Achieving the Balance Point: Current Flow & Galvanometer Reading

[werson9339]

Homework Statement

when the balance point is achieved , there're no current passing thru , so Galvanometer should show 0 . Am i right... so there's no current passing thru the 20 ohm resistor . so my working is ( 8.1/120 )(40) = 2.7 V ...
correct me if I'm wrong...

Homework Equations

The Attempt at a Solution

V_xy = V_ac = (9/10) x 10 = 8.1v
k = 8.1/120
V_xy = k (40) = 2.7v

(5/ (5+20+2) ) X E " = 2.7 V

E" = 14.58v[/B]

 

[BvU]

What is the problem statement ?

so there's no current passing thru the 20 ohm resistor

No. It only means there is no current passing through the galvanometer.

 

[BvU]

No current passing through the galvanometer means there is no potential difference over the thing. If we call the point between the 5 $\Omega$ and the 20 $\Omega$ resistors point z: V_AC = V_XZ.

For V_AC you have the right result (but perhaps a typo?) : V_AC / V_AB = 40/120 and V_AB = 9V * 9/10.

And for E' (single quote in the drawing) you also have the right result.