Comparing the Brightness of Incandescent Light Bulbs in Series
To determine which of two incandescent light bulbs connected in series will appear brighter, we must delve into the principles of electronic circuits, the behavior of resistors, and the relationship between power, voltage, and current.
1. Understanding Power Ratings
Incandescent light bulbs are rated for specific power and voltage. For this discussion, we have two bulbs with power ratings of 40 watts and 60 watts, connected across a 120V main supply. The power formula, (P V cdot I), indicates the relationship between power, voltage, and current. These bulbs are designed to operate at 120V, and their resistance can be calculated using (R frac{V^2}{P}.)
Calculating Resistance
For the 40W bulb:
[begin{aligned}R_{40} frac{120^2}{40} frac{14400}{40} 360 Omegaend{aligned}]For the 60W bulb:
[begin{aligned}R_{60} frac{120^2}{60} frac{14400}{60} 240 Omegaend{aligned}]When connected in series, the total resistance (R_{total}) is the sum of the individual resistances:
[R_{total} R_{40} R_{60} 360 Omega 240 Omega 600 Omega]Current Calculation
Using Ohm's law, we can find the current (I) flowing through the circuit:
[I frac{V_{total}}{R_{total}} frac{120V}{600 Omega} 0.2 A]2. Power Dissipation in Series Connection
The power dissipated by each bulb can be found using (P I^2 cdot R).
Power for the 40W Bulb
[begin{aligned}P_{40} (0.2)^2 cdot 360 0.04 cdot 360 14.4 Wend{aligned}]Power for the 60W Bulb
[begin{aligned}P_{60} (0.2)^2 cdot 240 0.04 cdot 240 9.6 Wend{aligned}]Based on these calculations, the 40W bulb will consume more power and therefore appear brighter.
3. Deeper Analysis of Bulb Behavior in Series Connection
Previous authors have made assumptions that can be misleading. They often assume Ohmic behavior (constant resistance) for the bulbs, implying (V I cdot R). However, the relationship between power, voltage, and resistance in a filament bulb is more complex, particularly for low voltage drop scenarios.
4. Realistic Voltage Drop and Power Calculation
Considering the actual behavior of filament bulbs:
The key equations for the voltage drop in a bulb are:
[V V_{rated} cdot V_{new}^{0.55}] [I I_{rated} cdot frac{V_{new}}{V_{rated}}^{0.55}] [P V_{new} cdot I_{new}]Given ratings of 120V for both bulbs, and current:
[I_{40} frac{40W}{120V} 0.333A quad I_{60} frac{60W}{120V} 0.500A]At 100V (since the total voltage is 120V and bulbs are in series), the current through the bulbs remains the same at 0.243A.
Using these values, the voltage and power dissipation:
[V_40 V_60 100Vcdot0.243A^{2/2.2} 100Vcdot0.59^2 34.81 V] [begin{aligned}P_40 V_40 cdot I_40 34.81V cdot 0.243A 8.44W P_60 V_60 cdot I_60 34.81V cdot 0.243A 8.44Wend{aligned}]These calculations indicate that the brightness of the bulbs is more affected by the specific voltage drop and current flow through them than by simplistic power ratings.
5. Conclusion and Final Thoughts
In a series circuit, the bulb with higher resistance (in this case, the 40W bulb) will appear brighter due to the greater voltage drop across it. This detailed analysis confirms the earlier finding.
While the traditional approach simplifies the process, a more nuanced understanding shows that bulb behavior is governed by complex relationships, requiring detailed calculations to accurately predict performance.
Final Answer: The 40W bulb will glow brighter than the 60W bulb when connected in series, due to the higher voltage drop and subsequent power dissipation.