1. Tutorial 2 Review Ohms law, KVL and KCL The Wheatstone Bridge
Source Transformation

2. 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.

3. Tutorial question-2, Q2
Find the power developed by the 50V source.

4. 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

5. 2. The Wheatstone Bridge
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

6. The Balanced Wheatstone Bridge
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

7. The Wheatstone Bridge continued
Hence

8. The Wheatstone Bridge: Example
Calculate R1 in a Wheatstone bridge when it is balanced and when R2 = 300Ω, R3 = 200Ω, R4 = 100Ω .

9. Graph of Voltage vs. Resistance
Unbalanced bridge will produce a voltage at Vab

10. 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

11. 3. Source Transformation
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

12. Thévenin Equivalent Circuit
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

13. Norton Equivalent Circuit
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)

14. 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,
determine the Thévenin equivalent voltage and resistance.
determine the Norton equivalent current and resistance
Voltage = 48V Resistance = 16Ω
Current = 3A Resistance = 16Ω

15. 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 = ×40 =48V
2. Short-circuit current of Iab (Isc)
V1 is not equal to previous V40Ω
Rtotal = ||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 -

16. Tutorial question-2, Q4
Find both Thevenin and Norton equivalent circuits for the circuit below

17. 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