Analyzing Bulbs in a Series Circuit and Predicting Fusing

This problem involves two electric bulbs with different power and voltage ratings connected in series across a voltage source. We need to determine which bulb is likely to fuse. A bulb fuses when the current flowing through it significantly exceeds its rated current, leading to excessive heating and damage to the filament. In a series circuit, the same current flows through all components. We will calculate the current in the series circuit and compare it to the rated current of each bulb.

Step-by-Step Solution

1. Calculate the Resistance of Each Bulb

The resistance of a bulb can be calculated using its power ($P$) and voltage ($V$) ratings with the formula $R = \frac{V^2}{P}$.

• Bulb A: (100 W, 100 V)

  \(R_A = \frac{V_A^2}{P_A} = \frac{(100 \, V)^2}{100 \, W} = \frac{10000}{100} = 100 \, \Omega\)

• Bulb B: (60 W, 100 V)

  \(R_B = \frac{V_B^2}{P_B} = \frac{(100 \, V)^2}{60 \, W} = \frac{10000}{60} = \frac{500}{3} \, \Omega\)

2. Calculate the Total Resistance of the Series Combination

In a series connection, the total resistance is the sum of individual resistances.

\(R_{total} = R_A + R_B = 100 \, \Omega + \frac{500}{3} \, \Omega = \frac{300}{3} \, \Omega + \frac{500}{3} \, \Omega = \frac{800}{3} \, \Omega\)

3. Calculate the Current Flowing Through the Series Circuit

The current ($I$) through the series combination is found using Ohm's Law, \(I = \frac{V_{total}}{R_{total}}\), where \(V_{total}\) is the applied voltage (200 V).

\(I_{series} = \frac{200 \, V}{\frac{800}{3} \, \Omega} = 200 \times \frac{3}{800} \, A = \frac{600}{800} \, A = \frac{3}{4} \, A = 0.75 \, A\)

This current of 0.75 A flows through both Bulb A and Bulb B as they are in series.

4. Calculate the Rated Current for Each Bulb

The rated current of a bulb is the current it safely handles under its rated voltage and power conditions. \(I_{rated} = \frac{P_{rated}}{V_{rated}}\).

• Bulb A Rated Current:

  \(I_{A,rated} = \frac{100 \, W}{100 \, V} = 1 \, A\)

• Bulb B Rated Current:

  \(I_{B,rated} = \frac{60 \, W}{100 \, V} = 0.6 \, A\)

5. Compare the Series Current with Rated Currents

Now, let's compare the calculated series current (0.75 A) with the rated current of each bulb:

• For Bulb A: Series current (0.75 A) < Rated current (1 A). Bulb A is operating below its rated current.
• For Bulb B: Series current (0.75 A) > Rated current (0.6 A). Bulb B is carrying more current than it is rated for.

Operating a bulb significantly above its rated current causes excessive heating of the filament, which can lead to it melting and fusing.

6. Calculate Voltage Across Each Bulb in the Series Circuit (Optional but helpful)

We can also calculate the voltage across each bulb using \(V = I_{series} \times R\).

• Voltage across Bulb A:

  \(V_A = I_{series} \times R_A = 0.75 \, A \times 100 \, \Omega = 75 \, V\)

  Bulb A is rated for 100V. It is operating at 75V.

• Voltage across Bulb B:

  \(V_B = I_{series} \times R_B = 0.75 \, A \times \frac{500}{3} \, \Omega = \frac{3}{4} \times \frac{500}{3} \, V = \frac{500}{4} \, V = 125 \, V\)

  Bulb B is rated for 100V. It is operating at 125V, which is above its rated voltage.

Operating a bulb above its rated voltage also leads to higher current and excessive power dissipation, increasing the risk of fusing.

7. Calculate Power Dissipated by Each Bulb in the Series Circuit (Optional but helpful)

We can also calculate the power dissipated by each bulb using \(P = I_{series}^2 \times R\).

• Power in Bulb A:

  \(P_A = (0.75 \, A)^2 \times 100 \, \Omega = (0.5625) \times 100 \, W = 56.25 \, W\)

  Bulb A is rated for 100W. It is dissipating 56.25W.

• Power in Bulb B:

  \(P_B = (0.75 \, A)^2 \times \frac{500}{3} \, \Omega = (0.5625) \times \frac{500}{3} \, W = 0.5625 \times 166.67 \, W \approx 93.75 \, W\)

  Bulb B is rated for 60W. It is dissipating approximately 93.75W, which is significantly higher than its rating.

Conclusion

Both the current analysis ($I_{series} > I_{B,rated}$) and the voltage analysis ($V_B > V_{B,rated}$) and the power analysis ($P_B > P_{B,rated}$) show that Bulb B is operating under conditions exceeding its safe limits. Bulb A is operating below its rated limits. Therefore, Bulb B will draw excessive current for its rating and will fuse first.

Bulb B has a higher resistance (\(500/3 \, \Omega \approx 166.67 \, \Omega\)) compared to Bulb A (100 Ω). In a series circuit, the component with higher resistance drops a larger portion of the total voltage and dissipates more power (for the same current $I$, $P=I^2R$). Although the total applied voltage (200V) is the sum of their individual ratings (100V + 100V = 200V), this is only true if they operate at their rated conditions, which they don't in this series combination across 200V. The 60W, 100V bulb (Bulb B) is designed to operate at 0.6A and 100V. When forced to carry 0.75A and experience 125V, it will overheat and fuse.

Revision Table: Bulb Circuit Analysis

Understanding how bulbs behave in different circuit configurations is key. Here's a summary of relevant concepts:

• Rated Values: Power and voltage marked on a bulb indicate its safe operating conditions.
• Resistance: Resistance is a fixed property for a given filament at a certain temperature (though resistance changes with temperature). It can be calculated from rated values.
• Series Circuit: Components are connected end-to-end. Current is the same through all components. Voltage is divided across components based on their resistance.
• Fusing: Occurs when current/voltage/power exceeds the component's limits, causing overheating and failure.

Additional Information: Electrical Component Ratings and Circuit Types

Electrical components like bulbs are designed to operate within specific voltage and current ranges. Exceeding these ratings can lead to damage or failure (like fusing).

• Power Rating: Indicates the power dissipated by the device under its rated voltage. A higher wattage bulb (at the same voltage) has lower resistance.
• Voltage Rating: The maximum voltage that can be safely applied across the device.
• Current Rating: The maximum current that can safely flow through the device.
• Parallel Circuit: In a parallel circuit, components are connected across the same voltage source. The total current is the sum of currents through each branch. If these bulbs were connected in parallel across 200V, both would likely fuse immediately as they are rated for only 100V.

In series circuits, the component with the proportionally higher resistance relative to its rating often experiences stress. In this case, the 60W (100V) bulb has a higher resistance than the 100W (100V) bulb. When connected in series, it draws a larger voltage drop and dissipates more power than its rating, even though the current is the same through both. This leads to its failure.