Example 2
Apply Thevenin’s
theorem to find current flowing through the 12 Ω resistor in figure 5 below.
theorem to find current flowing through the 12 Ω resistor in figure 5 below.
Solution
Finding Vth
Disconnect the
12Ω resistor as shown in figure 6 below and calculate Vth which is
the open circuit voltage when the circuit is viewed from terminals A and B.
12Ω resistor as shown in figure 6 below and calculate Vth which is
the open circuit voltage when the circuit is viewed from terminals A and B.
Note that point
A is at the same potentials as point C because there is no current flowing
through the 4Ω resistor and hence no voltage drop across it. As seen from
figure 6 above, Vth = drop across the 6Ω resistpr due to the current
set up by the 36V battery.
A is at the same potentials as point C because there is no current flowing
through the 4Ω resistor and hence no voltage drop across it. As seen from
figure 6 above, Vth = drop across the 6Ω resistpr due to the current
set up by the 36V battery.
I = 36/(3+6) =
4A
4A
Vth =
I x 6 = 4 x 6 = 24V
I x 6 = 4 x 6 = 24V
Finding Rth
Short the 36V
source and calculate Rth from terminals A and B as shown in figure 7
below.
source and calculate Rth from terminals A and B as shown in figure 7
below.
Rth =
4 + (6||3)
4 + (6||3)
= 4 + (6×3)/(6+3)
= 4 + 2 = 6Ω
Then the circuit
is reduced to Thevenin source of a 24V voltage source with a series resistance
of 6Ω. The 12Ω resistor (load resistance) is reconnected back into the circuit
in series with the source as shown in figure 8 below.
is reduced to Thevenin source of a 24V voltage source with a series resistance
of 6Ω. The 12Ω resistor (load resistance) is reconnected back into the circuit
in series with the source as shown in figure 8 below.
Current flowing
through the 12Ω resistor (I) is
through the 12Ω resistor (I) is
I = 24/(6+12)
= 24/18 = 1.33A
