Hello, I have set up 8 equations with 6 unknowns for the following circuit, but I cannot solve them.
Uq1= 18V
Uq2 = 16 V
Uq3 = 14 V
R = 5 ohms
(No Ratings Yet)
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I would solve it with superposition.
Because if the proportion current of each voltage source is first calculated for these alone, in which the respective other two voltage sources are considered to be “thankful” as a short circuit.
In this case, resistance networks of the form “balanced Wheatstone’s measuring bridge” arise again and again at which each resistor is 5 ohms. This would be a substitute resistance for the “Wheatstonesche Messbrücke” = 5 Ohm. And additionally a series resistance of 5 ohms. Thus, the total resistance of each superposition is 10 ohms (= 5 ohms + 5 ohms).
If I have it now per superposition (a,b,c for <=> Uq1,Uq2,Uq3) receivable it results:
I1 = I1a + I1b + I1c
I2 = I2b + I2a
I3 = I3a + I3b + I3c
The 1/2 total current flows in each of the branches of the Wheatstone measuring bridge, since all resistors are 5 ohms. This makes the calculation of the other 2 currents per superposition (a,b,c) easier in the calculation.
I1a = 18V/10 Ohm = 1.8A and (weight) I2a = I3a = 0.5*1.8A = 0.9A
I2b = 16V/10 Ohm = 1.6A and (weight) I1b = I3b = 0.5*1.6A = 0.8A
I3c = 14V/10 Ohm = 1.4A and (order) I1c = I2c = 0.5*1.4A = 0.7A
3 cases Significant add:
I1 = I1a + I1b + I1c = 1.8A + 0.7A = 1.7A
I2 = I2b + I2a + I2c = …
I3 = I3a + I3b + I3c = …
I didn’t draw a drawing with electric arrows. But the principle is recognizable.
Good luck!
You did not define directions of currents. Without them, you cannot discuss your solution. I’ll get this.