CN Tutorial 11

8/8/2019 CN Tutorial 11

1/13

Circuits & Networks 3rd E.C.

Examples:-

1. Calculate the equivalent resistance Rab in the circuit in Fig

below.

Solution :-

The 3and 6resistors are in parallel because they are connected to thesame two nodes c and b. Their combined resistance is

... [1]

Similarly, the 12 and 4 resistors are in parallel since they areconnected to the same two nodes dand b. Hence

................... [2]

Also the 1 and 5 resistors are in series; hence, their equivalentresistance is

1 + 5 = 6 Now equivalent circuit is as shown in fig (a)


2/13


3/13


Solution :-

The 6and 3 resistors are in parallel, so their combined resistance is

Thus our circuit reduces to that shown in Fig

Now apply Ohms law to get

Apply voltage division principle,

Now, from the fig (a), using current division principle,


4/13


The power dissipated in the 3resistor is,

3. Find the Node Voltages in given Fig.

Solution :-

The circuit in this example has three non reference nodes. We assignvoltages to the three nodes as shown in Fig and label the currents.


5/13


Using the elimination technique, we add Eqs.[1] and Eqs.[3]


6/13


4. Find the node voltages in the circuit given below.

Solution :-


7/13



8/13


We need four node voltages, v1, v2, v3, and v4, and it requires only four outof the five Eqs.

From Eq. (3), v2 = v1 20. Substituting this into Eqs. (1) and (2)respectively, gives

.. [ 6 ]

[ 7 ]

Equations (4), (6), and (7) can be cast in matrix form as


9/13


5. For the circuit shown in Fig. below, find i1 to i4 using mesh analysis.

Solution :-

Note that meshes 1 and 2 form a supermesh since they have anindependent current source in common.

Also, meshes 2 and 3 form another supermesh because they have adependent current source in common. The two supermeshes intersect andform a larger supermesh as shown.

Applying KVL to the larger supermesh,


10/13


6. Use mesh analysis to determine i1, i2, and i3 in Fig shownbelow.

Solution :-Current source between mesh 1 and 2 will be open circuited. So mesh 1 andmesh 2 will form a supermesh.

Apply KVL for supermesh,

6 = 2(i1 i3) + 4 (i2 i3) + 8i32i1 + 12i2 6i3 = 6 (1)

Apply KVL for mesh 3,

2(i3 i1) + 2i3 + 4(i3 i2) = 0- 2i1 - 4i2 + 8i3 = 0 (2)

at node 0 apply KCL,i1 = i2 + 3 (3)

Solving Eq. (1) & (2) we will get,4i2 + i3 =3 (4)

Put value of Eq. (3) in Eq. (1)7i2 3i3 = 0 (5)

Solving Eq. (4) & (5)i2 = 0.4737 A

put value of i2 in Eq. (1)

i1 = 3.4737Aput value of i2 in Eq. (4)

i3 = 1.1052 A


11/13


7. Find the node voltages for the circuit in Fig below.

Solution :-

According to supernode principle voltage source between v1 and v2 node isshortcircuited.

Node v1 and v2 makes supernode.

At supernode apply KCL,


12/13


At node v3 apply KCL,

Put Eq. (4) in Eq. (2)4 = V1 + 8V3 (5)

Solve Eq. (3) & Eq. (5)V1 = 4.97 V

Put Value of V1 in Eq.(5)V3 = -0.12 V

Put value of V1 & V3 in Eq. (4)V2 = 4.85 V

8. In the Circuit shown below solve for i1, i2, i3.

Solution :-


13/13


Between mesh 2 and 3 there is current source which will be open circuited

and both mesh will form supermesh.

Documents

CN Tutorial 11