circuit analysisthevenin
We are given the following circuit and told to calculate the equivalent voltage of everything to the left of the a and b terminals:
It omits the load resistor and lays the circuit out as such for mesh analysis:
It claims that voltage across terminals a and b, the voltage we are looking for, is the same as the potential in between the 4 ohm and 1 ohm resistor. Shouldn't that 1 ohm resistor cause the potential to change a little?
I'm not going to just give you the answer to your homework problem.
However, consider what it really means to be a voltage source. Let's say I have a 6V source that is good to 1 A. What voltage is it when nothing is connected? When I put a 60 Ω resistor accross it? A 30 Ω resistor accross it?
If I put the 6 V source and resistor inside a black box and brought out only the two leads, what difference could someone observe on the outside to distinguish the no load, 60 Ω and 30 Ω cases? Now consider this is the case you have. Your +6V and R are inside the box and the rest of the circuit on the outside.
The current through the load resistor can be obtained by finding the THEVENIN EQUIVALENT across it.If you are unaware about do check out thevenin intro or any other online material on it.It is very useful in solving networks for currents and voltages.
The thevenin equivalent across 120 ohm is :
From the above circuit the current through load can be easily calculated as:
I(Load) = [ 0.385Vs /(110.76 + 120) ] = 0.0016684 Vs ( where Vs is the supply voltage at input)
Now, to get the current drawn from the supply voltage Vs at the input,
Step 1: Find the equivalent resistance across the input source which involves star to delta conversion as you have mentioned.
The R(equivalent) = [ 37.9 + ( (56.8 + 80) || (25.3 + 180) ) ] = 120 ohms.
Step 2:
The current drawn from the input is : I(supply) = Vs/ Req = 0.00833 Vs.
Now that we have both currents in terms of the input supply voltage,
The ratio of load current to the current drawn from the supply is:
I(Load) / I(supply) = 0.001664 /0.00833 = 0.2
If the answer is 5 then it should be the inverse of the above as current through load is always less than supply current.
Edit: Since you wanted to solve it by mesh analysis here is an easier way:
Best Answer
You're trying to find the Thevenin equivalent voltage of the network.
This is the voltage that the network produces when the output is open. When the output is open, no current flows out or in to the
aterminal. Therefore no current flows through the 1 ohm resistor.Since no current is flowing through the 1 ohm resistor, the voltage across it is 0, by Ohm's law.