In the network shown below, points A, B and C are at the potential of 70 V, 0 V and 10 V respectively. The potential of point D is

10 V

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20 V

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30 V

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40 V

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Solution

The correct option is D
40 V

Applying KCL at junction D, $\frac{V_{A} - V_{D}}{10} = \frac{V_{D} - V_{B}}{20} + \frac{V_{D} - V_{C}}{30} \Rightarrow \frac{70 - V_{D}}{10} = \frac{V_{D} - 0}{20} + \frac{V_{D} - 10}{30} \Rightarrow 70 - V_{D} = \frac{V_{D}}{2} + \frac{V_{D}}{3} - \frac{10}{3} \Rightarrow \frac{11 V_{D}}{6} = \frac{220}{3} ∴ V_{D} = 40 V$

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In the network shown below, points A, B and C are at the potential of 70 V, 0 V and 10 V respectively. The potential of point D is

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