The maximum capacitance that can be formed from the given capacitors is 9000.01 μF (in parallel)
<h3>How to find Maximum Capacitance?</h3>
We are given the capacitances as;
3200 μF, 5800 μF, and 0.0100 μF
Now, if the capacitors are in parallel, the equivalent capacitance is;
C_eq = C₁ + C₂ + C₃
Thus;
C_eq = 3200 μF + 5800 μF + 0.0100 μF
C_eq = 9000.01 μF
Now, if the capacitors are in series, the equivalent capacitance is;
1/C_eq = 1/C₁ + 1/C₂ + 1/C₃
Thus;
1/C_eq = (1/3200) μF + (1/5800) μF + (1/0.0100) μF
C_eq = 0.00999995 μF
Therefore the maximum capacitance that can be formed from the given capacitors is 9000.01 μF (in parallel)
Complete question is;
Consider three capacitors, of capacitance 3200 μF, 5800 μF, and 0.0100 μF. What maximum capacitance can you form from these?
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Answer:
107 degrees
Explanation:
The lines intersect to create a triangle. The sum of the interior angles of a triangle is 180 degrees.
a + b+ c = 180
We deduce the unlabeled angle (in the lower right corner) by considering the line that goes from that angle to the a angle on the left side. The line itself has a 180 degree angle (on each side). And we know the exterior angle is 107 degrees... that leaves 73 for the interior angle.
We take the equation above again... by putting now the value of C (the un-named angle):
a + b + 73 = 180
We isolate the a+b part... to get a+b = 107, the same value as the exterior angle.
Answer:
I think this is a self question on the test, I can't make your decisions.
Awhh why is that the case?
