Engineering Voltage and current calculation of circuit

[PhysicsTest]

Homework Statement

Need to calculate the voltage and current for a circuit

Relevant Equations

KCL and KVL equations , V=IR

I am trying to find voltages and currents of the below circuit,

3.3 -22000*I2 - 2200(I2-I3) = 0
3.3 -24200*I2 + 2200*I3 = 0 -> eq1

-(10 - I3)*1500 - (I2 - I3)*2200 = 0
-15000 + 1500*I3 - 2200*I2 + 2200*I3=0
-15000 - 2200*I2 + 3700*I3=0 -> eq2

Solving equations 1 and 2 i get I2=0.38 Amps and I3 = 4.28 Amps

I assume i have completely gone wrong in calculations, can current I3 flow into the source. Please help.

 

[PhysicsTest]

If current 10 Amps flows through R3 then the voltage at vout1 shall be 1.5K*10 Volts = 15K volts, but if i see the simulation results it shows 20K Volts

It is all confusion.

 

[Baluncore]

Replace your .Transient analysis with a .DC operating point analysis. That will give you the voltages at all nodes and the currents in all components.

You may have to turn a passive component around to get that loop current flowing with the correct sign in the .DC report.

The arrow on the current source is conventional current. Your I3 arrow is in the opposite direction, which makes visualisation of the current flow more difficult.

You can replace R3 with a short circuit because it is in series with a current source.

 

[PhysicsTest]

You can replace R3 with a short circuit because it is in series with a current source.

This helped me to solve the problem
The updated circuit is

3.3 - 22000*I2 -2200*(10+I2) = 0
solving I2 = -0.90 Amps
Vout1=20K Volts it matches with the simulation results
--- Operating Point ---

V(n001): 3.3 voltage
V(vout1): 20000.3 voltage
I(I1): 10 device_current
I(R2): 9.09105 device_current
I(R1): -0.908955 device_current
I(V1): 0.908955 device_current

Still my question is why do i need to remove R3 to get result?

 

[Baluncore]

If current 10 Amps flows through R3 then the voltage at vout1 shall be 1.5K*10 Volts = 15K volts, but if i see the simulation results it shows 20K Volts

I think you were assuming the voltage across the current source was zero.
You were looking at the 10 amp through the 1k5 when it actually flows through the Thevenin resistance of Vout.

Another way to analyse it:
Node(2) has Thevenin resistance; Rth = (22k//2k2) = 2k000
3.3V * 2k2 / ( 22k + 2k2 ) = +0.300 volt;
Then add the (10A * 2k000) = 2000V;
To get node(Vout1), V(out) = +2000.300 V.

 

[osilmag]

If you would change your sign for loop2 like Baluncore wrote, I think you would arrive at the solution. The current source being in the positive direction.

(10-I3)*1500 + (I2-I3)*2200=0