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Chapter 17: Problem 65

The voltage across a \(10 \mu \mathrm{F}\) capacitor is \(50 t^{2} \mathrm{~V}\). The time, in seconds, it will take to store \(200 \mathrm{~J}\) of energy is most nearly a) \(0.15\) c) \(1.38\) b) \(0.21\) d) \(11.25\)

Short Answer

Expert verified
Answer: It takes approximately \(1.38\) seconds for the given capacitor to store \(200J\) of energy.

Step by step solution

01

Write down the given information

We are given a capacitor with capacity \(C = 10\mu F = 10 \times 10^{-6} F\) Voltage \(V = 50t^2 V\) Energy stored, \(E = 200J\)
02

Write down the energy formula for capacitors

The energy stored in a capacitor is given by \(E = \frac{1}{2}CV^2\)
03

Substitute the given values and solve for time t

Substituting the given values into the energy formula, we get \(200J = \frac{1}{2}(10\times 10^{-6} F)(50t^2 V)^2\) Now we need to solve for \(t\). Divide both sides by \(\frac{1}{2}(10\times 10^{-6})\): \(t^4 = \frac{200}{\frac{1}{2}(10\times 10^{-6})(50)}\) \(t^4 = 800\) Take the fourth root of both sides: \(t = \sqrt[4]{800}\) Using a calculator, we get: \(t \approx 1.38 s\) So, the answer is: c) \(1.38\)

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