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In the network shown, points A, B and C are at potentials of 70V, zero, and 10V respectively. The power drawn by the net work is

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200W

100W

400W

Zero

answer is A.

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Detailed Solution

$i = i_{1} + i_{2}; \frac{70 - V}{10} = \frac{V - 0}{20} + \frac{V 10}{30} \Rightarrow V = 40 volt Power P = \frac{V^{2}}{R} = \frac{(70 - 40)^{2}}{10} + \frac{(40 - 0)^{2}}{20} + \frac{(40 - 10)^{2}}{30} = 90 + 80 + 30 = 200 watt$

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