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Extension of Range
Shunts are used for the extension of range of Ammeters. So a good shunt should have the following properties:- 1- The temperature coefficient of shunt should be low 2- Resistance of shunt should not vary with time 3- They should carry current without excessive temperature rise 4- They should have thermal electromotive force with copper
- ‘Manganin’ is used for DC shunt and ‘Constantan’ as AC shunt.
Ammeter:- PMMC is used as indicating device. The current capacity of PMMC is small. It is impractical to construct a PMMC coil, which can carry a current greater than 100 mA. Therefore a shunt is required for measurement of large currents.
Rm = Internal resistance of movement (coil) in Ω Rsh = Resistance of shunt in Ω Im = Ifs = Full scale deflection current of movement in Amperes Ish = Shunt current in Amperes I = Current to be measured in Amperes Since the shunt resistance is in parallel with the meter movement, the voltage drop across shunt and movement must be same. I (^) sh Rsh = ImR m
sh sh m m I
R = I R
I (^) sh = I − I m
∴ We can write ( (^) m ) sh m m I I
R I R
sh
m m R
R
I
I
sh
m m R
R
I
I
m I
I
m
= is known as ‘multiplying power’
of shunt
Resistance of shunt R (^) sh = Rm ( m − 1 )
Or
m
sh m
I
I
R R
Multi Range Ammeter:- Let m 1 , m 2 , m 3 , m 4 be the shunt multiplying powers for current I 1 , I 2 , I 3 , I 4.
(^1) ( 1 − 1 )
m R Rm sh
(^2) ( 2 − 1 )
m R Rm sh
(^3) ( 3 − 1 )
m R Rm sh
(^4) ( 4 − 1 )
m R Rm sh
Voltmeter :-
For measurement of voltage a series resistor or a multiplier is required for extension of range. Im = Deflection current of movement Rm = Internal resistance of movement Rs = Multiplier resistance V = Full range voltage of instrument V = Im ( Rs + Rm )
m m m
m m s (^) I R
V
I
V I R
R = −
- For more than 500 V multiplier is mounted outside the case.
Multi Range Voltmeter:
m m
s R I
V
R 1 = 1 −
m m
s (^) I R
V
R 2 = 2 −
m m
s (^) I R
V
R 3 = 3 −
m m
s R I
V
R 4 = 4 −
_ For average value divide the reading by 1.11. For peak value multiply the voltage by 1.414. To get peak-to-peak ratio multiply the reading by 2.828. ** Thermocouple and hot wire instruments are used for measurement of true power and rms value of voltage & current_*.
Voltmeter & Ammeter by Moving Coil Instrument:- Same process as applied in PMMC.
Example 4:- A PMMC instrument gives full scale reading of 25 mA when a potential difference across its terminals is 75 mV. Show how it can be used (a) as an ammeter for the range of 0-100 A (b) as a voltmeter for the range of 0-750 V. Also find the multiplying factor of shunt and voltage amplification.
Solution 4:-
Instrument resistance = = − = Ω
− 3 25 * 10
min 75 * 10 3
3
Instrument current
Potentialdropacrosster als Rm
(a) Current to be measured I = 100 A
Multiplying power of shunt 4000 25 * 10
I m
I
m
Shunt resistance required for full scale deflection at 100 A
= = = Ω −
= −^ m m
R
R (^) sh m 7. 50 * 10 0. 75 3999
4
(b) Voltage to be measured V = 750 V
= − = − − 3 = 29 , 997 Ω 25 * 10
m 3 m
se (^) I R
V
R
Voltage amplification 10000 75 * 10
= (^) − 3 = Ans.
Example 5:- A moving coil instrument gives full scale deflection of 10 mA and potential difference across its terminals is 100 mV. Calculate (a) shunt resistance for full-scale deflection corresponding to 200 A (b) Series resistance for full reading corresponding to 1000 V.
