Find voltages 𝑉𝑎𝑐 and 𝑉𝑏𝑐 and the source current 𝐼𝑠3 .
First, note that voltage 𝑉𝑎 is directly across voltage source 𝐸1 .
Therefore, 𝑉𝑎 = 𝐸1 = 20 V
The same is true for voltage 𝑉𝑐 , which is directly across the
voltage source 𝐸3 . Therefore, 𝑉𝑐 = 𝐸3 = 8V
To find voltage 𝑉𝑏 , we must apply Kirchhoff's voltage law around
loop 1 as:
+𝐸1 − 𝐸2 − 𝑉3 = 0
so,
𝑉3 = 𝐸1 − 𝐸2 = 20 V − 5 V = 15 V
and, 𝑉𝑏 = 𝑉3 = 15 V
Voltage 𝑉𝑎𝑐 , which is actually the voltage across resistor 𝑅1 , can
then be determined as: 𝑉𝑎𝑐 = 𝑉𝑎 − 𝑉𝑐 = 20 V − 8 V = 𝟏𝟐 𝐕
Similarly, voltage 𝑉𝑏𝑐 , which is actually the voltage across resistor 𝑅2 , can then be determined as follows:
𝑉𝑏𝑐 = 𝑉𝑏 − 𝑉𝑐 = 15 V − 8 V = 𝟕 𝐕
𝑉
𝑉
7V
Current 𝐼2 can be determined using Ohm's law: 𝐼2 = 𝑅2 = 𝑅𝑏𝑐 = 4Ω = 1.75 A
2
2
The source current 𝐼𝑠3 can be determined using Kirchhoff's current law at node 𝑐 :
Σ𝐼𝑖 = Σ𝐼𝑜
𝐼1 + 𝐼2 + 𝐼𝑠3 = 0
𝑉1
𝐼𝑠3 = −𝐼1 − 𝐼2 = − − 𝐼2
𝑅1
with 𝑉1 = 𝑉𝑎𝑐 = 𝑉𝑎 − 𝑉𝑐 = 20 V − 8 V = 12 V
12 V
so that, 𝐼𝑠3 = − 10Ω − 1.75 A = −1.2 A − 1.75 A = −𝟐. 𝟗𝟓 𝐀
Given that the circuit of the figure is a balanced bridge.
a. Calculate the unknown impedance, Z𝑥 .
b. Determine the values of 𝐿𝑥 and 𝑅𝑥 if the circuit operates at a frequency of 1 kHz.
a. The expression for the unknown impedance is determined by:
Z𝑥
Z2 Z3
Z1
(10kΩ)(100Ω)
=
30kΩ∠ − 20∘
= 33.3Ω∠20∘
= 𝟑𝟏. 𝟑 + 𝒋𝟏𝟏. 𝟒 𝛀
=
b. From the preceding result, we have, 𝑅𝑥 = 𝟑𝟏. 𝟑 𝛀
𝑋
11.4Ω
𝐿
and, 𝐿𝑥 = 2𝜋𝑓
= 2𝜋(1000 Hz) = 𝟏. 𝟖𝟏 𝐦𝐇
A station operating at 33 kV is divided into sections A and B. Section A consists of three generators 15 MVA each
having a reactance of 15% and section B is fed from the grid through a 75 MVA transformer of 8% reactance.
The circuit breakers have each a rupturing capacity of 750 MVA. Determine the reactance of the reactor in ohm
to prevent the breakers being overloaded if a symmetrical short circuit occurs on an outgoing feeder connected to
A.
Assume the base MVA as 75.
75
The p.u. reactance of each generator = 0.15 × 15 = 𝑗0.75
The p.u. reactance of transformer = 𝑗0.08 p.u.
Let 𝑥% be the reactance of the reactor for base of 75
MVA. The equivalent circuit for a fault of 𝐴 is as shown
in diagram.
The per cent impedance between the fault point and the
neutral bus is
0.25(𝑋 + 0.08) 0.25𝑋 + 0.0200
=
0.25 + 𝑋 + 0.08
𝑋 + 0.33
Now
S.C. MVA
750
187𝑋 + 15.00
112𝑋
𝑋
Base MVA
p.u. impedance
75(𝑋 + 0.33)
=
0.25𝑋 + 0.02
= 75𝑋 + 24.75
= 9.75
= 0.08705 p.u.
