Download Tutorial-2 (Week-5)

Tutorial 2
1. Review Ohms law, KVL and KCL
2. The Wheatstone Bridge
3. Source Transformation
Tutorial question-2, Q1
Given the circuit below, show your working to find:
The voltage vo
The current i1 and i2
The power developed by the current source.
Tutorial question-2, Q2
Find the power developed by the 50V source.
Tutorial question-2, Q3
For the circuit shown calculate:
a) the total current delivered through terminals a and b
b) the power generated by the voltage source
c) the current in the 48W resistor
2. The Wheatstone Bridge
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We use an “Ohmmeter” to measure an unknown resistance
The heart of the simplest Ohmmeter is a so-called
“Wheatstone Bridge” circuit
If R1 was a variable resistor, we can adjust it until Vab = 0
The Balanced Wheatstone Bridge
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When Vab = 0, a special condition occurs: the bridge is
said to be “balanced”, i.e. Va = Vb
This implies that ig = 0, hence from KCL, i4 = i3 and i2 = i1
Further, from Ohm’s Law & KVL; i4R4 = i2R2 and i3R3 = i1R1
The Wheatstone Bridge continued
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Hence
i1 R1 i3 R3

i2 R2 i4 R4
R3
R1 
R2
R4
R1R4  R2 R3
The Wheatstone Bridge: Example

Calculate R1 in a Wheatstone bridge when it is
balanced and when R2 = 300Ω, R3 = 200Ω, R4 =
100Ω .
R1R4  R2 R3
R3
R1 
R2
R4
R3
200
R1 
R2 
300  600
R4
100
Graph of Voltage vs. Resistance
Unbalanced bridge will produce a voltage at Vab
Tutorial question-1 Q9
In the Wheatstone bridge circuit above, R1 = 1000Ω, R2 = 2000Ω,
R3 is adjusted so that the voltmeter reads 0V (i.e. the bridge is
"balanced"), at this point R3 = 2000Ω. Given this information state
the value of Rx
Answer = R2xR3/R1 = 4000Ω
3. Source Transformation
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Source transformations are useful method of circuit analysis
It is theoretically possible to replace any given arbitrary linear
circuit containing any number of sources and resistances with
either a Thévenin equivalent or Norton equivalent circuit
A simple source transformation allows a voltage source with
series resistance to be replaced by a current source with a
parallel resistance and vice versa.
Both circuits behave in the same way to external loads.
a
b
Thévenin Equivalent Circuit
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All linear circuits can be modelled by an independent
voltage source and a series resistor
The voltage of the source is the open circuit voltage of
the network across ‘a’ and ‘b’
The resistance is determined from the short circuit
current
Norton Equivalent Circuit
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All linear circuits can be modelled by an independent
current source and a parallel resistor
The current of the source is the short circuit current
through ‘a’ and ‘b’
The resistance is determined from the open circuit
voltage (same as Thévenin resistance)
Tutorial question-1 Q10
This question relates to source transformations. The circuit above
contains an independent voltage source and three resistors. By
making a series of source transformations, or otherwise,
1. determine the Thévenin equivalent voltage and resistance.
2. determine the Norton equivalent current and resistance
1. Voltage = 48V
2. Current = 3A
Resistance = 16Ω
Resistance = 16Ω
1. open-circuit voltage of Vab(Voc)
Note: a, b are open, no current in
the 8Ω resistor, no voltage drop as
well.
Vab(Voc)= V40Ω = I40Ω×R
60
=
10+40
×40 =48V
2. Short-circuit current of Iab (Isc)
V1 is not equal to previous V40Ω
Rtotal = 10 + 40||8 = 16.67Ω
Itotal = 60/Rtotal = 3.6 A
V1 = 60V-V10Ω = 60-10*3.6 = 24V
Isc = V1/8 = 24/8 = 3A
3. Rth = Voc/Isc = 48/3 = 16Ω
+
V1
-
Isc
Tutorial question-2, Q4

Find both Thevenin and Norton equivalent circuits
for the circuit below
Tutorial question-2, Q5
In the circuit shown below, by applying suitable source
transformations, find the current i0 flowing in the 2.7kW resistor.
Thevenin convert
Thevenin convert