20.10 SHUNTS AND MULTIPLIERS
20.10.1 Converting Galvanometer into Ammeter
1. Galvanometer
(a) A moving-coil galvanometer is an electrical instrument which composes of a coil and several other components. The coil, which can be rotated, has resistance $R_g$.
(b) When a current of magnitude $I_g$ flows through it, the coil would rotate, causing the pointer which is attached to it to deflect through full scale deflection (f.s.d.).
(c) The current that produces f.s.d. could be just several microamperes or milliamperes.
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| Figure 20.60 A low-resistance shunt resistor ($R_S$) is connected in parallel with the galvanometer coil. This allows most of the current to bypass the galvanometer, enabling it to measure larger currents safely without damage. |
2. Converting into an ammeter
(a) We can convert the galvanometer into an ammeter so that we can use it to read current up to several amperes.
(b) To do that, we connect a resistor of very low resistance $R_s$ parallel to the coil of the galvanometer, as shown in Figure 20.60. This resistor is known as a shunt.
(c) The values of $R_s$ and $R_g$ will determine the maximum current that can be read from the scale of the ammeter.
3. Principle
(a) Suppose that we want to convert the galvanometer into an ammeter whose scale can be read up to a maximum of I amperes. This means that the reading at f.s.d. is going to represent this maximum current I.
(b) In actual fact the current which flows through the coil of the galvanometer to produce f.s.d. is only $I_g$ and not I. Hence, we have to divert current of amount $I-I_g$ to the shunt so that only current of amount $I_g$ flows through the coil.
4. Value of Rs
When current $I_g$ flows through the coil, we have
p.d. across shunt = p.d. across coil
$(I-I_g)R_s=I_gR_g$
$\boxed{R_s=\left(\frac{I_g}{I-I_g}\right)R_g}$
Notice that $I >> I_g$. For example, we may have $I = 2.0 \ A$, $I_g = 0.10 \ mA$. Hence we have the approximation
$\boxed{R_s \cong \left(\frac{I_g}{I}\right)R_g}$
20.10.2 Multi-range Ammeter
1. Instead of just reading one maximum current value at f.s.d., the ammeter can be adapted to read several different maximum values. In other words, the ammeter can be converted into a multi-range ammeter.
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| Figure 20.61 A galvanometer with multiple terminals (N, P, Q, R) uses different shunts to measure various current ranges. |
2. Figure 20.61 shows how to convert a galvanometer into a multi-range ammeter. The adapted galvanometer has one common negative terminal N and three other terminals P, Q and R. For example, the meter will show a reading of 2.5 A at f.s.d. if we make use of terminals N and P.
20.10.3 Converting Galvanometer into Voltmeter
1 Converting into a voltmeter
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| Figure 20.62 A high-resistance multiplier ($R_M$) is connected in series with a galvanometer to measure voltage. |
A moving-coil galvanometer can be converted into a voltmeter. This conversion can be achieved by connecting a resistor of high resistance RM in series to the galvanometer, as shown in Figure 20.62. This resistor is known as a multiplier.
2 Principle
(a) Suppose we want to convert the galvanometer into a voltmeter whose scale can be read up to a maximum voltage of V volts. This means that the reading at f.s.d. is going to represent this maximum voltage V.
(b) To produce f.s.d., a current Ig must flow through the coil of the galvanometer. Since RM is connected in series to the galvanometer, the current flowing through RM must be Ig too.
3 Value of RM
When f.s.d. is achieved, we have
p.d. across XY = p.d. across Rg + p.d. across RM
$V = V_g + V_M = I_gR_g + I_gR_M$
$\boxed{R_M = \frac{V}{I_g}− R_g}$
20.10.4 Multi-range Voltmeter
1. We can make the scale reading at f.s.d. to represent several different maximum voltages. This can be achieved by connecting several multipliers, RM1, RM2, RM3 of different resistances in series to the coil of the galvanometer, as shown in Figure 20.63.
