Page 328 - Basic Electrical Engineering

Inductive branch 3,

Current I = V/Z 3
3

This current will lag the applied voltage by an angle ϕ ,
3

Choose a current scale and draw to the scale the current vectors with the

voltage as the reference axis. Add vectorially any two currents, say I and I .
2
1
The vector sum of I and I is OE as shown in Fig. 3.30 (a). Add vectorially
2
1
OE with the other branch current, i.e., with I to get the sum of the three
3
currents as OF. Convert this length OF to amperes using the current scale
choosen earlier.

An alternate method is to show the three currents with the voltage as the
horizontal reference axis as shown in Fig. 3.30 (b). Calculate the sum of the
horizonal components and vertical components of the currents and then

determine the resultant.
The branch currents with their phase angles with respect to V has been

shown (not to the scale) separately in Fig. 3.30 (b).
The resultant current I can be found out by resolving the branch currents I ,
1
I , and I into their X and Y components as shown in Fig. 3.30 (b).
2
3

X component of I (OL) = I cos ϕ 1
1
1
X component of I (OM) = I cos ϕ 2
2
2
X component of I (ON) = I cos ϕ 3
3
3