Solution 5:-
Instrument resistance = = − = Ω
− 10 10 * 10
min 100 * 10 3
3
Instrument current
Potentialdropacrosster als Rm
(a) Shunt resistance required for full scale deflection corresponding to 200 A
= Ω −
−
4
3
m
m sh I
I
R
R
(b) Series resistance required for full scale deflection corresponding to 1000 V
= − = − − 10 = 99 , 990 Ω 10 * 10
m 3 m
se (^) I R
V
R
Example 6:- A moving coil instrument having internal resistance of 50 ΩΩΩΩ indicates full scale deflection with a current of 10 mA. How can it be made to work as (i) a voltmeter to read 100 V on full scale (ii) an ammeter of 1 A, on full scale?
Solution 6:- Resistance of the instrument coil Rm = 50 Ω
Current flowing through the instrument for full-scale deflection I (^) m = 10 mA = 0. 01 A
(i) Series resistance required to measure 100 V
= − = − 50 = 9950 Ω
- 01
m m
se (^) I R
V
R
(ii) Shunt required to measure 1 A current
= Ω −
m
m sh I
I
R
R
Example 7:- A moving coil instrument has a resistance of 2 ΩΩΩΩ and it reads upto 250 V when a resistance of 5000 ΩΩΩΩ is connected in series with it. Find the current range of the instrument when it is used as ammeter with the coil connected across a shunt resistance of 2 milli ΩΩΩΩ.
Solution 7:- Resistance of the instrument coil Rm = 2 Ω
Current flowing through the instrument for full-scale deflection
R Seriesresis ce
Fullscalereading I m
m = + tan
- 04998 A 49. 98 mA 2 5000
Shunt resistance Rsh = 2 * 10 −^3 Ω
Current through shunt resistance A R
I R
I
s
m m sh (^) 2 * 10 49.^98
3
3 = = − =
−
Current range of instrument = Full scale deflection current = I (^) m + I = 0. 04998 + 49. 98 = 50 A
Example 8:- A moving coil ammeter gives full scale deflection with 15 mA and has a resistance of 5 ΩΩΩΩ. Calculate the resistance to be calculated in (a) Parallel to enable the instrument to read upto 1 A (b) Series to enable it to read up to 10 V. [UPTU 2002]
Solution 8:- Instrument resistance Rm = 5 Ω
Full scale deflection current I (^) m = 15 mA = 15 * 10 −^3 A
(a) Current to be measured I = 1 A
Shunt resistance to be connected in parallel
= Ω −
−
m
m sh I
I
R
R
(b) Voltage to be measured V = 10 V Series resistance required
- Instrument and meters can be standardized so that there is saving in costs. Replacement of damaged instruments is easy.
- Single range instruments can be used to cover large current or voltage ranges, when used with suitable multi range instrument transformers.
- The metering circuit is isolated from the high voltage power circuits. Hence isolation is not a problem and the safety is assured for the operators
- There is low power consumption in metering circuit.
- Several instruments can be operated from a single instrument transformer.
Ratios of Instrument Transformer:
Some definitions are:
- Transformation ratio: It is the ratio of the magnitude if the primary phasor to secondary phasor.
Transformation ratio:
Nominal Ratio : It is the ratio of rated primary winding current (voltage) to the rated secondary winding current (voltage).
Turns ratio : This is defined as below.
for a C.T.
for a P.T.
for a C.T.
for a P.T.
Burden of an Instrument Transformer:
The rated burden is the volt ampere loading which is permissible without errors exceeding the particular class of accuracy.
Current Transformer equivalent circuit:
X1 = Primary leakage reactance R1 = Primary winding resistance X2 = Secondary leakage reactance
for a C.T.
for a P.T.
Ratio :The CT ratio is the ratio of primary current input to secondary current output at full load. For example, a CT with a ratio of 300:5 is rated for 300 primary amps at full load and will produce 5 amps of secondary current when 300 amps flow through the primary. If the primary current changes the secondary current output will change accordingly. For example, if 150 amps flow through the 300 amp rated primary the secondary current output will be 2.5 amps (150:300 = 2.5:5).
