E1.1 Circuit Analysis Problem Sheet 1 (Lectures 1 & 2)

1 Ver Analysis of Circuits (2014) Circuit AnalysisProblem Sheet 1 (Lectures 1 & 2)Key: [A]= easy .. [E]=hard1. [A] One of the following circuits is a series Circuit and the other is a parallel Circuit . Explain whichis which.(a)(b)2. [B] Find the power absorbed by by each of the subcircuitsAandBgiven that the voltage andcurrent are 10 V and 2 A as [B] For each of the four circuits below, find the power absorbed by the voltage source (PV), thepower absorbed by the current source (PI) and the total power absorbed (PV+PI).(a)(b)(c)(d)4. [B] Determine the voltageVXin the following [B] Determine the currentIXin the following [B] What single resistor is equivalent to the three resistor sub- Circuit shown below? Problem Sheet 1 Page 1 of 2 Ver Analysis of Circuits (2014)7. [B] What single resistor is equivalent to the three resistor sub- Circuit shown below?

2 8. [C] What single resistor is equivalent to the five resistor sub- Circuit shown below?9. [A] If a resistor has a conductance of 8 S, what is its resistance?10. [B] Determine the voltage across each of the resistors in the following Circuit and the power dissipatedin each of them. Calculate the power supplied by the voltage [B] Determine the current through each of the resistors in the following Circuit and the power dissip-ated in each of them. Calculate the power supplied by the current [B] DetermineR1so thatY= [B] ChooseR1andR2so thatY= +R2= 10 M .14. [D] You have a supply of resistors that have the values{10,12,15,18,22,27,33,39,47,56,68, 82} 10n for all integer values ofn. Thus, for example, a resistor of 390 is available and the next highervalue is 470 . Show how, by combining two resistors in each case, it is possible to make networkswhose equivalent resistance is (a) 3 k , (b) 4 k and (c) as close as possible to k.

3 Determine isthe worst case percentage error that might arise if, instead of combining resistors, you just pick theclosest one Sheet 1 Page 2 of 2 Ver Analysis of Circuits (2014) Circuit AnalysisProblem Sheet 1 - Solutions1. Circuit (a) is a parallel Circuit : there are only two nodes and all four components are connectedbetween (b) is a series Circuit : each node is connected to exactly two components and the same currentmust flow through For subcircuitBthe voltage and current correspond to the passive sign convention ( the currentarrow in the opposite direction to the voltage arrow) and so the power absorbed byBis given byV I= 20 deviceAwe need to reverse the direction of the current to conform to the passive sign the power absorbed byAisV I= 20 must always be true, the total power absorbed by all components is The power absorbed is positive if the voltage and current arrows go in opposite directions and negativeif they go in the same direction.

4 So we get: (a)PV= +4,PI= 4, (b)PV= +4,PI= 4, (c)PV= 4,PI= +4, (a)PV= 4,PI= +4. In all cases, the total power absorbed isPV+PI= We can find a path (shown highlighted below) from the bottom to the top of theVXarrow thatpasses only through voltage sources and so we just add these up to get the total potential difference:VX= ( 3) + (+2) + (+9) = +8 If we add up the currents flowingoutof the region shown highlighted below, we obtainIX 5 1+2 =0. HenceIX= 4 The three series resistors are equivalent to a single resistor with a value of 1 + 5 + 2 = 8 k .7. The three series resistors are equivalent to a single resistor with a value of11/1+1/5+1/2= k .8. We can first combine the parallel 2 k and 3 k resistors to give2 32+3= k. This is then in series withthe 4 k resistor which makes k in all. Now we just have three resistors in parallel to give a totalof11/1+1/5+1.

5 9. The resistance is18 10 6= 125 k Solution Sheet 1 Page 1 of 2 Ver Analysis of Circuits (2014)10. [Method 1]: The resistors are in series and so form a potential divider. The total series resistance is7 k, so the voltages across the three resistors are 14 17= 2 V, 14 27= 4 V and 14 47= 8 V. Thepower dissipated in a resistor isV2R, so for the three resistors, this gives221= 4 mW,422= 8 mW and824= 16 mW.[Method 2]: The total resistance is is 7 k so the current flowing in the Circuit is147= 2 mA. The voltageacross a resistor isIRwhich, in for these resistors, gives 2 1 = 2 V, 2 2 = 4 V and 2 4 = 8 power dissipated isV Iwhich gives 2 2 = 4 mW, 4 2 = 8 mW and 8 2 = 16 mW. Thecurrent through the voltage source is 2 mA, so the power it is supplying isV I= 14 2 = 28 is, inevitably, equal to the sum of the power disspipated by the three resistors: 4 + 8 + 16 = [Method 1]: The resistors are in parallel and so form a current divider: the 21 mA will divide inproportion to the conductances: 1 mS, mS and mS.

