Question

in the fig. shown the potential of junction O is :

Answer 1

Q : In the figure shown, the potential of junction O is :
Final Answer : 3V.

Steps and Understanding:
1) Let the current be I(1),I(2) and I(3) as shown in figure.
We apply Kirchoffs Loop rule,

2) V = IR (Ohms Law)
Current flows from high potential to low potential.

For Calculation see pic.

Answer 2

Let the Potential at the Junction of O be m V.

Potential difference across AO = + (6 - m)
Resistance = 6 Ω

By Ohm's law,
V₁ = I × R₁
⇒ I₁ = V₁/R₁
∴ I₁ = (6 - m)/6
∴ I₁ = (6 - m)/6 A.

Now,
Potential Difference across BO = (m - 2) V
Resistance = 2 Ω

∴ I₂ = (m - 2)/2 A.

Potential Difference across CO = (m - 3) V
Resistance = 3 Ω

V₃ = I₃ × R₃
∴ I₃ = (m - 3)/3


Now,
Using the Kirchhoff Law,
I₁ = I₂ + I₃
∴ (6 - m)/6 = (m - 2)/2 +  (m - 3)/3
⇒ (6 - m)/6 = [3(m - 2) + 2(m - 3)]/6
⇒ 6 - m = 3m - 6 + 2m - 6
⇒ 6 - m = 5m - 12
⇒ 5m + m = 12 + 6
⇒ 6m = 18
∴ m = 3 V


Hence, the Potential at the Junction of the O is 3 V.


Hope it helps.