(a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.
The power in an AC circuit is given by P = VI cos φ, where cos φ is the power factor. To supply a given power at a given voltage, if the power factor is small, the current must be increased accordingly. This increased current leads to large power loss (I²R) in the transmission lines.
The power factor cos φ = R/Z. To improve it (tend to 1), we need to make Z tend to R. The current can be resolved into Ip (power component) in phase with voltage, and Iq (wattless component) perpendicular to voltage. By connecting a capacitor of appropriate value in parallel, the leading wattless current from the capacitor neutralizes the lagging wattless current Iq. These currents cancel each other, improving the power factor.
Explanation
This question tests understanding of power factor in AC circuits. Part (a) explains why low power factor is undesirable in power transmission - it requires higher current for the same power delivery, leading to greater I²R losses in transmission lines. Part (b) explains the method of power factor correction using capacitors. The capacitor provides a leading wattless current that cancels the lagging wattless current from inductive loads, effectively improving the power factor.