Question

MULTISIM 4.36 a) Use the mesh-current method to find the branch PSPICE currents in, in, and i, in the circuit in Fig. P4.36. b) Repeat (a) if the polarity of the 140 V source is reversed. Figure P4.36 7512 150 12 Ic 80 V in 182000 140 V 1252 250 12 w w

Answer 1

From the diagram

$i_b=i_a+i_c$

a)

75n 200 80V /25 m i'a 250n

Now, applying mesh analysis in loop 1

$-80+75i_a+200i_b+125i_a=0$

$-80+75i_a+200(i_a+i_c)+125i_a=0$

$400i_a+200i_c=80$

Divide by 10 on both sides

$40i_a+20i_c=8$                                       ....(1)

Now, applying mesh analysis in loop 2

$140-250i_c-200i_b-150i_c=0$

$200i_a+600i_c=140$

Divide by 10 on both sides

$20i_a+60i_c=14$                                    ....(2)

On solving equation (1) and (2), we have

${\color{Red} i_a=0.1\;A}$

${\color{Red} i_c=0.2\;A}$

${\color{Red} i_b=i_a+i_c=0.1+0.2=0.3\;A}$

b)

75 150L l'a 200A 8OU 140 V 250n

Polarity of 140 V source is reversed

Now, applying mesh analysis in loop 1

$-80+75i_a+200i_b+125i_a=0$

$-80+75i_a+200(i_a+i_c)+125i_a=0$

$400i_a+200i_c=80$

Divide by 10 on both sides

$40i_a+20i_c=8$                                       ....(1)

Now, applying mesh analysis in loop 2

$-140-250i_c-200i_b-150i_c=0$

$200i_a+600i_c=-140$

Divide by 10 on both sides

$20i_a+60i_c=-14$                                    ....(2)

On solving equation (1) and (2), we have

${\color{Red} i_a=0.38\;A}$

${\color{Red} i_c=-0.36\;A}$

${\color{Red} i_b=i_a+i_c=0.38-0.36=0.02\;A}$

Here negative sign represents that the direction of current is opposite to the direction we have assumed in the figure.