In the following DC circuit, current in R₃

Question

Question

This question was previously asked in

ISRO (VSSC) Technician Electrician 2017 Official Paper

Answer 1

By re-drawing the dc circuit, we get

Now applying nodal analysis at node ‘A’, we get

$\frac{V_{A} - 7.5}{5} + \frac{V_{A}}{5} + \frac{V_{A}}{5} = 0$

V_A- 7.5 + V_A+ V_A = 0

3 V_A = 7.5

∴ V_A = 2.5 V

Hence current flowing through ‘R₃’ branch is

$I = \frac{V_{A}}{5} = \frac{2.5}{5}$

∴ I = 0.5 A