=
∴ Actual value of reactance in ohms =
0.08705×332
75
= 𝟏. 𝟐𝟔𝟒 𝐨𝐡𝐦𝐬
A 6.6kV, 5MVA star connected alternator has a reactance of 1.5 ohm per phase and negligible resistance. MerzPrice protection scheme is used which operates when the out of balance current exceeds 25% of the full load
current. The neutral of the generator is grounded through a resistance of 8 ohms. Determine the proportion of the
winding which remains unprotected against earth fault. Show that the effects of the alternator reactance can be
ignored.
First, we will show that the effect of the alternator reactance can be neglected. Since the reactance of the
winding is ∝ square of the number of turns, i.e., 𝑥G ∝ 𝑁 2 , let 𝑥% be the winding that remains unprotected. The
number or turns will be 𝑥𝑁/100, i.e., the reactance will be proportional to 𝑥 2 𝑁 2 /1002 .
1.5𝑥 2 𝑁 2
∴ If 1.5 ohm is the reactance of the winding, the reactance of the un-protected winding will be 1002 . Since
this reactance is to be added vectorially with resistance of 8 ohms and 𝑥 being small, the effect of the reactance
can be neglected.
The phase voltage will be,
6600
√3
= 3810 volts.
𝑥
The voltage of the unprotected portion = 3810 100
𝑥 1
amps
100 8
5000
=
= 437.37 amps
√3 × 6.6
The fault current = 3810
The out of balance current required for the operation of the relay = 437.37 × 0.25 = 109.34 amps
Now,
or,
3810𝑥
800
4.7625𝑥
𝑥
= 109.34
= 109.34
= 𝟐𝟐. 𝟗𝟓%.
The difference amplifier in the figure has an input voltage 𝑣𝑎 = 3 V. What range of 𝑣𝑏 will ensure that the opamp operates in the linear region?
The limits on the output voltage are determined by the power supply rail voltages. In this example, the supply
voltages are +15 V and -15 V.
Therefore, the output voltage must be, −15 V < 𝑣𝑜 < +15 V
𝑅
𝑅
𝑅
Since 𝑅𝐴 = 𝑅 𝐶 the input voltage 𝑣𝑏 can be calculated, 𝑣𝑏 = 𝑅𝐴 𝑣𝑜 + 𝑣𝑎
𝐵
𝐷
𝐵
Substituting 𝑅𝐴 = 2.2 kΩ and 𝑅𝐵 = 5.1 kΩ into the equation for 𝑣𝑏 yields for the upper and lower limits of the
output voltage, 𝑣𝑜 = +15 V: 𝑣𝑏 = 9.47 V
And, 𝑣𝑜 = −15 V: 𝑣𝑏 = −3.47 V
Then the input voltage range for 𝑣𝑏 to ensure linear operation of the amplifier is, −𝟑. 𝟒𝟕 𝐕 < 𝒗𝒐 < 𝟗. 𝟒𝟕 𝐕
In a 25kVA. 1100/400 V, single phase transformer, the iron and copper losses at full load are 350 and 400 watts,
respectively. Calculate the efficiency on unity power factor at half load. Determine the kVA at which maximum
efficiency occurs.
Transformer rating = 25kVA
Iron losses, 𝑃𝑖 = 350 W
Full load copper losses, 𝑃𝑐 = 400 W
Load power factor cos 𝜙 = 1
1
Fraction of the load 𝑥 = 2 = 0 ⋅ 5
Efficiency of transformer, 𝜂𝑥
𝑥 × 𝑘𝑉𝐴 × 103 × cos 𝜙
=
× 100
𝑥 × 𝑘𝑉𝐴 × 103 × cos 𝜙 + 𝑃𝑖 + 𝑥 2 𝑃𝑐
0.5 × 25 × 103 × 1
=
× 100 = 𝟗𝟔. 𝟓𝟐 %
0.5 × 25 × 103 × 1 + 350 + (0.5)2 × 400
𝑃
350
Output kVA corresponding to maximum efficiency = Rated kVA √𝑃𝑖 = 25 × √400 = 23.385 kVA
𝑐
Output power or load on maximum efficiency = output kVA for max efficiency × 𝑝. 𝑓.
= 23.385 × 1 = 𝟐𝟑. 𝟑𝟖𝟓 𝐤𝐖
What is the resolution of a 8 -bit ADC operating at 10 V range.
Here, Operating Range, 2mp = 10 V
No. of bit, 𝑛 = 8
So, Quantization Level, 𝐿 = 2𝑛 = 28 = 256
2 mp
10
So, resulation, Δ = L = 256 = 0.0390625 V = 𝟑𝟗. 𝟎𝟔𝟐𝟓 𝐦𝐕
293