The charge q on a capacitor is given by 𝑞 = 𝑞0(1 − 𝑒−𝑘𝑡), where0q is the initialcharge, k is a constant, and t is the time. How long does it take to recharge thecapacitor to 90% of capacity if 𝑘 = 15?
Question
The charge q on a capacitor is given by 𝑞 = 𝑞0(1 − 𝑒−𝑘𝑡), where0q is the initialcharge, k is a constant, and t is the time. How long does it take to recharge thecapacitor to 90% of capacity if 𝑘 = 15?
Solution
To solve this problem, we need to set q/q0 (the fraction of the capacitor's capacity that's charged) equal to 0.9 (since we want to know when it's 90% charged), and solve for t.
The equation becomes:
0.9 = 1 - e^(-15t)
We can start solving for t by first subtracting 0.9 from both sides:
0.1 = e^(-15t)
Next, we take the natural logarithm (ln) of both sides to get rid of the exponential on the right side:
ln(0.1) = -15t
Finally, we solve for t by dividing both sides by -15:
t = ln(0.1) / -15
Using a calculator, we find that t ≈ 0.1549 seconds. So it takes approximately 0.1549 seconds to recharge the capacitor to 90% of its capacity.
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