Understanding Bulb Brightness in Series Circuits

This problem involves comparing the brightness of two incandescent bulbs when connected in series to a power supply. The brightness of a bulb depends on the actual power it dissipates, which is determined by the current flowing through it and its resistance. We need to calculate the resistance of each bulb first, then analyze the circuit behavior when they are connected in series.

Bulb Resistance Calculation

The power rating of an appliance tells us how much power it consumes under specific voltage conditions. The relationship between power ($P$), voltage ($V$), and resistance ($R$) is given by the formula $P = V^2 / R$. We can rearrange this to find the resistance of each bulb, assuming the resistance remains constant.

• Bulb 1: Rated Power $P_1 = 200\,\text{W}$ at $V_{rated} = 220\,\text{V}$.
  • Resistance $R_1 = \frac{V_{rated}^2}{P_1}\)
  • $R_1 = \frac{(220\,\text{V})^2}{200\,\text{W}} = \frac{48400\,\text{V}^2}{200\,\text{W}} = 242\,\Omega$
• Bulb 2: Rated Power $P_2 = 100\,\text{W}$ at $V_{rated} = 220\,\text{V}$.
  • Resistance $R_2 = \frac{V_{rated}^2}{P_2}\)
  • $R_2 = \frac{(220\,\text{V})^2}{100\,\text{W}} = \frac{48400\,\text{V}^2}{100\,\text{W}} = 484\,\Omega$

From the calculations, we see that the $100\,\text{W}$ bulb (Bulb 2) has a higher resistance ($R_2 = 484\,\Omega$) than the $200\,\text{W}$ bulb (Bulb 1, $R_1 = 242\,\Omega$).

Series Circuit Current Calculation

When the two bulbs are connected in series across a supply voltage $V_{supply} = 110\,\text{V}$, the total resistance in the circuit is the sum of their individual resistances:

$ R_{total} = R_1 + R_2 $ $ R_{total} = 242\,\Omega + 484\,\Omega = 726\,\Omega $

The current ($I$) flowing through a series circuit is the same for all components and can be calculated using Ohm's Law ($V = IR$):

$ I = \frac{V_{supply}}{R_{total}} $ $ I = \frac{110\,\text{V}}{726\,\Omega} \approx 0.1515\,\text{A} $

Actual Power and Brightness Comparison

The brightness of a bulb is determined by the actual power it dissipates under the operating conditions. In a series circuit, the power dissipated by each resistor (bulb) can be calculated using the formula $P_{actual} = I^2R$.

• Actual Power dissipated by Bulb 1 ($P_{actual,1}$): $ P_{actual,1} = I^2 R_1 $ $ P_{actual,1} = \left(\frac{110}{726}\,\text{A}\right)^2 \times 242\,\Omega $ $ P_{actual,1} \approx (0.1515\,\text{A})^2 \times 242\,\Omega \approx 0.02295 \times 242\,\text{W} \approx 5.55\,\text{W} $
• Actual Power dissipated by Bulb 2 ($P_{actual,2}$): $ P_{actual,2} = I^2 R_2 $ $ P_{actual,2} = \left(\frac{110}{726}\,\text{A}\right)^2 \times 484\,\Omega $ $ P_{actual,2} \approx (0.1515\,\text{A})^2 \times 484\,\Omega \approx 0.02295 \times 484\,\text{W} \approx 11.1\,\text{W} $

Comparing the actual power dissipated: $P_{actual,2} (11.1\,\text{W})$ is greater than $P_{actual,1} (5.55\,\text{W})$. Therefore, the $100\,\text{W}$ bulb (Bulb 2) will glow brighter than the $200\,\text{W}$ bulb (Bulb 1).

Brightness Factor Determination

To find out by what factor the $100\,\text{W}$ bulb glows brighter than the $200\,\text{W}$ bulb, we calculate the ratio of their actual powers:

$ \text{Factor} = \frac{P_{actual,2}}{P_{actual,1}} $ $ \text{Factor} = \frac{11.1\,\text{W}}{5.55\,\text{W}} = 2 $

Thus, the $100\,\text{W}$ bulb glows brighter by a factor of $2$. This is because, in a series circuit, the current is the same through both bulbs, and power is proportional to resistance ($P = I^2R$). The bulb with higher resistance (the $100\,\text{W}$ rated bulb) dissipates more power and hence glows brighter.