A coil of inductance L and a resistor R are connected in series to an…

The correct answer is Option (3) → increasing the resistance R

For a coil and a resistor in series, the power factor is the ratio of the resistance to the impedance:

    \( \displaystyle \cos\phi=\frac{R}{\sqrt{R^{2}+\omega^{2}L^{2}}} \)

It moves towards 1 when R becomes large compared with the inductive reactance \( \omega L \). It falls when \( \omega L \) becomes large compared with R.

Option (1) increasing the frequency of the source — Incorrect. A higher frequency raises the inductive reactance \( \omega L \). So the impedance grows while R stays the same, and the power factor falls.

Option (2) inserting a soft iron core into the coil — Incorrect. The iron core increases the inductance L of the coil. The reactance \( \omega L \) goes up, and the power factor goes down.

Option (3) increasing the resistance R — Correct. Write the power factor as \( 1/\sqrt{1+(\omega L/R)^{2}} \). A larger R makes \( \omega L/R \) smaller. So Z comes closer to R, and \( \cos\phi \) increases.

Option (4) increasing the rms voltage of the source — Incorrect. The power factor depends only on R, L and \( \omega \). A higher voltage drives a larger current but leaves the phase angle between the current and the voltage unchanged.

NCERT (2026-27):
Example 7.7 – “(b) Suppose in a circuit, current I lags the voltage by an angle φ. Then power factor cosφ =R/Z. We can improve the power factor (tending to 1) by making Z tend to R.”