Timeline for Is it safe to assume it to be a balanced Wheatstone bridge?
Current License: CC BY-SA 4.0
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| Jun 5 at 13:41 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Jun 4 at 18:23 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Jun 1 at 10:30 | comment | added | Philip Wood | Yes. If the 4 ohm resistor wasn't there (between points C and D, let's say), we know from potential divider theory applied to the top three resistors and to the bottom three resistors that the pd between C and D would be zero. Therefore there'll be no current through any conductor that we connect between C and D. So this 'bridging' conductor (resistor!) might as well not be there. This is exactly how we justify the balance condition for a Wheatstone bridge – in which, originally, the bridging component was a (low resistance) galvanometer. There is no need for a Kirchhoff's laws treatment. | |
| May 30 at 4:31 | comment | added | Physics_enthus | Can we say that the equivalent resistance of the given circuit is independent of the values of 3 and 4 ohm resistors? | |
| May 30 at 4:17 | comment | added | Physics_enthus | You said - ... as both ends of the 4 ohm are at the same potential.. , we know that they are at same potential after doing page long calculation and solving a system of four equations in four variables. My question is can we just straightaway remove those 3 and 4 ohm resistors, without actually calculating the potential at junctions? | |
| May 29 at 18:23 | history | answered | Philip Wood | CC BY-SA 4.0 |