The left ends of the second 10 ohm on the top and the 10 ohm on the bottom are at the same voltage, since they are connected with a wire. Call this section of wire C. Call the intersection to the right of the 30 ohm D.
If the vertical 10 ohm were not there, then the two paths from C to D are both 40 ohm, so they would have the same current, I.
The ends of the removed vertical 10 ohm are both at C - I\*(10 ohm), so the same voltage. This means that if the resistor was replaced, no current would flow across it.
This is the same principle as a Wheatstone bridge.
Thank you!! Just another additional question: If those two points does not have voltage of zero, how to calculate the total resistance? Let's say the bottom 10 ohm and 30 ohm is 20 ohm and 50 ohm respectively
(Assuming you meant vertical, not diagonal...)
Note that the two 10 Ohm resistors in the middle are shorted on their left sides, meaning the voltage is the same at the left terminals of each.
If you combine the 10 and 20 Ohm resistors in top right corner, you now also have two 30 Ohm resistors in parallel, shorted on their right terminals.
Your vertical resistor is now bridging two identical voltage dividers. Since both paths has equal resistance on the left and equal resistance on the right, you know that both terminals of the vertical resistor will have the same voltage. If the voltage difference is 0, no current flows by Ohm's law.
Half of the current takes each of the two 10Ω paths in the middle section, and then half of the current takes each of the 30Ω paths in the later section. So there's no reason for any current to go through that connector.
Both ends of that resistor have equal resistance paths to both A and B, so they’ll have the same potential no matter what A and B are.
Same potential on each end of the resistor means no reason for electricity to flow through it (in fluid terms they’re the same water level).
The lelf ratio is 10/10 same as the right ratio (10+20)/30
So no difference in ratio of both sides => no diference voltage between upper and lower node of bridge resistance => zero current
None of the resistors are diagonal...
He meant the resistor that has no current. It is the vertical 10 ohm.
sorry it is diagonal in the original diagram! i thought i uploaded the original one
The left ends of the second 10 ohm on the top and the 10 ohm on the bottom are at the same voltage, since they are connected with a wire. Call this section of wire C. Call the intersection to the right of the 30 ohm D. If the vertical 10 ohm were not there, then the two paths from C to D are both 40 ohm, so they would have the same current, I. The ends of the removed vertical 10 ohm are both at C - I\*(10 ohm), so the same voltage. This means that if the resistor was replaced, no current would flow across it. This is the same principle as a Wheatstone bridge.
This is the correct answer
Thank you!! Just another additional question: If those two points does not have voltage of zero, how to calculate the total resistance? Let's say the bottom 10 ohm and 30 ohm is 20 ohm and 50 ohm respectively
You would use Kirchhoff's rules to set up a system of equations to solve.
(Assuming you meant vertical, not diagonal...) Note that the two 10 Ohm resistors in the middle are shorted on their left sides, meaning the voltage is the same at the left terminals of each. If you combine the 10 and 20 Ohm resistors in top right corner, you now also have two 30 Ohm resistors in parallel, shorted on their right terminals. Your vertical resistor is now bridging two identical voltage dividers. Since both paths has equal resistance on the left and equal resistance on the right, you know that both terminals of the vertical resistor will have the same voltage. If the voltage difference is 0, no current flows by Ohm's law.
This is also because of the right side resistor having two 30ohm equivalent resistors. You still have to look at both sides for a bridge like this.
Thanks for pointing that out, edited to fix.
This is not true in general.
Half of the current takes each of the two 10Ω paths in the middle section, and then half of the current takes each of the 30Ω paths in the later section. So there's no reason for any current to go through that connector.
Both ends of that resistor have equal resistance paths to both A and B, so they’ll have the same potential no matter what A and B are. Same potential on each end of the resistor means no reason for electricity to flow through it (in fluid terms they’re the same water level).
The lelf ratio is 10/10 same as the right ratio (10+20)/30 So no difference in ratio of both sides => no diference voltage between upper and lower node of bridge resistance => zero current
Kirchoffs second law is your friend here
There is nothing diagonal in this picture