Hawkins Electrical Guide v. 05 (of 10) / Questions, Answers, & Illustrations, A progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications - N. Hawkins

1. Active pressure, to overcome resistance;
2. Self-induction pressure to overcome inductance.

Fig. 1,313.—Diagram illustrating the active, and self-induction pressures, or the two components of the impressed pressure in circuits containing resistance and inductance. The active pressure is the volts required to overcome the resistance of the circuit. In the figure only the portion from A to C is considered as having resistance (the rest being negligibly small) except at R, a resistance equivalent to that of the inductive coil is inserted next to the non-inductive coil, so P_a will give the total "ohmic drop" or active pressure, that is, the pressure necessary to force any equivalent direct current from A to C. This active pressure P_a or component of the impressed pressure is in phase with the current. The other component or self-induction pressure P_i that is the reactance drop necessary to overcome the reverse pressure of self-induction and is at right angles to the current and 90° ahead of the current in phase. It is registered by a voltmeter between B and C, less the pressure due to ohmic resistance of the inductive coil. The impressed pressure P_im then or total pressure required to force electricity around the circuit not including the resistance R, (which is removed from the circuit when the reading of the impressed pressure is taken), is equal to the square root of the sum of the squares of the two components, that is, P_im = √(P_{a}² + P_{i}²).

The active pressure is in phase with the current.

The self induction pressure is at right angles to the current and 90 degrees ahead of the current in phase.

Ques. Why is the active pressure in phase with the current?

Ans. The pressure used in overcoming resistance is from Ohm's law, E = RI. Hence, when the current is zero, E is zero, and when the current is a maximum E is a maximum. Hence, that component of the impressed pressure necessary to overcome the resistance must be in phase with the current.

Ques. Why is this?

Ans. Since the reverse pressure of self induction is 90° behind the current, the component of the impressed pressure necessary to overcome the reverse pressure of self induction, being opposite to this, will be represented as being 90° ahead of the current.