 D \left(\dfrac{A}{A + B}\right)\)	\(d\) = fault distance (m), \(D\) = route length (m), \(A, B\) = ratio arms (\(\Omega\))
Slide-Wire Murray-Loop Equation	\(d = 2 D \left(\dfrac{\ell'}{\ell + \ell'}\right) = 2 D \left(\dfrac{100 - s}{100}\right)\)	\(d\) = fault distance (m), \(D\) = route length (m), \(\ell, \ell'\) = segment lengths (cm), \(s\) = scale reading (cm)