Page 220 - Basic Electrical Engineering
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As the same current I is flowing from P to Q, the voltage drop across PQ
2
which is the same as voltage drop across the 10 Ω resistor is calculated as
V = V = I R = 0.99 × 4.44 = 4.39 V
2
PQ
10 Ω
The battery voltage of 24 V is dropped across the series resistance of 7 Ω,
across the combination of 7 Ω and 4.44 Ω resistors as
7 × I + 7 × I + 4.44 × I = 7 × 1.8 + 11.44 × 0.99 = 24 V
2
2
This problem can be solved in another way like, total current, I = 1.8 A.
Drop across AC = 7 × 1.8 = 12.6 V.
Voltage across CD = 24 − 12.6 = 11.4 V
Voltage across EQ = Voltage across CD = 11.4 V
Voltage across
Example 2.40 Calculate the equivalent resistance between the terminals A
and B of the network shown in Fig. 2.141.
Figure 2.141
Solution:
