Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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.
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
