Three capacitors are connected in series. The equivalent capacitance of this combination is 3.18 µF. Two of the individual capacitances are 6.54 µF and 8.19 µF. What is the third capacitance (in µF)?
If 1/C1+1/C2+1/C3=1/Ceq and Ceq=3.18 mF, then replace the given variables as such:
1/6.54+1/8.19+1/C:unknown=1/3.18
Now it becomes an algebra problem:
Multiply both sides of the eqn by 3.18, then you'll have:
3.18/6.54 +3.18/8.18+3.18/Cunknown=1
Simplify:
.49+.39+3.18/C=1
1-(.49+.39)=3.18/C
.12=3.18/C
Solve for C:
C unknown= 3.18/.12= ~ 25 microF
To find the third capacitance in the series combination of capacitors, we can use the formula for the equivalent capacitance in a series circuit:
1/Ceq = 1/C1 + 1/C2 + 1/C3
Given that the equivalent capacitance Ceq is 3.18 µF, and the individual capacitances C1 and C2 are 6.54 µF and 8.19 µF respectively, we can substitute these values into the formula:
1/3.18 µF = 1/6.54 µF + 1/8.19 µF + 1/C3
Now let's solve for C3:
1/C3 = 1/3.18 µF - 1/6.54 µF - 1/8.19 µF
To simplify the calculation, we can convert the fractions into a common denominator:
1/C3 = (8.19 - 6.54 - 3.18) / (6.54 * 8.19 * 3.18) µF
1/C3 = -1.53 / 169.540316 µF
Now we can take the reciprocal of both sides to isolate C3:
C3 = 1 / (-1.53 / 169.540316) µF
C3 ≈ -110.786 µF
The third capacitance, based on the calculation, is approximately -110.786 µF. However, it's important to note that a capacitance value cannot be negative. Therefore, there might be an error in the given information or the calculation process. Double-check the information and calculations to find the correct value for the third capacitance.