Using Thevenin’s
theorem, calculate the current flowing through the 4Ω resistor in circuit shown
in figure 9 below.
Solution
Finding Vth
Disconnect the
4Ω resistor and redraw the circuit as shown in figure 10 below.
4Ω resistor and redraw the circuit as shown in figure 10 below.
Since 10A
current passes through the 2Ω resistor, voltage drop across it is 10 x 2 = 20V.
Hence VB is 20V with respect to the common ground. The resistor 3Ω
and 6Ω are now in series across the 12V battery. Hence voltage drop across the
6Ω resistor = (12 x 6)/ (3 + 6) = 8V.
current passes through the 2Ω resistor, voltage drop across it is 10 x 2 = 20V.
Hence VB is 20V with respect to the common ground. The resistor 3Ω
and 6Ω are now in series across the 12V battery. Hence voltage drop across the
6Ω resistor = (12 x 6)/ (3 + 6) = 8V.
Therefore VA
= 8V with respect to the common ground.
= 8V with respect to the common ground.
Vth =
VBA = VA – VB
VBA = VA – VB
= 20 – 8 = 12V (with B at a higher potential).
Finding Rth
To find Rth
i.e. equivalent resistance of the
network as we looked back into the open-circuited terminals A and B. For this
purpose, we will replace both the voltage and current sources. Since the
voltage has no internal resistance, it would be replaced by a short circuit
i.e. zero resistance. However, current source would be removed and replaced by
an ‘open’ i.e. infinite resistance. In the case, the circuit becomes as shown
in figure 11 below.
i.e. equivalent resistance of the
network as we looked back into the open-circuited terminals A and B. For this
purpose, we will replace both the voltage and current sources. Since the
voltage has no internal resistance, it would be replaced by a short circuit
i.e. zero resistance. However, current source would be removed and replaced by
an ‘open’ i.e. infinite resistance. In the case, the circuit becomes as shown
in figure 11 below.
= 2 + 2 = 4Ω
Hence,
Thevenin’s equivalent circuit consist of a voltage source of 12V and a series
resistance of 4Ω as shown in figure 12 below.
Thevenin’s equivalent circuit consist of a voltage source of 12V and a series
resistance of 4Ω as shown in figure 12 below.
When the 4Ω
resistor is connected across terminals A and B, the circuit is shown as figure
13 below.
resistor is connected across terminals A and B, the circuit is shown as figure
13 below.
The current
through the 4Ω resistor is
through the 4Ω resistor is
I = 12/(4+4)
= 1.5A from B to A.
