A 270 𝜇𝐹 capacitor is charged at a steady rate of 150 𝜇𝐶𝑠−1. Find the time (in seconds) it will taketo raise the potential difference across the capacitor to 15 V.
Question
A 270 𝜇𝐹 capacitor is charged at a steady rate of 150 𝜇𝐶𝑠−1. Find the time (in seconds) it will taketo raise the potential difference across the capacitor to 15 V.
Solution
The formula that relates charge (Q), capacitance (C), and voltage (V) in a capacitor is Q = CV.
We are given: C = 270 μF = 270 x 10^-6 F (converting microfarads to farads) V = 15 V The rate of charge = 150 μC/s = 150 x 10^-6 C/s (converting microcoulombs to coulombs)
We need to find the time it takes to charge the capacitor to 15 V.
First, we can find the total charge needed to get the capacitor to 15 V by rearranging Q = CV to Q = CV = (270 x 10^-6 F) * (15 V) = 4.05 x 10^-3 C.
Next, we can find the time it takes to reach this charge by dividing the total charge by the rate of charge: time = Q / rate = (4.05 x 10^-3 C) / (150 x 10^-6 C/s) = 27 seconds.
So, it will take 27 seconds to raise the potential difference across the capacitor to 15 V.
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