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| Figure 20.63 Different series resistors (multipliers) are connected to a galvanometer to obtain multiple voltage ranges at full-scale deflection. |
2. The circuit has one common negative terminal and one terminal for each multiplier, like P for RM3, Q for RM2 and R for RM1. If we use the negative terminal together with, say, the 5 V terminal (P), then the reading at f.s.d. would represent a voltage of 5 V.
Worked Example: Galvanometer to Ammeter Conversion (Step-by-Step Solution)
EXAMPLE 20.20
A coil in a moving-coil galvanometer has resistance 5.0 Ω. When a current of 0.10 A flows through the coil, the meter indicates f.s.d. The meter is to be converted into an ammeter which can be used to read current up to a maximum of 10 A. Determine the resistance of the shunt which has to be connected to the meter.
Answer
Given:
Resistance of galvanometer coil, $R_g=5.0 \ \Omega$
Full-scale deflection current, $I_g = 0.10 \ A$
Maximum current to be measured, $I = 10 \ A$
Concept:
To convert a galvanometer into an ammeter, a low resistance called a shunt is connected in parallel with the galvanometer.
The shunt allows most of the current to bypass the galvanometer.
Using current relation:
$I = I_g + I_s$
where $I_s$ is the current through the shunt.
Since both branches are in parallel:
$I_gR_g = I_sR_s$
Step 1: Calculate Shunt Current
$I_s = I − I_g = 10 − 0.10 = 9.9 \ A$
Step 2: Calculate Shunt Resistance
$I_g R_g = I_s R_s$
$R_s= \frac{I_gR_g }{I_s}$
$R_s = \frac{0.10 \times 5.0}{ 9.9}$
$R_s ≈ 0.0505 \ \Omega$
The required shunt resistance is approximately 0.051 Ω.
EXAMPLE 20.21
The resistance of the coil in a moving-coil galvanometer is 20 Ω. A current of 2.0 mA flowing through the coil can produce f.s.d. Determine the resistance of a multiplier which has to be connected to the coil in order to convert the galvanometer into a voltmeter whose scale can be read up to a maximum of 2.0 V.
Answer
Given:
Resistance of galvanometer coil, $R_g = 20 \ \Omega$
Full-scale deflection current, $I_g = 2.0 \ mA = 0.002 \ A$
Maximum voltage to be measured, $V = 2.0 \ V$
Concept:
To convert a galvanometer into a voltmeter, a high resistance called a multiplier is connected in series with the galvanometer.
The total resistance required is given by Ohm’s Law:
$V = IR$
So, total resistance:
$R_{total} = \frac{V}{I_g}$
Step 1: Calculate Total Resistance
$R_{total} = \frac{2.0}{0.002} = 1000 \ \Omega$
Step 2: Calculate Multiplier Resistance
The total resistance is the sum of galvanometer resistance and multiplier:
$R_{total} = R_g + R_M$
So,
$R_M= R_{total} − R_g$
$R_M = 1000 − 20 = 980 \ \Omega$
The required multiplier resistance is 980 Ω.
EXAMPLE 20.22
The resistance of a coil in a moving-coil galvanometer is 20 Ω. A current of 1.0 mA flowing through the galvanometer produces full-scale deflection (f.s.d.). Determine the resistance of a shunt which is to be connected in parallel to the coil in order to convert the galvanometer into an ammeter that can read current up to 1.0 A.
Given:
Resistance of galvanometer, $R_g = 20 \ \Omega$
Full-scale deflection current, $I_g = 1.0 \ mA = 0.001 \ A$
Maximum current, $I = 1.0 \ A$
Concept:
To convert a galvanometer into an ammeter, a shunt resistor (Rs) is connected in parallel with the galvanometer.
