Light Bulb Circuit Simulation
This simulation demonstrates the electrical characteristics of a 100W light bulb connected to a 220V AC supply:
220V
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Resistance:
484 Ω
\( R = \frac{V^2}{P} \)
Peak Voltage:
311 V
\( V_p = V_{rms}\sqrt{2} \)
RMS Current:
0.45 A
\( I = \frac{P}{V} \)
AC Voltage (220V RMS)
Physics Explanation:
Given:
- Power rating (P) = 100 W
- Voltage (V) = 220 V (RMS)
(a) Resistance of the bulb:
\( P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P} \)
\( R = \frac{(220)^2}{100} = 484 \, \Omega \)
(b) Peak voltage of the source:
\( V_p = V_{rms} \times \sqrt{2} \)
\( V_p = 220 \times 1.414 = 311 \, \text{V} \)
(c) RMS current through the bulb:
\( P = I_{rms} \times V_{rms} \Rightarrow I_{rms} = \frac{P}{V} \)
\( I_{rms} = \frac{100}{220} = 0.45 \, \text{A} \)
Key observations:
- The bulb's resistance is constant (for a given temperature)
- AC power uses RMS values for equivalent DC power calculations
- Peak voltage is higher than RMS voltage by √2 factor