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gracy
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Why R1/R4 = R2/R5 and not R1/R2=R4/R5 ?gneill said:R1/R4 = R2/R5
Why R1/R4 = R2/R5 and not R1/R2=R4/R5 ?gneill said:R1/R4 = R2/R5
Isn't it one and the same expression?gracy said:Why R1/R4 = R2/R5 and not R1/R2=R4/R5 ?
You mean at A and B?gneill said:you could easily determine the potentials at those two middle nodes with a voltage divider formula used on each pair.
I think these two are same in the sense that equality will still hold,but let's saycnh1995 said:Isn't it one and the same expression?
Yes.gracy said:I think these two are same in the sense that equality will still hold,but let's say
R1/R4 = R2/R5=x
Then R1/R2=R4/R5=y
Am I correct?
Yes but the equations are true only for a "balanced" bridge. For this circuit, you'll need KCL and KVL.gracy said:So what in this case?Can I apply either of the two equations?
gneill said:the potential that a potential divider produces at its center node depends upon ratios.
No,I don't think we can apply either of them.We have to use the one @gneill mentioned.But I am still confused ,how I will the decide the ratios in other questions of this type.I think we should look for resistors/capacitors in series considering there is no resistance /and capacitors in between.As in here we assumed R3 is not there.gracy said:So what in this case?Can I apply either of the two equations?
That's because the "ratios" aren't equal. If they were equal, Va would be equal to Vb i.e. Vab would be 0.gracy said:No,I don't think we can apply either of them.
And bridge would be balanced?cnh1995 said:Vab would be 0.
Yes. That's why the middle resistor is equivalent to an open switch in the balanced condition. It doesn't matter if it is there or not.gracy said:And bridge would be balanced?
gracy said:And bridge would be balanced?
And that's what we want.But with this formula R1/R2=R4/R5cnh1995 said:Yes.
Actually, with that formula, we 'check' if the bridge is balanced or not. If it is, the calculations become simpler. If it isn't, we have to use the methods gneill mentioned earlier.gracy said:And that's what we want.But with this formula R1/R2=R4/R5
It is not possible
that's why we can't use it,right?
This onecnh1995 said:with that formula
Right.gracy said:This one
R1/R4 = R2/R5 ?
How to do that?How I will know this capacitor should come in between i.e which capacitor would form C3 ?Which of them would be C1,C2,C4?And moreover there is no voltage in my actual problem .gneill said:match the components from your circuit to their equivalent locations on this new bridge diagram.
You'll need some practice for "spotting" the bridge in the circuit. Same goes for star-delta conversions in more complex circuits.gracy said:How to do that?How I will know this capacitor should come in between i.e which capacitor would form C3 ?Which of them would be C1,C2,C4?And moreover there is no voltage in my actual problem .
Always assume a voltage source between the points between which the equivalent resistance or capacitance is asked. Equivalent resistance or capacitance between two points is actually the equivalent resistance or capacitance "seen" by a voltage source connected between those points.gracy said:moreover there is no voltage in my actual problem .
In the given circuit,after assuming the voltage source,if you trace the path of current through the capacitor branches and observe the 'splitting' and 'reunion' of the currents carefully, you'll realize its a bridge.gracy said:How to do that?How I will know this capacitor should come in between i.e which capacitor would form C3 ?Which of them would be C1,C2,C4?And moreover there is no voltage in my actual problem .
cnh1995 said:Always assume a voltage source between the points between which the equivalent resistance or capacitance is asked. Equivalent resistance or capacitance between two points is actually the equivalent resistance or capacitance "seen" by a voltage source connected between those points.
Applying these I drew a diagram (Sorry it is not neat)cnh1995 said:After assuming the voltage source,if you trace the path of current through the capacitor branches and observe the 'splitting' and 'reunion' of the currents carefully, you'll realize its a bridge.
You can label your components in any order you wish. If you are going to refer to them in formulas you should show their labels on your circuit diagram.gracy said:Applying these I drew a diagram (Sorry it is not neat)
View attachment 92753
As my bridge is not horizontal it is vertical hence formula to check whether it is balanced or not would be
##\frac{C1}{C2}##=##\frac{C3}{C4}## and not C1/C4 = C2/C5
Right!Putting values
##\frac{2}{4}##=##\frac{1}{2}##
##\frac{3}{6}##=##\frac{1}{2}##
Hence the bridge is balanced.
Right?
No. You can rotate either of them to make them look alike.gracy said:Will there be any change in formulas
You can remember it this way-ratio of resistors on one side of the bridge=ratio of resistors on the other side of the bridge. But it is important to spot the bridge first.gracy said:But I am still confused ,how I will the decide the ratios in other questions of this type?
It doesn't matter which should be numerator or denominator. Just pick any ratio on one side, use the same on the other side.gracy said:But then I get confused which one will be denominator and which one will be numerator.
You want to see if the two voltage dividers in the bridge circuit will produce the same voltage division. So a simple procedure is to make ratios of the components that make up each voltage divider. Compare the ratios. You can't go wrong if you select corresponding pairs (use symmetry and common sense).gracy said:But then I get confused which one will be denominator and which one will be numerator.
You mean if I have selected C1 as numerator for one potential divider I shall select the one which is adjacent to it but at the other side of bridge as numerator of second potential divider,Right?gneill said:You can't go wrong if you select corresponding pairs (
You can call the two resistors on one side as 'upper' and 'lower'. So, upper/loewer on one side=upper/lower on the other side.gracy said:Let say if I would have taken
C1/C2=C4/C3
After putting the values
I could have thought the bridge is not balanced.
Its just a memory trick..To understand the bridge balance conceptually, you should analyse it using voltage divider.cnh1995 said:You can call the two resistors on one side as 'upper' and 'lower'. So, upper/loewer on one side=upper/lower on the other side.
Right. Once you're fluent, you won't need this trick.gracy said:
upper/lower on one side=upper/lower on the other side.And if bridge is of this sort
upper left/upper right on one side=lower left/lower right on the other side