For the four capacitors in the circuit shown in the figure below, CA = 1.00 µF, CB = 4.10 µF, CC = 2.10 µF, and CD = 3.10 µF. What is the equivalent capacitance between points a and b?

Question

For the four capacitors in the circuit shown in the figure below, CA = 1.00 µF, CB = 4.10 µF, CC = 2.10 µF, and CD = 3.10 µF. What is the equivalent capacitance between points a and b?

www.webassign.net/katzpse1/27-p-030.png

Answer 1

To find the equivalent capacitance between points a and b in the given circuit, we can use the concept of series and parallel capacitors.

Looking at the circuit, we observe that CA and CB are connected in parallel (they share the same nodes a and b), and their equivalent capacitance (CAB) can be calculated using the formula:

1/CAB = 1/CA + 1/CB

Substituting the given values:

1/CAB = 1/1.00 µF + 1/4.10 µF

Now, let's simplify by finding the least common denominator:

1/CAB = (4.10 + 1)/(4.10 * 1.00) µF

1/CAB = 5.10 µF / 4.10 µF

Now, let's calculate the equivalent capacitance CAB:

CAB = 4.10 µF * 5.10 µF / 4.10 µF

CAB = 5.10 µF

So, the equivalent capacitance between points a and b (CAB) is 5.10 µF.