Of the two methods of loop and node variable analysis
Which of the following is true about the circuit shown below?
Number of junction point for the given circuit is 6.
For the circuit shown below, the current through the 1 Ω resistor is
Using source transformation, the current source is converted to a voltage source as shown below.
Applying nodal analysis at node (x), we get
An electric circuit with 8 branches and 4 nodes will have
Number of loop equation
= b - ( n - 1)
= 8 - ( 4 - 1 )
= 8 - 3 = 5
For the circuit shown below, the current through the 10 V battery is
Using source transformation, first we convert the current source into an equivalent voltage source
Applying KVL in loop-1, we get
Applying KVL in loop-2, we get
On solving equations (i) and (//). we get
i1 = 2.36 A and i2 = 4.91 A
Hence, the current through the battery of 10 V is
i2 = 4.91 A
Find the value of the currents I1 and I2 flowing clockwise in the first and second mesh respectively.
The two mesh equations are:
5I1 - 3I2 = 10
-3I1 + 7I2 = -15
Solving the equations simultaneously, we get I1 = 0.96A and I2 = -1.73A.
The value of V in volts for the circuit shown below is
The equivalent resistance across the 5 A current source is
The number of KVL and KCL equations for the circuit shown below are
Number of nodes, n = 5
Number of branchs, b = 6
Hence, number of KCL equation
= n - 1 = 5 - 1 = 4
and, number of KVL equations = b - ( n - 1)
= 6 - (5-1) = 6- 4 = 2
The current l flowing in the given figure is
The value of dependent source for the circuit shown below is
Applying KVL in the loop,
5 = 2i — 2i + i or i = 5 A
∴ Value of dependent source
= 2i = 2 x 5 = 10 volt