At full-scale deflection, both branches have the same voltage:
$I_g R_g = I_s R_s$
Total current is:
$I =I_g + I_s$
Step 1: Calculate Shunt Current
$I_s = I − I_g = 1.0 − 0.001 = 0.999 \ A$
Step 2: Calculate Shunt Resistance
$R_s= \frac{I_gR_g}{I_s}$
$R_s= \frac{0.001 \times 20}{0.999}$
$R_s = \frac{0.02}{0.999} ≈ 0.0200 \ \Omega$
Final Answer:
The required shunt resistance is approximately 0.020 Ω.
EXAMPLE 20.23
The resistance of a coil in a moving-coil galvanometer is 20 Ω. When a current of 10 mA flowing through the galvanometer produces full-scale deflection (f.s.d.), a shunt of resistance 0.05 Ω is connected in parallel to the coil to convert the galvanometer into an ammeter. Determine the maximum current that can be read from the scale of the ammeter.
Given:
Resistance of galvanometer, $R_g = 20 \ \Omega$
Full-scale deflection current, $I_g = 10 \ mA = 0.01 \ A$
Shunt resistance,$R_s = 0.05 \ \Omega$
Concept:
A galvanometer is converted into an ammeter by connecting a shunt resistor in parallel.
At full-scale deflection, voltage across both branches is equal:
$I_g R_g = I_s R_s$
Total current:
$I =I_g + I_s$
Step 1: Calculate Voltage across Galvanometer
$V = I_gR_g = 0.01 \times 20 = 0.2 \ V$
Step 2: Calculate Current through Shunt
$I_s= \frac{V}{R_s} = \frac{0.2 }{0.05} = 4.0 \ A$
Step 3: Calculate Maximum Current
$I =I_g + I_s$
$I = 0.01 + 4.0 = 4.01 \ A$
The maximum current that can be read is approximately 4.01 A.
EXMAPLE 20.24
The resistance of a coil in a moving-coil galvanometer is 20 Ω. A current of 5.0 mA flowing through the galvanometer produces full-scale deflection (f.s.d.). Determine the resistance of a multiplier which must be connected to the galvanometer to convert it into a voltmeter that can read up to a maximum voltage of 10 V.
Given:
Resistance of galvanometer, $R_g= 20 \ \Omega$
Full-scale deflection current, $I_g = 5.0 \ mA = 0.005 \ A$
Maximum voltage, $V = 10 \ V$
Concept:
To convert a galvanometer into a voltmeter, a multiplier resistor (RM) is connected in series.
Total resistance required:
$R_{total} = \frac{V}{I_g}$
And:
$R_{total} = R_g+ R_M$
Step 1: Calculate Total Resistance
$R_{total} = \frac{10}{0.005} = 2000 \ \Omega$
Step 2: Calculate Multiplier Resistance
$R_M = R_{total} − R_g$
$R_M = 2000 − 20 = 1980 \ \Omega$
The required multiplier resistance is 1980 Ω.
EXAMPLE 20.25
A 0.0200-Ω ammeter is placed in series with a 10.00-Ω resistor in a circuit.
(a) Draw a circuit diagram of the connection.
(b) Calculate the resistance of the combination.
(c) If the voltage is kept the same across the combination as it was through the 10.00-Ω resistor alone, what is the percent decrease in current?
(d) If the current is kept the same through the combination as it was through the 10.00-Ω resistor alone, what is the percent increase in voltage?
(e) Are the changes found in parts (c) and (d) significant? Discuss.
