If a circuit delivers fixed real power and power factor is improved, the current will:

If a circuit delivers fixed real power and power factor is improved, the current will:

• A. Increase
• B. Stay the same
• C. Decrease
• D. Become infinite

Answer: (C)

Improving power factor while delivering the same real power reduces the current. Real power is P = V I cos(phi), where cos(phi) is the power factor. If the voltage V and the real power P are fixed, the current is I = P / (V cos(phi)). So as the power factor cos(phi) increases (moving toward 1), the denominator grows and the current decreases. You can see this with a quick example: with a fixed voltage and P, a poorer power factor means more current is needed to deliver the same real power. As the pf improves, the reactive portion drops (Q = P tan(phi) shrinks), so less total current is required to push the same amount of real power through the circuit. Hence the current decreases.

Improving power factor while delivering the same real power reduces the current.

Real power is P = V I cos(phi), where cos(phi) is the power factor. If the voltage V and the real power P are fixed, the current is I = P / (V cos(phi)). So as the power factor cos(phi) increases (moving toward 1), the denominator grows and the current decreases.

You can see this with a quick example: with a fixed voltage and P, a poorer power factor means more current is needed to deliver the same real power. As the pf improves, the reactive portion drops (Q = P tan(phi) shrinks), so less total current is required to push the same amount of real power through the circuit. Hence the current decreases.