# Direct current

1. Nov 19, 2014

### werson9339

1. The problem statement, all variables and given/known data
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....

2. Relevant equations

3. 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

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2. Nov 19, 2014

### BvU

What is the problem statement ?
No. It only means there is no current passing through the galvanometer.

3. Nov 19, 2014

### 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: VAC = VXZ.

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

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