By applying
Thevenin’s theorem to figure 1 below,
Thevenin’s theorem to figure 1 below,
Calculate:
i.
The equivalent e.m.f. of the
network when viewed from terminals A and B.
ii.
The equivalent resistance of
the network when looked into from terminals A and B.
iii.
Current in the load resistance
RL of 15Ω.
Current in the load resistance
RL of 15Ω.
Solution
i.
Equivalent e.m.f. of the
network when viewed from terminals A and B.
Disconnect the
load resistance (RL) and calculate the open circuit voltage (Voc)
as shown in figure 2 below.
load resistance (RL) and calculate the open circuit voltage (Voc)
as shown in figure 2 below.
Open circuit resistance (R) = 3+12+1 =
16Ω
16Ω
Open circuit current (I) = 24/16 = 1.5A
Therefore voltage across terminals A and
B which is called the open circuit voltage (Voc) = 1.5 x 12 = 18V.
B which is called the open circuit voltage (Voc) = 1.5 x 12 = 18V.
i.
Equivalent resistance of the
network when looked into from terminals A and B.
Short the
voltage source and leave only its internal resistance behind as shown in figure
3 below.
voltage source and leave only its internal resistance behind as shown in figure
3 below.
Then,
resistance of the circuit as looked into from points A and B is;
resistance of the circuit as looked into from points A and B is;
Rth
= 12||(3+1)
= 12||(3+1)
= (12×4)/(12+4)
= 48/16 = 3Ω
i.
Current in the load resistance
(RL) of 15Ω
Reconnect the load resistance (RL)
across terminals A and B where it was initially disconnected from as shown in
figure 4 below.
across terminals A and B where it was initially disconnected from as shown in
figure 4 below.
Current in the
load resistance (I) = Vth/(Rth+RL)
load resistance (I) = Vth/(Rth+RL)
= 18/(3+15)
= 1A
