In the adjoining diagram $R_{1} = 10 Ω, R_{2} = 20 Ω, R_{3} = 40 Ω, R_{4} = 80 Ω$ and $V_{A} = 5 V, V_{B} = 10 V, V_{C} = 20 V, V_{D} = 15 V$ . The current in the resistance $R_{1}$ will be

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Answer 1

Let the potential at the common point of all branches be $V$ .
Then, using nodal analysis:
$\frac{5 - V}{1 0} = \frac{V - 1 0}{2 0} + \frac{V - 2 0}{4 0} + \frac{V - 1 5}{8 0} V = \frac{1 3 5}{1 5}, I = \frac{5 - V}{1 0} = - 0.4 A$
Since, negative sign is there, that means the current will be away from $O$ .

Answer 2

Let the potential at the common point of all branches be

V

.

Then, using nodal analysis:

\frac{5 - V}{1 0} = \frac{V - 1 0}{2 0} + \frac{V - 2 0}{4 0} + \frac{V - 1 5}{8 0} V = \frac{1 3 5}{1 5}, I = \frac{5 - V}{1 0} = - 0.4 A

Since, negative sign is there, that means the current will be away from

O

.