6 The total conductance is mS, so thethree resistor currents are 21 12 mA, 21 6 mA and 21 3 mA. The powerdissipated in a resistor isI2 Rwhich gives 122 1 = 144 mW, 62 2 = 72 mW and 32 4 = 36 mW.[Method 2]: The equivalent resistance of the three resistors is11/1+1/2+1/4=47k . Therefore thevoltage across all components in the parallel Circuit is 21 47= 12 V. The current through a resistorisVRwhich gives121= 12 mA,122= 6 mA and124= 3 mA. The power dissipated in a resistor isV Iwhich gives 12 12 = 144 mW, 12 6 = 72 mW and 12 3 = 36 mW. The power supplied by thecurrent source is 12 21 = 252 mW which as expected equals 144 + 72 + The resistors form a potential divider, soYX=4R1+4. So we want4R1+4=14 R1+ 4 = 16 R1= 12 The resistors form a potential divider, soYX=R2R1+R2. So we wantR2R1+R2=110andR1+R2=10 M.

7 Substituting one into the other and cross-multiplying gives 10R2= 10 M R2= 1 M .Substituting this into the simpler of the two initial equations givesR1= 10 1 = 9 M .14. (a) 3 k = k + k = k||33 k, (b) 4 k = k + 100, (c) k = k||33 make an exhaustive search for creating a resistance ofR, you need to consider two possibilities:(i) for two resistors in series, the largest of the two resistors must be in the range [12R, R] or (ii) fortwo resistors in parallel , the smallest resistor must be in the range [R,2R]. In both cases there areat most four possibilities, so you need to consider up to eight possibilities in all. So, for example, forR= k, we would consider the following possibilities: (i) k + k = k, k + k = k, k + k = k, k + k = k and (ii) k||33 k = k, k||15 k = k, k||10 k = k, k|| k = k.

8 The choice with least error is the one given we are interested in % errors, we need to consider the ratio between resistor values. The largestratio between successive resistors is the series is1512= (this includes the wraparound ratio of10082= ). The worst-case percentage error will arise if our target resistance is the mean of thesetwo values, The percentage error in choosing either one is Sheet 1 Page 2 of 2 Ver Analysis of Circuits (2015) Circuit AnalysisProblem Sheet 2 (Lectures 3 & 4)Key: [A]= easy .. [E]=hard1. [B] CalculateVXandIXin the following Circuit using (a) nodal Analysis and (b) simplifying thecircuit by combining parallel [B] CalculateVXandIXin the following Circuit using (a) nodal Analysis and (b) simplifying thecircuit by combining parallel [C] CalculateVXin the following Circuit using (a) nodal Analysis and (b) [C] CalculateVXin the following [C] CalculateVXin the following [C] CalculateVXin the following Sheet 2 Page 1 of 3 Ver Analysis of Circuits (2015)7.

9 [C] CalculateVXin the following Circuit . The value of the dependent current source is 99 time thecurrent flowing through the 1 V voltage [C] In the following Circuit calculateVXin terms ofVandIusing (a) nodal Analysis and (b) [C] CalculateVXandIXin the following Circuit using (a) nodal Analysis and (b) [C] Determine an expression forIXin terms ofVin the following Circuit . Determine the value ofVthat will makeIX= [C] CalculateVXin the following Circuit using (a) nodal Analysis and (b) [C] CalculateVXin the following Circuit which includes a dependent voltage Sheet 2 Page 2 of 3 Ver Analysis of Circuits (2015)13. [C] Find the equivalent resistance of the network shown [D] Prove that ifVAB= 0 , thenR= 4 k in the following Circuit . The Circuit is used to detectsmall changes inRfrom its nominal value of 4 k.

10 Find an expression forVABas a function changes inVABof 10 mV can be detected, what is the smallest detectable change [D] CalculateVXin the following Circuit . You can either use nodal Analysis directly or else simplifythe Circuit a little to reduce the number of [D] CalculateVXin the following Circuit which includes a floating dependent voltage Sheet 2 Page 3 of 3 Ver Analysis of Circuits (2015) Circuit AnalysisProblem Sheet 2 - SolutionsNote:In many of the solutions below I have written the voltage at nodeXas the variableXinstead ofVXin order to save writing so many [Nodal Analysis ] KCL at nodeVXgivesVX 141+VX2+VX4= 0 which simplifies to 7VX 56 = 0 fromwhichVX= 8.[ parallel resistors] We can merge the 2 and 4 resistors to make one of2 42+4=43 as shown we have apotential divider, soVX= 14 8 both cases, we can now calculateIX=VX4= 2 A.