Current Electricity and Electrical InstrumentHard

Question

In the network shown, points A, B and C are potentials of 70 V, zero and 10 V respectively.

Options

A.Point D is at a potential of 40 V
B.The currents in the sections AD, DB, DC are in the ratio  3: 2: 1
C.The currents in the sections AD, DB, DC are in the ratio 1: 2: 3
D.The network draws a total power of 200 W.

Solution


Let potential of point D is x. by KCL at point D.
I1 + I2 + I3 = 0
+ + = 0
⇒    6x − 420 + 3x + 2x − 20 = 0
⇒    11x = 440
⇒     x = 40 volt
∴    I1 = = −  3A,                      I2 = = 2A
        i3 = = 1A
        P = i2R
        P = 32 × 10 + 22 × 20 + 12 × 30
        P = 200 W

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