Question

A 75,000 -W light bulb (yes, there are such things!) operates at \(I_{\mathrm{rms}}=200 . \mathrm{A}\) and \(V_{\mathrm{rms}}=440 . \mathrm{V}\) in a \(60.0-\mathrm{Hz} \mathrm{AC}\) circuit. Find the resistance, \(R,\) and self- inductance, \(L,\) of this bulb. Its capacitive reactance is negligible.

Short answer

Answer: To find the resistance (R) and self-inductance (L) of the bulb, follow these steps: 1. Calculate the impedance (Z) using the formula: \(Z=\frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}}\). 2. Approximate the resistance (R) using the equation: \(R \approx Z\). 3. Calculate the inductive reactance (X_L) using the formula: \(X_L = \sqrt{Z^2 - R^2}\). 4. Finally, find the self-inductance (L) using the equation: \(L = \frac{X_L}{2 \pi f}\). Performing these calculations will give you the resistance and self-inductance of the light bulb.

Step-by-step solution

Calculate Impedance, \(Z\)

Using power, rms current, and rms voltage, we will calculate the impedance \(Z\). The formula for power is given by: $$ P = I_{\mathrm{rms}} V_{\mathrm{rms}}\cos\varphi $$ where \(\cos\varphi\) is the power factor. Since the capacitive reactance is negligible, the power factor would be almost equal to 1. Thus, we have: $$ P \approx I_{\mathrm{rms}}V_{\mathrm{rms}} $$ Now, we can find the impedance \(Z\) using the formula: $$ Z=\frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}} $$

Calculate Resistance, \(R\)

Knowing the impedance \(Z\) found in step 1, we can now find the resistance of the bulb. Using the power formula and power factor, we get: $$ R = Z \cos\varphi \approx Z $$

Calculate Inductive Reactance, \(X_L\)

To find the self-inductance \(L\), we first need to calculate the inductive reactance \(X_L\). The inductive reactance can be calculated using the impedance \(Z\) and resistance \(R\) found in the previous steps, through the following formula: $$ X_L = \sqrt{Z^2 - R^2} $$

Find Self-Inductance, \(L\)

Finally, to find the self-inductance \(L\), we will use the inductive reactance \(X_L\) calculated in step 3. The formula for inductive reactance is given by: $$ X_L = 2 \pi f L $$ From this formula, we can find the self-inductance \(L\) as: $$ L = \frac{X_L}{2 \pi f} $$ Calculating each step will give us the desired resistance \(R\) and self-inductance \(L\) of the bulb.

Key concepts

Impedance Calculation

Impedance (\textbf{Z}) is the measure of opposition that an AC circuit presents to the flow of current. It can be thought of as the AC equivalent of resistance in a DC circuit, but impedance also takes into account the reactance from inductors and capacitors. To calculate impedance, you can use the root mean square (RMS) values of voltage (\textbf{V_{rms}}) and current (\textbf{I_{rms}}).

The formula to find impedance is:\[Z = \frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}}\]
Once you have both the RMS voltage and current, impedance is found by simply dividing the voltage by the current. In the example exercise, with values for a high-power light bulb in an AC circuit, the impedance is a key value needed to further explore the characteristics of the bulb's electrical behavior. Notably, in this case, since capacitive reactance is negligible, the power factor is assumed to be approximately 1.

Resistive Reactance

Resistive reactance refers to the resistance component of an AC circuit's overall impedance. In most simple terms, it is the resistance that does not change with frequency and affects both the amplitude and phase of the AC circuit. Unlike reactances which depend on frequency, the resistive part of impedance acts the same way as resistance in a DC circuit.

To find the resistive reactance (\textbf{R}), you typically use the impedance (\textbf{Z}) and the power factor (\textbf{cos\textphi}). For non-reactive components where the capacitive or inductive effects can be ignored, resistive reactance is equal to the impedance, given by:\[R = Z \times \cos\textphi \text{(or approximately equal to Z if cos\textphi is close to 1)}\]
The exercise showcases that with knowledge of the impedance and the assumption of a negligible capacitive reactance, you can directly conclude that the resistance of the bulb is approximately equal to its impedance.

Inductive Reactance

Inductive reactance (\textbf{X_L}) arises in a circuit that contains inductance, typically due to coils or inductors, and varies with the frequency of the AC supply (\textbf{f}). The formula to compute the inductive reactance of a component in an AC circuit is:\[X_L = 2 \textpi f L \]

where \textbf{L} is the self-inductance of the inductor. Inductive reactance causes the current to lag behind the voltage by a 90-degree phase difference. To determine the reactance in the example exercise, one would subtract the resistive component squared from the impedance squared and take the square root:\[X_L = \sqrt{Z^2 - R^2}\]

Understanding the concept of inductive reactance is fundamental when analyzing circuits that incorporate elements such as coils or inductors, especially in power distribution and radio-frequency applications.

Self-inductance

Self-inductance (\textbf{L}) is a property of a conducting loop or coil that quantifies its ability to resist changes in current. When the current in a coil changes, it induces a voltage within itself, opposing the change in current, a phenomenon known as self-inductance. It is a key factor in the study of inductors in AC circuits.

To find the self-inductance from the inductive reactance (\textbf{X_L}), you can apply the following equation which relates reactance, inductance, and the frequency of the AC current (\textbf{f}):\[L = \frac{X_L}{2 \textpi f}\]
This formula is extrapolated in the exercise to solve for \textbf{L} after finding the inductive reactance. Understanding self-inductance is crucial in the design of various electrical components, including transformers, inductors, and even the discussed light bulb's filament, which behaves inductively.