JEE Advance - Physics (2013 - Paper 2 Offline - No. 12)
If the direct transmission method with a cable of resistance 0.4 $$\Omega$$ km$$-$$1 is used, the power dissipation (in %) during transmission is
20
30
40
50
Explanation
P = Vi
$$\therefore$$ $$i = {P \over V} = {{600 \times {{10}^3}} \over {4000}} = 150$$ A
Total resistance of cables,
R = 0.4 $$\times$$ 20 = 8$$\Omega$$
$$\therefore$$ Power loss in cables = i2R
= (150)2 (8)
= 180000 W = 180 kW
This loss is 30% of 600 kW.
Comments (0)