Current Transformer: Cautions:
Inspect the physical and mechanical condition of the CT before installation.
Check the connection of the transformer requirements for the instrument or the system requirements before connecting the CT.
Inspect the space between the CT phases, ground and secondary conductor for adequate clearance between the primary and secondary circuitry wiring.
Verify that the shorting device on the CT is properly connected until the CT is ready to be installed. The secondary of the CT must always have a burden (load) connected when not in use. NOTE: A dangerously high secondary voltage can develop with an open-circuited secondary.
Construction of Current Transformer:
Current transformers are constructed in various ways. In one method there are two separate windings on a magnetic steel core. The primary winding consists of a few turns of heavy wire capable of carrying the full load current while the secondary winding consist of many turns of smaller wire with a current carrying capacity of between 5/ amperes, dependent on the design. This is called the wound type due to its wound primary coil.
Construction of Current Transformer:
Wound Type
Another very common type of construction is the so-called “window,” “through” or donut type current transformer in which the core has an opening through which the conductor carrying the primary load current is passed. This primary conductor constitutes the primary winding of the CT (one pass through the “window” represents a one turn primary), and must be large enough in cross section to carry the maximum current of the load.
Construction of Current Transformer:
Another distinguishing feature is the difference between indoor and outdoor construction.
Window-type
15kV Outdoor CT 15kV Indoor CT
Equivalent Circuit of C.T.
Fig. 1 Equivalent circuit of C.T.
Phasor diagram
Fig 2. Phasor diagram
A section of Phasor diagram
Fig 3. A section of Phasor diagram
Relationship in current transformer:
Fig 1 represents the equivalent circuit and Fig 2 the phasor diagram of a current transformer. n= turns ratio = (No. of secondary winding turns)/(No. of primary winding turns) rs = resistance of the secondary winding; xs = reactance of the secondary winding; re = resistance of external burden i.e. resistance of meters, current coils etc. including leads; xe = reactance of external burden i.e. reactance of meters, current coils etc. including leads; Ep = primary winding induced voltage ; Es = secondary winding induced voltage; Np = No. of primary winding turns; Ns = No. of secondary winding turns; Vs = Voltage at the secondary winding terminals; Is = secondary winding current; Ip = primary winding current; ϴ = phase angle of transformer; Φ = working flux of the transformer; δ = angle between secondary winding induced voltage and secondary winding current = phase angle of total burden including impedance of secondary winding
tan 1 ( s^ e ) s e
x x r r
As
Phase angle The angle by which the secondary current phasor, when reversed, differs in phase from primary current, is known as the “phase angle” of the transformer. +ve if secondary reversed current leads the primary current -ve if secondary reversed current lags behind the primary current. The angle between Is and Ip is θ. Therefore, the phase angle is θ. From the phasor diagram,
As θ is very small, we can write
Now Io is very small as compared to nIs, and, therefore we can neglect the term Io sin(δ+α)
Errors in current transformers:
- Turns ratio and transformation ratios are not equal.
- The value of transformation ratio is not constant.
It depends upon:
- Magnetizing and loss components of exciting current,
- The secondary winding load current and its power
This introduces considerable errors into current measurement
I m = I 0 cos α and I e = I 0 sin α
0 0
cos( ) tan s sin(^ )
bc bc I ob oa ab nI I
= = =
0
0
c o s( )
s sin (^ )
I
ra d
n I I
0 cos(^ )
s
I
rad
nI
0 cos^ cos^0 sin^ sin^ m cos^ e sin s s
I I I I rad nI nI
≈ ≈
1 8 0 c o s s in
( m^ e ) d e g
s
I I
r e e
n I
δ − δ
In power measurement it is necessary that the phase of secondary winding current shall be displaced by exactly 180° from that of the Primary current. Here, phase difference is different from 180° by an angle θ. Hence due to C.T. two types of errors are introduced in power measurements. Due to actual transformation ratio being different from the turns ratio. Due to secondary winding current not being 180° out of phase with the primary winding current.
