Answered

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A 6.0 μF and a 8.0 μF capacitor are connected in series across an 8.0-V DC source. What is the
voltage across the 6.0 μF capacitor?

Sagot :

Answer: 4.59

Step-by-step explanation:When capacitors are connected in series, the total capacitance (C_total) can be calculated using the formula:

1/C_total = 1/C1 + 1/C2

Given that C1 = 6.0 μF and C2 = 8.0 μF, we can find the total capacitance:

1/C_total = 1/6 + 1/8

1/C_total = (4/24) + (3/24)

1/C_total = (7/24)

C_total = 24/7 ≈ 3.429 μF

In a series circuit, the voltage across each capacitor depends on the ratio of their capacitances to the total capacitance, multiplied by the total voltage. So, the voltage across the 6.0 μF capacitor (V_6) can be found as follows:

V_6 = (C_total / C1) * V_total

V_6 = (3.429 / 6.0) * 8.0

V_6 ≈ 4.59 V

Therefore, the voltage across the 6.0 μF capacitor in this series circuit is approximately 4.59 V.

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