Given:
Resistance of ammeter, $R_a = 0.0200 \ \Omega$
Resistance of resistor, $R = 10.00 \ \Omega$
Concept:
When resistors are connected in series, total resistance is:
$R_{total} = R + R_a$
Ohm’s Law: $V = IR$
(a) Circuit Diagram:
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| Figure 20.64 An ammeter (0.0200 Ω) is connected in series with a 10.00 Ω resistor, so the same current flows through both components. |
(b) Total Resistance
$R_{total} = 10.00 + 0.0200 = 10.02 \ \Omega$
(c) Percent Decrease in Current (Voltage Constant)
Original current:
$I_1= \frac{V}{10.00}$
New current:
$I_2= \frac{V}{10.02}$
Percent decrease:
$= \left(\frac{I_1-I_2}{I_1}\right)100\% $
$= \left(1-\frac{10.00}{10.02}\right)100\% $
$= (1 − 0.9980) \times 100\% $
$≈ 0.20\%$
(d) Percent Increase in Voltage (Current Constant)
Original voltage:
$V_1=I \times 10.00$
New voltage:
$V_2=I \times 10.02$
Percent increase:
$= \left(\frac{V_2-V_1}{V_1}\right)100\% $
$= \left(\frac{10.02-10.00}{10.00}\right)100\% $
$= (0.02/10.00) \times 100\% $
$= 0.20\%$
(e) Discussion:
The changes in both current and voltage are only about 0.20%, which is less than 0.5% and even smaller than the measurement uncertainty; therefore, the effect of the ammeter resistance can be considered negligible.
EXAMPLE 20.26
A 1.20-MΩ voltmeter is connected in parallel with a 80.0-kΩ resistor in a circuit.
(a) Draw a circuit diagram of the connection.
(b) Determine the equivalent resistance of the combination.
(c) If the voltage across the combination remains the same as it was across the resistor alone, calculate the percentage increase in current.
(d) If the current through the combination remains the same as it was through the resistor alone, calculate the percentage decrease in voltage.
(e) Comment on whether the changes are significant.
Given:
Resistance of voltmeter, $R_v = 1.20 \ M \Omega = 1,200,000 \ \Omega$
Resistance of resistor, $R = 80.0 \ k \Omega = 80,000 \ \Omega$
Concept:
For resistors in parallel:
$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R_v}$
Ohm’s Law: $V = IR$
(a) Circuit Diagram:
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| A voltmeter with internal resistance $Rv=1.20 \ M \Omega$ is connected in parallel with a resistor $R=80.0 \ k \Omega$ across a power source, so both components share the same voltage. |
(b) Equivalent Resistance
$\frac{1}{R_{eq}} ≈ 0.000012500 + 0.00000083333 = 0.0000133333$
$R_{eq} ≈ \frac{1}{0.0000133333} ≈ 75,000 \ \Omega$
$R_{eq} ≈ 75.0 \ k \Omega$
(c) Percentage Increase in Current (Voltage Constant)
Original current:
$I_1= \frac{V}{80,000}$
New current:
$I_2= \frac{V}{75,000}$
Percentage increase:
$= \left(\frac{I_1-I_2}{I_1}\right)100\% $
$= \left(\frac{80,000}{75,000}-1\right)100\% $
$= (1.067-1 ) \times 100\% $
$≈ 0.67\%$
(d) Percentage Decrease in Voltage (Current Constant)
Original voltage:
$V_1=I \times 80,000$
New voltage:
$V_2=I \times 75,000$
Percentage decrease:
$= \left(\frac{V_2-V_1}{V_1}\right)100\% $
$= \left(\frac{80,000-75,000}{80,000}\right)100\% $
$= (5,000/80,000) \times 100\% $
$= 6.25\%$
(e) Discussion:
The observed changes, approximately 6–7%, are relatively significant and cannot be neglected. This indicates that the voltmeter has a noticeable loading effect on the circuit. Such an effect occurs because the internal resistance of the voltmeter is not sufficiently large compared to the resistance of the circuit element being measured, thereby altering the current distribution and reducing measurement accuracy.
Applications of Shunts and Multipliers
1 Application of Shunts
A shunt is widely used to extend the range of a galvanometer so that it can measure large currents. By connecting a low resistance in parallel with the galvanometer, most of the current is diverted through the shunt while only a small portion flows through the coil.
