You have three capacitors, capacitor 1 has a capacitance 6.78 mF, capacitor 2 has capacitance 2.51 mF and capacitor 3 has a capacitance of 3.84 mF. Calculate the capacitance of the circuit in the following configurations: Configuration 1 (all in series): C = Answer Question 3[input] mF

Question

BlackTom AI · Accepted Answer

To find the total capacitance when capacitors are in series, we use the formula 1/C_total = 1/C1 + 1/C2 + 1/C3.

First, compute the reciprocals:
- 1/C1 = 1/6.78 mF ≈ 0.1475 (1/mF)
- 1/C2 = 1/2.51 mF ≈ 0.3984 (1/mF)
- 1/C3 = 1/3.84 mF ≈ 0.2604 (1/mF)

Next, sum these reciprocals: 0.1475 + 0.3984 + 0.2604 ≈ 0.8063 (1/mF).

Finally, take the reciprocal to obtain the total capacitance: C_total = 1 / 0.8063 ≈ 1.240 mF.

Therefore, the combined capacitance for configuration 1 (all in series) is approximately 1.240 mF.

PHYS1221-PHYS1231-Physics 1B T1 2025

You have three capacitors, capacitor 1 has a capacitance 6.78 mF, capacitor 2 has capacitance 2.51 mF and capacitor 3 has a capacitance of 3.84 mF. Calculate the capacitance of the circuit in the following configurations: Configuration 1 (all in series): C = Answer Question 3[input] mF

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