In the circuit shown, A is joined through a 6 μ F capacitor to a node P. Two 5 μ F capacitors hang in parallel from P down to a node M, and a 2 μ F runs along the top from P to a node R, where R is joined to M by a plain wire. From M a 2 μ F goes down to B, and from R the top wire runs to the right…

Question

In the circuit shown, A is joined through a 6 μ F capacitor to a node P. Two 5 μ F capacitors hang in parallel from P down to a node M, and a 2 μ F runs along the top from P to a node R, where R is joined to M by a plain wire. From M a 2 μ F goes down to B, and from R the top wire runs to the right and down through a 4 μ F to B. The equivalent capacitance between A and B is

Accepted Answer

(4) 2.4 μ F. The wire from R to M makes R and M one node, which is what unlocks the circuit. Between P and that node: the two 5 μ F in parallel give 10 μ F, and the 2 μ F from P to R is across the same two points, so 10+2=12 μ F. Between that node and B: the 2 μ F from M and the 4 μ F from R are in parallel, 2+4=6 μ F. These two blocks are in series: (12×6)/(12+6)=4 μ F from P to B. Finally the 6 μ F from A to P is in series with that: (6×4)/(6+4)=2.4 μ F.

Step-by-step solution

Why the other options are wrong