A \(300 .-\Omega\) resistor is connected in series with a \(4.00-\mu F\) capacitor and a source of time-varying emf providing \(V_{\mathrm{rms}}=40.0 \mathrm{~V}\) a) At what frequency will the potential drop across the capacitor equal that across the resistor? b) What is the rms current through the circuit when this occurs?

The resistor value: \(R = 300\,\Omega\) The capacitor value: \(C = 4.00\,\mu F = 4 \times 10^{-6}\,F\) The rms value of the source voltage: \(V_{rms} = 40.0\,V\)

We know that in a series RC circuit, the potential drop across the resistor is given by: \(V_R = IR\) And across the capacitor: \(V_C = I \frac{1}{\omega C}\) Where \(\omega = 2\pi f\), \(f\) is the frequency, and \(I\) is the current through the circuit. We are asked to find the frequency at which \(V_R = V_C\), so we have: \(IR = I \frac{1}{\omega C}\)

We can cancel the \(I\) from both sides, and then solve for the frequency \(\omega\): \(R = \frac{1}{\omega C}\) \(\omega = \frac{1}{RC}\) Now we can plug in the values of \(R\) and \(C\) to calculate the frequency \(\omega\): \(\omega = \frac{1}{(300\,\Omega)(4 \times 10^{-6}\,F)}\) \(\omega = 833.33\, rad/s\) Now, convert the angular frequency to the regular frequency: \(f = \frac{\omega}{2\pi} = \frac{833.33}{2\pi} = 132.6\, Hz\) So, the frequency at which the potential drop across the capacitor equals that across the resistor is \(132.6\, Hz\).

We can now find the rms current when this occurs. First, we need to calculate the impedance \(Z\) of the circuit, which can be given by: \(Z = \sqrt{R^2 + X_C^2}\) Where \(X_C = \frac{1}{\omega C}\) is the capacitive reactance. Using the previously calculated \(\omega\): \(X_C = \frac{1}{(833.33\, rad/s)(4 \times 10^{-6}\,F)} = 300\,\Omega\) Now, we can calculate the impedance \(Z\): \(Z = \sqrt{(300\,\Omega)^2 + (300\,\Omega)^2} = 424.26\,\Omega\) Finally, we can find the rms current using Ohm's Law: \(I_{rms} = \frac{V_{rms}}{Z} =\frac{40\,V}{424.26\,\Omega} = 0.0943\,A\) Thus, the rms current through the circuit when the potential drop across the capacitor equals that across the resistor is \(0.0943\,A\) or \(94.3\, mA\).