The discussion focuses on calculating the equivalent capacitance between points A and B in a bridge circuit involving capacitors of 3μF, 5μF, 4μF, and 2μF. The user initially calculated the equivalent capacitance as 4μF but found a discrepancy with the textbook answer of 10/3μF. The conversation emphasizes the importance of correctly identifying nodes and understanding that the circuit does not allow for simple series or parallel reductions due to its bridge configuration. Participants suggest using advanced circuit analysis methods such as Y-Delta transformations and Kirchhoff's laws to solve the problem.
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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:
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:
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 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:
Its just a memory trick..To understand the bridge balance conceptually, you should analyse it using voltage divider.cnh1995 said:
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