A factory takes the same average power from the mains at the same rms voltage in two cases, first at a power factor of 1 and then at a power factor of 0.5. The power lost as heat in the resistance of the supply line is \( P_L \) in the first case. In the second case it is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 4P_L \)
The average power taken from an ac supply is \( P=VI\cos\phi \). For the same power at the same rms voltage, a smaller power factor \( \cos\phi \) needs a larger current. The heat produced in the supply line of resistance \( R_l \) is \( I^{2}R_l \). So it grows as the square of the current.
The currents drawn in the two cases are:
\( \begin{aligned} I_1 &= \frac{P}{V\times1} \\ &= \frac{P}{V} \end{aligned} \)
\( \begin{aligned} I_2 &= \frac{P}{V\times0.5} \\ &= 2I_1 \end{aligned} \)
The power lost in the line in the second case is:
\( \begin{aligned} P_L' &= I_2^{2}R_l \\ &= (2I_1)^{2}R_l \\ &= 4P_L \end{aligned} \)
Therefore, halving the power factor makes the line loss four times as large.
NCERT (2026-27):
Example 7.7 – “Solution (a) We know that”
\( P=IV\cos\phi \)
“where cosφ is the power factor. To supply a given power at a given voltage, if cosφ is small, we have to increase current accordingly. But this will lead to large power loss (I2R) in transmission.”