Question

A circuit contains a source of time-varying emf, which is given by \(V_{\mathrm{emf}}=120.0 \sin [(377 \mathrm{rad} / \mathrm{s}) t] \mathrm{V},\) and a capacitor with capacitance \(C=5.00 \mu \mathrm{F}\). What is the current in the circuit at \(t=1.00\) s? a) 0.226 A b) 0.451 A c) \(0.555 \mathrm{~A}\) d) \(0.750 \mathrm{~A}\) e) \(1.25 \mathrm{~A}\)

Short answer

Answer: The current in the circuit at t=1.00s is 0.555 A.

Step-by-step solution

List the given values

We have the given values: Voltage function: \(V_{\mathrm{emf}}=120 \sin [(377 \: \mathrm{rad/s}) \: t] \: \mathrm{V}\) Capacitance: \(C= 5.00 \: \mu \mathrm{F}\) Time: \(t=1.00 \: \mathrm{s}\)

Write down the formula for current in a capacitor

The formula for the current in a capacitor is given by: \(I(t) = C \frac{dV(t)}{dt}\)

Differentiate the voltage function with respect to time

The derivative of the voltage function with respect to time is: \(\frac{dV}{dt} = \frac{d}{dt}(120 \sin [(377 \: \mathrm{rad/s}) t])\) Using chain rule, we can differentiate this equation: \(\frac{dV}{dt} = 120 \cdot (377 \: \mathrm{rad/s}) \cdot \cos [(377 \: \mathrm{rad/s}) t]\)

Find the current at given time interval (t = 1.00 s)

Now, we'll substitute the capacitance, \(C\), and time, \(t\), into the current formula: \(I(1.00 \: \mathrm{s}) = (5.00 \: \mu \mathrm{F}) \: (120 \cdot (377 \: \mathrm{rad/s}) \cdot \cos [(377 \: \mathrm{rad/s})(1.00 \: \mathrm{s})])\) \(I(1.00 \: \mathrm{s}) = (5.00 \times 10^{-6} \: \mathrm{F}) \: (120 \cdot (377 \: \mathrm{s^{-1}}) \cdot \cos (377 \: \mathrm{rad})\)

Calculate the value for current

After performing the calculations, we get: \(I(1.00 \: \mathrm{s}) = 0.555 \: \mathrm{A}\) The current in the circuit at 1.00 s is \(0.555 \: \mathrm{A}\). The correct answer is (c).