Electric circuits help needed

Daniel S

Hi everyone,
I'm not a regular poster on this part of the forum, but I can't think of any other options at the moment. I'm a mechanical engineering student and I've an electrical engineering exam coming up (thankfully it's just the one module in the whole course) and I'm struggling with it. I can't find any solutions so I was hoping someone could have a quick look over mine and tell me if I've gone wrong anywhere:

Q1:

VT=10 Volts
RT=220+(470//560)+100=575.53 Ohms
IT=10/575.53=0.017375288 Amps

Using Voltage divider:
VR4=(VT*R4)/(RT)=(10x100)/(575.53)= 1.74 Volts

VA=10volts-VR1=10-10(220)/575.53= 6.18 Volts

Using Current Divider:
IR2=(IT*(1/R2))/(1/R2+1/R3)=(0.017375288x(1/470))/(1/470+1/560)=0.009446758 Amps

Q2:


From 10v source on its own:
VT=10 Volts
RT=330+(100//470)+560=972.46 Ohms
IT=10/972.46=0.010283199 Amps

VA=10-10(330)/972.46=6.61 Volts

IR3=0.010283199 Amps as the 10mA source is replaced with an open circuit, right?

From 10mA source:
IT=0.01A
IR3=(0.01(1/470))/(1/(560+330)+1/470)=0.006544117 Amps

Overall/real values:

VA= 6.61 Volts
IR3=IR3(from 10v source)+IR3(from 10mA source)=0.010283199+0.006544117= 0.01627316 Amps

Q3:




From 10v source on its own:
VT=10 Volts
RT=330+(780//100)=418.64 Ohms
IT=10/418.64=0.023887079 Amps

VR1=10(330)/418.64= 7.88 Volts

Isolating the resistors in parallel:
10-7.88=2.12 Volts

VR2=2.12(560)/780=1.52 Volts

VA=2.12-1.52= 0.6 Volts

VB= 0 Volts as it's the same node as ground, right?

On the front of the exam paper: "Please note that when you are asked for the voltage at a particular node, this should be taken to mean the voltage difference that exists between that node and ground (0V)".

IR2=(0.023887079(1/780))/(1/780+1/100)=0.00271444 Amps

From 5v source:

VT=5 Volts
RT=(780//330)+100=331.89 Ohms
IT=10/331.89=0.015065232 Amps

VR4=5(100)/331.89=1.51 Volts

5-1.51=3.49 Volts

VR3=3.49(220)/780=0.98 Volts

VA=5-0.98=4.02 Volts

VB=0 Volts as it's at the end of the 5v circuit isn't it?

IR2=0.015065232(1/780)/(1/780 + 1/330)= 0.004478852 Amps

So overall:


VA=4.02-0.6=3.42 Volts

VB=0

IR2=0.00271444-0.004478852=0.001764412 Amps

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Sorry, I know it's really really long, but if someone could check over it they'd be a lifesaver (college course saver anyway!).

Thanks,

Dan

Daniel S

In case anyone else has a similar question, a guy on answers.yahoo.com gave me an answer:

Roger on answers.yahoo.com wrote:

Your answers to the first question are right. for total current It, the voltage across resistor R4 and the voltage at point A.
I through R2 = (Va - V R4)/R2 = (6.26- 1.74)/470 = 9.6 x 10^-3 amps (9.6 mA)
I through R3 = (Va - V R4)/560 = 8.07 mA
I total = 8.07 + 9.6 mA = 17.67 mA which is 10V/RT, RT = 575.5 ohms

Question 2 when using supper position you must replace current sources with opens and voltage sources with shorts.
V a from the 10 Volt source is found by
10 V = It(R1 + R3 + R4)
It = 10/(330 + 470+ 560) = 7.35 mA (7.35 x 10^-3 A)
Va = 10 - It(330) = 7.575 Volts


Now find Vab with the current source
I R3 = 10mA(R1+R3)/(R1 +R4 +R3) = 10mA(890/1,360) = 6.544 mA
I R4 = -10mA(R3/(R1 +R4 +R3) = -10mA(470/1,360) = -3.456 mA
V a = I R3(470) - I R4(560)= 3.0757 - 1.935 Volts = 1.14 volts
the total voltage across R3 is the sum of 7.575 + 1.14 = 8.715 Volts
Add the two IR3 which are 6.544 mA and 7.35 mA to get the total I R3 = 13.89 mA

Question 3
Replace the 5 volt source with a short, V b = 0 volts
I total = 10/Rt where Rt = 330 + 100||780 ohms = 418.64 ohms
I total = 23.89 mA ( 2.389 x 10^-2 amps)
I R3 = 23.89(100/(780 +100)) = 2.714 mA
V a = I R3(R3) = (2.714 mA)(220) = 0.597 volts
V ab = 0.97 volts

now replace the 10 volt source with a short

Vb = -5 volts
I total = -5/Rt where Rt = 100 + 330 || 780 = 331.89 ohms
I total = - 15.07 mA
I R2 = -(330/1,110)(15.07) = -4.48 mA
Va = -220(4.48 mA) = -0.986 volts
Vab = -0.986 -(-5) = 4.014 Volts

Vab = 4.014 +0.97 = 4.98 volts
I R2 = -4.48 mA + 2.71 mA = -1.77 mA