
#1
Oct1809, 05:37 AM

P: 3

In the network given below find the voltage VX .
Please check the link for the circuit diagram. And Kindly explain the method? http://i33.tinypic.com/o7u748.jpg 



#2
Oct1809, 05:43 AM

P: 86

I can help. but I am mindful of your need to learn, so I want to see some effort. What have you tried so far?




#3
Oct1809, 09:21 AM

P: 3

because which i learnt is v.easy circuits 



#4
Oct1809, 09:25 AM

Sci Advisor
P: 4,003

Find Voltage
What is the current in the 10 ohm resistor?
So what is the current in the 5 ohm resistor? What is the current in the 15 ohm resistor? What else do you know that would help here? 



#5
Oct1809, 05:11 PM

P: 86

This network may be simplified. Notice first of all that a 5 ohm resistor is in series with a 10 ohm resistor in the lefthand (LH) branch of the cct. The total resistance of the LH branch is then 5+10=15 ohms.
This 15 ohms is in parallel with the middle branch, also 15 ohms... the combined resistance of these 2 branches is then 15/2=7.5 ohms. Now we see that an equivalent resistance of 7.5 ohms is in series with the 30 ohm resistance, making 37.5 ohms in all. Hence whatever potential is supplied by the power source is divided by 5 to discover the potential on the lower equivalent resistance of 7.5 ohms, because 7.5 is 1/5 of the resistance in the equivalent series cct. The potential of 1/5 the supply is applied to a series cct (the LH branch) comprised of a 5 ohm and a 10 ohm resistor. 2/3 of the 1/5 is expressed across the 10 ohm resistance. By multiplication we see that 2/3 x 1/5 = 2/15. Now we know that the 30v measured across the 10 ohm resistor is 2/15 ofthe supply voltage. We can proceed by recognising that if 30V = (2/15 )*(Vsupply) then supply voltage must be: Vs=30V/(2/15) or 225V. Everything else should be easy enough from here. However if you need more help to understand what is going on, please speak up. (I hope I have not overhelped you, it really is good to bang your head aganst the wall until you find the door... that way you don't forget where the door is and you get some confidence that you can find solutions to physics problems by yourself.) I really hope that once you have understood this circuit you will go out of your way to find lots more homework problems and keep practising until they become easy and you can show your classmates how to do them. If you do this there will be great rewards for you, if you don't it will be a great, missed, opportunity. 



#6
Oct1909, 11:30 AM

P: 3

Thanks buddy... for this useful post.... But after finding equivalent resistance which formula you have used?




#7
Oct1909, 08:07 PM

Sci Advisor
P: 4,003

You can do this with OHMS law and some mental arithmetic.
Just work it out one bit at a time. What is the current in the 10 ohm resistor? I = V / R It has 30 volts across it so the current is 30 volts / 10 ohms = 3 amps So what is the current in the 5 ohm resistor? It is in series with the 10 ohm resistor so it also has 3 amps flowing in it. What is the voltage across the 10 ohm plus the 5 ohm resistors? V = I * R = 3 amps * 15 ohms = 45 volts What is the current in the 15 ohm resistor? I = V / R = 45 / 15 = 3 amps So the 15 ohm resistor will also have 3 amps flowing in it. What is the current in the 30 ohm resistor? It is the total of 3 amps plus 3 amps = 6 amps What is the voltage across the 30 ohm resistor? V = I * R = 6 amps * 30 ohms = Now, add this voltage to the 45 volts we got earlier to get Vs. 


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