This principle is applied in the construction of ammeters used in electrical circuits, power systems, and laboratory measurements where high currents need to be measured safely.
Shunts are also used in digital ammeters and current sensors to measure current indirectly by detecting the voltage drop across the shunt resistor.
2 Application of Multipliers
A multiplier is used to increase the voltage measuring range of a galvanometer. By connecting a high resistance in series, the galvanometer can measure much higher voltages than its original design.
This principle is used in voltmeters for measuring potential difference in electrical circuits, electronic devices, and experimental setups.
Multipliers are essential in designing multi-range voltmeters, allowing a single instrument to measure different voltage ranges by selecting different resistors.
3 Practical Uses in Electrical Instruments
Both shunts and multipliers are fundamental components in analog measuring instruments. They allow a basic galvanometer to be adapted into versatile measuring devices such as ammeters and voltmeters.
These components are commonly used in electrical laboratories, industrial equipment, and electronic testing devices to ensure accurate measurement of current and voltage.
4 Importance in Circuit Protection
Shunts help protect sensitive galvanometer coils from excessive current by diverting most of the current away. Similarly, multipliers prevent excessive current from flowing through the galvanometer when measuring high voltages.
Thus, both components play an important role not only in measurement but also in protecting instruments from damage.
Conclusion of Shunts and Multipliers
Shunts and multipliers are essential components used to extend the functionality of a moving-coil galvanometer. By using a shunt (a low resistance connected in parallel), the galvanometer can be converted into an ammeter capable of measuring large currents. On the other hand, by using a multiplier (a high resistance connected in series), the galvanometer can be converted into a voltmeter for measuring higher voltages.
These modifications allow a sensitive instrument like a galvanometer to be adapted for a wide range of practical applications in electrical measurements. The use of appropriate resistance values ensures accurate readings while protecting the instrument from damage due to excessive current or voltage.
Furthermore, the concepts of shunts and multipliers are fundamental in the design of multi-range ammeters and voltmeters, enabling a single device to measure different ranges efficiently. This makes them highly valuable in laboratories, industrial systems, and electronic devices.
In conclusion, understanding shunts and multipliers not only helps in mastering measurement techniques but also provides a strong foundation in electrical instrumentation and circuit design.
Frequently Asked Questions (FAQ) – Shunts and Multipliers
1. What is a shunt in electrical instruments?
A shunt is a low resistance connected in parallel with a galvanometer to divert most of the current. This allows the galvanometer to measure large currents by converting it into an ammeter.
2. What is a multiplier?
A multiplier is a high resistance connected in series with a galvanometer. It is used to extend the voltage range, allowing the galvanometer to function as a voltmeter.
3. Why is a shunt connected in parallel?
A shunt is connected in parallel so that it can carry most of the current, protecting the sensitive galvanometer coil from excessive current.
4. Why is a multiplier connected in series?
A multiplier is connected in series to limit the current flowing through the galvanometer when measuring high voltages, ensuring safe and accurate readings.
5. What is full scale deflection (f.s.d.)?
Full scale deflection (f.s.d.) is the maximum deflection of the galvanometer pointer, corresponding to the maximum current that the instrument can safely measure.
6. How does a shunt increase the current range?
A shunt increases the current range by allowing only a small portion of the total current to pass through the galvanometer, while the rest flows through the shunt resistor.
7. How does a multiplier increase the voltage range?
A multiplier increases the voltage range by adding resistance in series, which reduces the current through the galvanometer even when measuring high voltages.
8. What is a multi-range ammeter?
A multi-range ammeter is an instrument that can measure different ranges of current by using multiple shunts and selecting them through different terminals.
9. What is a multi-range voltmeter?
A multi-range voltmeter uses multiple multipliers to allow measurement of different voltage ranges with a single instrument.
10. Why are shunts and multipliers important?
They are important because they allow a sensitive galvanometer to be used for a wide range of current and voltage measurements while ensuring accuracy and protecting the instrument.
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