Ratio error and phase angle error Ratio Error is defined as:
Percentage ratio error = ((nominal ratio – actual ratio)/(actual ratio))x
Phase angle
Approximate formulas for errors: The usual instrument burden is largely resistive with some inductance. Therefore, δ is positive and small. Hence, sin δ = 0 and cos δ = 1. Therefore equations (4) and (8) can be written as:
(10) and (11)
But, and, therefore, eqns. (10) and (11) can be rewritten as
And
Problem No. Two current transformers of the same nominal ratio 500/5 A, are tested by Silsbee’s method. With the current in the secondary of the transformer adjusted at its rated value, the content in the middle conductor ∆I = 0.05e-j126.9° A expressed with respect to current in the secondary of standard transformer as the reference. It is known that standard transformer has a ratio correction factor (RCF) of 1.0015 and phase error +8’. Find RCF and phase angle error of transformer under test.
n 100
K R
R
= ×
1 8 0 c o s s i n ( m^ e ) d e g s
I I
r e e n I
e s
I
R n I
m deg s
I
ree nI
θ
^
I (^) p ≈ nIs
e 1 e
p p
n I I
R n n
I I
m d e g p
I
r e e I
^
Π ^
Exciting Current = Im = magnetizing component and α = 0. Magnetizing component of no load current Im =(magnetizing mmf)/(Primary winding turns) =80/1=80A
Secondary winding current = 5 A Reflected secondary winding current = n×Is=200×5 =1000A From the phasor diagram shown in Fig. 1
Fig. 1
= 1003.2 A.
Actual transformation ratio R= Ip/Ir = 1003.2/5 =2000.
Phase angle θ = tan1 (Im/InIs) = tan1 (80/1000)° = 4°34’ We Know that, Es = 4.44fΦm Ns Φm = Es/(4.44 f Ns) = 5/(4.44×50×200) = 0.1126 mwb . Area of core = 1000 Sq. mm. = 1000 μ sq.m Maximum flux density Bm=(0.1126mwb)/1000 μ sq.m = 0.1126 Wb/Sq.m.
Problem No. 2
Im
θ
IP
nIS
IS
I (^) p = +
A current transformer with a bar primary has 300 turns in its secondary winding. The resistance and reactance of secondary circuits are 1.5Ω and 1.0 Ωrespectively including the transformer winding. With 5 A flowing in the secondary winding, the magnetizing mmf is 100 A and the iron losses is 1.2W. Determine the ratio and phase angle error. Solution : Primary winding turns Np= 1; Secondary winding turns Ns = 300; Turns ratio = Ns/Np = 300/1 = 300. Secondary circuit burden impedance = = 1.8 Ω For secondary winding circuit: cos δ = 1.5/1.8 = 0.833 and sin δ = 1.0/1.8 = 0.555. Secondary induced voltage Es = 5 × 1.8 = 9.0 V. Primary induced voltage Ep = Es/n = 9.0/300 = 0.03 V. Loss component of current referred to primary winding Ie = iron loss/(Ep) = 1.2/0.03 = 40 A. Magnetizing current Im = (magnetizing mmf)/(primary winding turns) = 100/1 = 100 A
Actual ratio R =
= 300+(100 ×0.555+40 × 0.833)/5 = 317.
In the absence of any information to the contrary we can take nominal ratio to be equal to the turns ratio, or Kn = n = 300
Percentage ratio error
Phase angle θ =
= 180/π ((100 × 0.833 - 40 × 0.555)/(300 × 5)) = 2.34°.
Problem No. 3 A 100/5 A, 50 Hz CT has a bar primary and a rated secondary burden of 12.5 VA. The secondary winding has 196 turns and a leakage inductance of 0.96 mH. With a purely
(1.5)^2 +(1.0)^2
n 100
K R
R
= ×
