Tutorial 2 1.Review Ohms law, KVL and KCL 2.The Wheatstone Bridge 3.Source Transformation

Tutorial 2

1. Review Ohms law, KVL and KCL 2. The Wheatstone Bridge3. Source Transformation

Tutorial question-2, Q1

Given the circuit below, show your working to find:The voltage vo

The current i1 and i2The 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 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

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 = i1Further, from Ohm’s Law & KVL; i4R4 = i2R2 and i3R3 = i1R1

The Wheatstone Bridge continued

Hence

44

33

22

11

Ri

Ri

Ri

Ri 2

4

31 RR

RR

3241 RRRR

The Wheatstone Bridge: Example

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

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2002

4

31 RR

RR

24

31 RR

RR 3241 RRRR

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 RxAnswer = R2xR3/R1 = 4000Ω

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

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

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)

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 Resistance = 16Ω2. Current = 3A Resistance = 16Ω

Tutorial question-1 Q10

1. open-circuit voltage of Vab(Voc)

Note: a, b are open, no current in the 8Ω resistor, no voltage drop as well.

×40 =48V

2. Short-circuit current of Iab (Isc)V1 is not equal to previous V40Ω

Rtotal = 10 + 40||8 = 16.67ΩItotal = 60/Rtotal = 3.6 AV1 = 60V-V10Ω = 60-10*3.6 = 24VIsc = V1/8 = 24/8 = 3A

3. Rth = Voc/Isc = 48/3 = 16Ω

Isc

+

-

V1

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