Question

please solve these 2 problems. I will rate you up

In the circuit shown below, let V. - 3 V.1. - 1 mA, R-3k, Ry - 2.5 kA, R = 1 kq Use mesh analysis to find mesh currents I, and 12. Also, find Vı. R1 V1 R2 W 2 is len 3kR 12 R3 100 IMA Problem 5 In the circuit shown below, let V, -8 V, 1. - 2 mA, R2 = 1 k 2, R2 = 400, R3-1.5 k-2, R4 -2.5 k-2, Rs = 600 2. Use mesh analysis to find mesh currents 11, 12, and 13. Also, find V1, V2, and V3. R1 V1 12 R2 R3 w w 11 Is Vs 8V 0 V2 V3 13 R4 O R5 w W

Answer 1

1

Let's rewrite the given circuit with polarity of current through resistors.

RI V1 W R2 WV 1511L VLE+ + 17 + AE () RS T + YS

Let's assume right hand side loop is mesh-2 and left hand side loop is mesh-1.

We can see in mesh-2 mesh current is flowing I_{​​​​​​2} in clockwise direction and the source current I_{​​​​​​s} is not sharing with any other mesh flowing opposite to I_{​​​​​​2} . Since in a wire there can be only one current therefore

-1mA

Now apply KVL for mesh-1 we get

11R1 + 11 – 12) R3 = V.

Put the given values we get

3kΩΙ1 + (Ι1 + 1mA)1kΩ = 3V = 41kΩ = 3 – 1 = 2V I = 0.5mA

Now the current through resistor is

R3

11 – 12 = (0.5m A-(-1m A)) = 1.5m A

Therefore potential

Vi = (11 – 12) R3 = 1.5m A x 12 Vi = 1.5V

therefore answer to first question is I1=0.5mA and I2=-1.0mA and potential V1=1.5V.

.2

Now here in this problem since source current is sharing with two mesh so can not solve as we did in previous problem. So what we do here let's consider the potential across source current is v_{​​​​​s}

R1 VE + 12 + R2 + 1 R3 - + 11 Is VS Vz V3 8v Vis + 1 + R4 R5 + 3 13

Then we apply KVL in the three loops.

Mesh-1

11R1 + (11 – 12) R2 + (11 – 13)R4 = 8V 11R1 + R2 + R4) – 12R2 – 13 R4 = 8V............... (1)

For Mesh-2

(12 - 11)R2 + 13R3 = V1s → -11R2 + 12(R2 + R3) = V1................(2)

For mesh-3

(13 – 11)R4 + 13R5 = -V1s → -I1R4 + 13(R4 + R5) = -V1............ (3

Add equation (2) and (3) we get

-I1(R2 + R4) + 12(R2 + R3) + 13(R4 + R5) = 0........

Now consider the source current and apply the Kirchhoff's current law at the highlighted node. We get

I2+1, = 13............ ...(5)

Put this value of Ι3 in equation (1) and (4) we get respectively.

11(R1 + R2 + R4) – 12R2 - (12 +13)R4 = 8V = 11(R1 + R2 + R4) – 12(R2 + RA) - I.R4 = 8V..............(6)

And

-I (R2+R4)+12 R2+R3)+(I2+I, (R1+R5) = 0 → -11R2 + R4) + 12(R2 + R3 + R4 + R5) + Is(R4 + R5) 0............. (7

Let's first put the values of resistance in equation (6) and (7) remember 1k-ohm=1000 ohm

From (6)

11(1000 + 400 + 2500) – 12(400 + 2500) – 2 x 10- (2500) = 8V 390011 – 290012 5...........

And from (7)

- 11(400+2500) +12(400 + 1500 +2500+600) +2 x 10-3(2500+600) = 0 - 290011 + 500012 = -6.2 ............(9)

Solve equation (8) and (9) we get

11 = 1.5m A

And

0.4mA

Put values in equation (5) we get

13 = 12 + 1) = -0.4mA + 2m A 13 = 1.6mA

Now

V1 = Vs – 11R1 V1 8V - 1.5m A x 1k2 Vi = 6.5V

And

V2 = (11 – 13)R4 = V2 = (1.5m A - 1.6m A) x 2.5k V2 = -0.25V

And

V3 = (13) R5 → V3 = 1.6m A x 6002 V3 = 0.96V

Hence we have found all quantities.