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Two electic bulbs marked 25 W -220 V and...

Two electic bulbs marked `25 W -220 V` and `100 W - 220 V` are connected in series to a 440 V supply. Which of the bulbs will fuse?

A

both

B

100W

C

25W

D

neither

Text Solution

AI Generated Solution

The correct Answer is:
To determine which bulb will fuse when the two bulbs are connected in series to a 440 V supply, we can follow these steps: ### Step 1: Understand the Bulb Ratings We have two bulbs: - Bulb 1: 25 W, 220 V - Bulb 2: 100 W, 220 V ### Step 2: Calculate the Current Ratings of Each Bulb The current through each bulb can be calculated using the formula: \[ I = \frac{P}{V} \] where \( P \) is the power rating and \( V \) is the voltage rating. **For Bulb 1:** \[ I_1 = \frac{25 \, \text{W}}{220 \, \text{V}} = 0.1136 \, \text{A} \] **For Bulb 2:** \[ I_2 = \frac{100 \, \text{W}}{220 \, \text{V}} = 0.4545 \, \text{A} \] ### Step 3: Calculate the Resistance of Each Bulb The resistance of each bulb can be calculated using the formula: \[ R = \frac{V^2}{P} \] **For Bulb 1:** \[ R_1 = \frac{220^2}{25} = 1936 \, \Omega \] **For Bulb 2:** \[ R_2 = \frac{220^2}{100} = 484 \, \Omega \] ### Step 4: Calculate the Total Resistance in Series Since the bulbs are connected in series, the total resistance \( R_{\text{eq}} \) is: \[ R_{\text{eq}} = R_1 + R_2 = 1936 \, \Omega + 484 \, \Omega = 2420 \, \Omega \] ### Step 5: Calculate the Total Current Through the Circuit Using Ohm's law, the total current \( I \) when a 440 V supply is connected can be calculated as: \[ I = \frac{V}{R_{\text{eq}}} = \frac{440 \, \text{V}}{2420 \, \Omega} = 0.1818 \, \text{A} \] ### Step 6: Compare the Current Ratings Now we compare the calculated current \( I \) with the rated currents of the bulbs: - \( I_1 = 0.1136 \, \text{A} \) - \( I_2 = 0.4545 \, \text{A} \) - \( I = 0.1818 \, \text{A} \) ### Step 7: Determine Which Bulb Will Fuse - For Bulb 1: The rated current \( I_1 \) (0.1136 A) is less than the current \( I \) (0.1818 A), so Bulb 1 can handle this current. - For Bulb 2: The rated current \( I_2 \) (0.4545 A) is greater than the current \( I \) (0.1818 A), meaning Bulb 2 is rated for a higher current than what is flowing through it. Since Bulb 1 has a lower rated current than the current flowing through it, it will fuse. ### Conclusion The bulb that will fuse is the **25 W bulb**. ---
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