Electric power is sent from a power station to a town through transmission cables. When the power is sent at 11 kV, the power lost as heat in the cables is 18 kW. The same power is now sent at 33 kV through the same cables. The power lost in the cables is now
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 kW
When a fixed power \( P \) is sent at voltage \( V \), the current in the cables is \( I=P/V \). The heat loss in cables of resistance \( R_c \) is then \( P_c=I^{2}R_c \), which is \( P^{2}R_c/V^{2} \). Here \( P \) and \( R_c \) do not change, so \( P_c \) varies as \( 1/V^{2} \).
\( \begin{aligned} \frac{P_c'}{P_c} &= \left(\frac{V}{V'}\right)^{2} \\ &= \left(\frac{11}{33}\right)^{2} \\ &= \frac{1}{9} \end{aligned} \)
\( \begin{aligned} P_c' &= \frac{18}{9} \\ &= 2\ \mathrm{kW} \end{aligned} \)
Therefore, three times the voltage cuts the loss to one-ninth.
NCERT (2026-27):
Electrical Energy, Power – “… The power dissipated in the connecting wires, which is wasted is Pc with”
\( P_c=I^{2}R_c=\frac{P^{2}R_c}{V^{2}} \qquad (3.35) \)
Electrical Energy, Power – “… Thus, to drive a device of power P, the power wasted in the connecting wires is inversely proportional to V2.”