Question

A \(150-\Omega\) resistor is in series with a \(0.75-\mathrm{H}\) inductor in an ac circuit. The rms voltages across the two are the same. (a) What is the frequency? (b) Would each of the rms voltages be half of the rms voltage of the source? If not, what fraction of the source voltage are they? (In other words, \(V_{R} / \ell_{m}=V_{L} / \mathcal{E}_{m}=?\) ) (c) What is the phase angle between the source voltage and the current? Which leads? (d) What is the impedance of the circuit?

Short answer

Question: Determine the following for an AC circuit with a resistor and an inductor connected in series, given that the voltage across both the resistor and inductor are equal: (a) the frequency of the AC source, (b) the ratio of rms voltages across inductor and resistor to the source voltage, (c) the phase angle between the source voltage and the current, and (d) the impedance of the circuit. Answer: (a) The frequency is approximately \(31.83 \, \text{Hz}\), (b) the ratio of the rms voltages is approximately \(0.67\), (c) the phase angle between the source voltage and the current is approximately \(36.87^\circ\), and the current leads the source voltage, (d) the impedance of the circuit is approximately \(225 \, \Omega\).

Step-by-step solution

Finding the frequency

To find the frequency, we will use the equation: $$\frac{V_R}{V_L} = \frac{R}{\omega L}$$ Where \(V_R\) is the voltage across the resistor, \(V_L\) is the voltage across the inductor, \(R\) is the resistance, \(\omega\) is the angular frequency, and \(L\) is the inductance. Since \(V_R\) and \(V_L\) are equal, we can put them on the same side and obtain the following equation: $$\omega = \frac{R}{L}$$ Now, we can plug in the values of \(R\) and \(L\) and solve for \(\omega\). $$\omega = \frac{150}{0.75} = 200 \, \text{rad/s}$$ To find the frequency \(f\), we recall that \(\omega = 2 \pi f\). $$f = \frac{\omega}{2 \pi} = \frac{200}{2 \pi} \approx 31.83 \, \text{Hz}$$ So, the frequency of the AC source is approximately \(31.83 \, \text{Hz}\).

Finding the ratio of rms voltages (VR / Im = VL / Em)

Since the rms voltages across the resistor and inductor are equal, the equation becomes: $$\frac{V_R}{\ell_{m}} = \frac{V_L}{\mathcal{E}_{m}} = \frac{R}{Z}$$ Where \(Z\) is the impedance of the circuit, which can be calculated using the equation: $$Z = \sqrt{R^2 + (\omega L)^2}$$ $$Z = \sqrt{150^2 + (200 \times 0.75)^2} \approx 225 \, \Omega$$ Now, we can plug in the values of \(R\) and \(Z\) to find the ratio: $$\frac{V_R}{\ell_{m}} = \frac{V_L}{\mathcal{E}_{m}} = \frac{150}{225} = \frac{2}{3} \approx 0.67$$ So, each of the rms voltages is \(0.67\) or \(67 \%\) of the rms voltage of the source.

Finding the phase angle between source voltage and current

The phase angle \(\phi\) between the source voltage and the current can be calculated using the equation: $$\tan(\phi) = \frac{\omega L}{R}$$ $$\phi = \tan^{-1} \left(\frac{200 \times 0.75}{150}\right) \approx 36.87 ^\circ$$ As the impedance of the inductor is higher than the impedance of the resistor, the current leads the source voltage in an LR-circuit. Therefore, the phase angle is approximately \(36.87^\circ\), and the current leads the source voltage.

Finding the impedance of the circuit

We've already calculated the impedance of the circuit in Step 2: $$Z = \sqrt{R^2 + (\omega L)^2} \approx 225 \, \Omega$$ So, the impedance of the circuit is approximately \(225 \, \Omega\). To summarize, (a) the frequency is approximately \(31.83 \, \text{Hz}\), (b) the ratio of the rms voltages is approximately \(0.67\), (c) the phase angle between the source voltage and the current is approximately \(36.87^\circ\), and the current leads the source voltage, (d) the impedance of the circuit is approximately \(225 \, \Omega\).