Question

14)(4 points) Analyze the circuit in problem 9-12 using PSPICE as follows: Use a Bias Point analysis to find Ix, Iy, VA, VB and Vc in problem 12 above. Be sure to add text to the schematic as indicated in the Sample Report. Only show the result of Ix, ly, VA, VB and Ve on the schematic. Adjust the placement of each value on the schematic so that it is moved slightly away from the component to avoid crowding. Do not show any current, voltage, or power values on the schematic other than those required.




Please show all work and to follow directions. Only answer Problem 14

Answer 1

Solution :

Q9.

A с 20 50 IX IY 200 V 40 70 B

Here, Total resistance seen by the 200 v voltage source will be the equivalent resistance between A and B. And the current flowing through the Equivalent resistance will be I_x [ As 20 ohm is in series with the resistance ].

So, Resistance Between A,B = R_AB = 20 + { 40 || (50+70) } = 20 + { 40 || (120) } = 20 + 40*120/(40+120) = 20 + 30 = 50 Ohm.

[ As, 50 and 70 ohm are in series with each other and 40 ohm is parallel ( represented by ' || ' symbol ) with them and finally 20 ohm is in series with the whole circuit. ]

So, Ix will be 200/50 A = 4 A. [ As per ohm's law, V = I* R. or, I(current ) = V(voltage) / R(resistance) ].

Here, Current flows through Ix = 4 A, and in divided in 2 , one go to the 40 ohm resister and the other to the 50 series 70 = 120 ohm resistor.

By, Current divisor rule, Iy will be,

Source Current * { Resistance of the opposite lane / (Resistance of the opposite lane + Resistance of the same lane) }

= 4 A * { 40 / (40+120) } = 4 A * 40 / 160 = 1 A.

So, Answer  

a. 50 ohm.

b. 4 A.

c. 1 A.

Q.10.

Using Source transformation,

R1 R2 200 500 V1 -200 V R3 4002 R4 7002

R2 5002 11 10 A R3 R1 200 R4 7002 -400

R2 50Ω Rx 11 10A 13.3333Ω R4 70Ω

Rx R2 13.3333 5002 R4 -700 V1 133.333 V

Ry 133.3333 _V1 —133.333 V

So, We converted v1 to i1, where i1 = v1/r1 = 200/20 = 10 A,

next we find R1 and R3 parallel, that makes 13.333 ohm in equivalent as Rx.

Now, we convert I1 to V1, where V1 = I1 * Rx = 133.333 V

Now, Rx , R2 and R4 are in series makes 133.333 ohm equivalent as Ry.

Now Current through Ry = V1 / Ry = 133.33/133.33 = 1 A.

Which is same as Iy.

Q11.

Mesh Analysis,

KVL in Loop Y,

0 = (40+50+70)*Iy - 40*Ix

or, 40*Ix = 160 Iy

or, Ix = 4 * Iy ....................................eq.1

KVL in Loop X,

200 = Ix*(20+40) - Iy*40

or, 200 = 60 * Ix - 40* Iy

Putting eq. 1 to action,

or, 200 = 60* 4 *Iy - 40*Iy

or, 200 = 240 *Iy - 40* Iy

or, 200 = 200 Iy

Or, Iy = 1 A

Putting value in eq.1

Ix = 4*Iy = 4*1 = 4 A.

Q.12.

Nodal Analysis,

Here,

A B с 20 50 11 13 200 V 70 12 D

Applying KCL at node B,

I1 + I2 + I3 = 0

Or. (Va - Vb)/20 + (Vd - Vb) / 40 + (Vd - Vb) / (50+70) = 0

We know, Va = 200 V, Vd = 0 V

Or, (200-vb) / 20 + (-Vb/40) + (-Vb) / 120 = 0

Or, 10 -( vb / 20) + (-Vb/40) + (-Vb) / 120 = 0

Or, 10 - Vb / 12 = 0

Or, Vb = 120 V.

So, I3 = (0-120) / (50+70) = -1 A

So, Vc = Voltage across C to D = -1 * 70*-1 = +70 Volts.

Iy = - I3 here,( opposite direction) = +1 A

Ix = I1 = (Va - Vb )/ 20 = (200 - 120) /20 = 80/20 A = 4 A

Simulation,

I: 4.00 A V: 120 V • V: 70.0 V R1 R2 tivo Ix 2002 Vb 5022 I: 1.00 A Iy -V1 —200 V R3 34002 SR4 7002

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