Finding Short
Circuit Current (ISC)
Circuit Current (ISC)
Disconnect the
5Ω resistor and put a short across terminals A and B as shown in figure 14
below.
5Ω resistor and put a short across terminals A and B as shown in figure 14
below.
As seen in
figure 14 above, the 10Ω resistor has been short circuited. The battery
sees a parallel combination of 4Ω and 8Ω
in series with a 4Ω resistance. The total resistance seen by the battery is;
figure 14 above, the 10Ω resistor has been short circuited. The battery
sees a parallel combination of 4Ω and 8Ω
in series with a 4Ω resistance. The total resistance seen by the battery is;
= 4 + (4||8)
= 6.67Ω
Therefore the
current as a result of the 20V battery is 20/6.67 = 3A
current as a result of the 20V battery is 20/6.67 = 3A
The short
circuit current is the current flowing through CAB. This is calculated using
the current division principle at point C.
circuit current is the current flowing through CAB. This is calculated using
the current division principle at point C.
Finding Norton’s
Equivalent Resistance (Ri)
Equivalent Resistance (Ri)
Disconnect the
battery leaving behind the internal resistance which is zero in this case as
shown in figure 15 below.
battery leaving behind the internal resistance which is zero in this case as
shown in figure 15 below.
The Norton’s
equivalent resistance (Ri) is the effective resistance of the
network shown in figure 15 above looking through terminals A and B.
equivalent resistance (Ri) is the effective resistance of the
network shown in figure 15 above looking through terminals A and B.
Ri =
[(4||4) + 8]|| 10
[(4||4) + 8]|| 10
= 5Ω
Therefore, the
Norton’s equivalent circuit is shown in figure 16 below
Norton’s equivalent circuit is shown in figure 16 below
To calculate the current through the 5Ω resistor as
required from figure 13 above, reconnect the 5Ω across terminals A and B as
shown in figure 17 below.
required from figure 13 above, reconnect the 5Ω across terminals A and B as
shown in figure 17 below.
