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i want to know that wether my answer is right or wrong and if it is where?

  • 1
    \$\begingroup\$ Why do you remove the 5Ohm resistor? \$\endgroup\$
    – po.pe
    Feb 27 '18 at 7:49
  • \$\begingroup\$ @po.pe, I think the idea is to reduce the circuit, minus the 5 Ohm resistor, to a Thévenin equivalent source (by Thévenin's theorem) with terminals A and B; and then predict the Voltage that would be dropped by a 5 Ohm resistor connected across that source. There's other ways to get the same answer, but maybe not if "by Thévenin's theorem" is quoted from a homework assignment. \$\endgroup\$ Jun 22 at 2:12

Looks correct. I didn't use thevinin's theorem, but I got the same answer as you by doing some algebra on the node voltages. I2, I1, etc below are the currents into node A, through the 2 ohm, 1 ohm, etc resistors respectively. They must sum to zero, since current that enters the node must also leave it somewhere. A is the voltage at point A. I've set the voltage at point B to zero. (since you have to do that somewhere).

I2 = (20-A)/2

I1 = (-10-A)

I5 = -A/5

I4 = (12-A)/4

I1+I2+I4+I5 = 0

10 - A/2 -10 -A -A/5 +3 -A/4 = 0

10 - 10A/20 -10 -4A/20 +3 -5A/20 = 0

60 - 10A -20A -4A -5A = 0

60 = 39A

A = 60/39 = 1.538


A rather faster way to do it, practically in ones head, is to notice that all the supplies are in parallel. So convert them all to their Norton equivalent.

By inspection, the first source is 10A//2ohms, the second is -10A/1ohm, and the third source 3A//4ohms.

That's a total current into node A of 3A, with a total shunt resistance of 2//1//5//4 (OK, I used the calculator for that bit!) which is a resistance of 0.5128 ohms, which at 3A is 1.538v.

It's always nice to do a problem using several different methods and get the same answer.

Confession, my first answer was 11.79v. I wonder how many other people didn't notice the 10v source was reversed compared to the other two!


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