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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 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 = i1 Further, from Ohm’s Law & KVL; i4R4 = i2R2 and i3R3 = i1R1 The Wheatstone Bridge continued 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 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) 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