Case Study
Passage with linked questions
Case Set 1
Case AnalysisPassage
Riya is a Class 12 student studying ac circuits in her physics lab. She connects a 100 Ω resistor to a 220 V, 50 Hz ac source and observes the current waveform on an oscilloscope. She notices that the current and voltage waveforms are perfectly aligned — they reach their peaks and zero crossings at the same time. Her teacher explains that this is unique to resistors among the three basic circuit elements. Riya then calculates the power consumed and compares it to what would happen if the same resistor were connected to a 220 V dc source. She finds that the heating effect in both cases is identical when rms values are used for the ac case, confirming the concept her teacher had explained about the equivalence of rms ac and dc for power calculations.
Question 1: What is the phase difference between voltage and current in a pure resistor connected to an ac source?
- The phase difference between voltage and current in a pure resistor is zero — they are in phase.
- Both voltage and current reach their maximum, minimum, and zero values at the same instant, as confirmed by the oscilloscope traces aligning perfectly.
Question 2: Calculate the rms current through the 100 Ω resistor connected to 220 V ac. What is the net power consumed over a full cycle?
- Rms current: I = V/R = 220/100 = 2.2 A. This follows from Ohm's law, which applies equally for rms values in a purely resistive ac circuit.
- Net power consumed: P = I²R = (2.2)² × 100 = 484 W. Alternatively P = V²/R = 220²/100 = 484 W. Power factor = 1 for a pure resistor.
Question 3: Explain why the average current over one complete ac cycle is zero, yet Joule heating is not zero. Define rms current and show how it makes the ac power formula identical in form to the dc power formula.
- Average current is zero because the positive and negative half-cycles of a sinusoidal current are equal and opposite, canceling over one complete cycle.
- Joule heating depends on i²R. Since i² is always positive regardless of current direction, the average of i² is not zero — it equals im²/2.
- Rms current I = √(average of i²) = im/√2 = 0.707 im. It is the square root of the mean of the square of instantaneous current over one cycle.
- Substituting: P = (im²/2)R = I²R — identical in form to dc power P = I²R, so the rms current is the equivalent dc current that produces the same average Joule heating.