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A 100 watt bulb working on 200 volt and ...

A 100 watt bulb working on 200 volt and a 200 watt bulb working on 100 volt have

A

resistance in the ratio of `4:1`

B

maximum current ratings in the ratio of `1:4`

C

resistance in the ratio of `2:1`

D

maximum current ratings in the ratio of `1:2`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the resistance and current ratings for both bulbs and then compare them. Let's break it down step by step. ### Step 1: Calculate the resistance of the 100-watt bulb (Bulb 1) We know the power (P) and voltage (V) for Bulb 1: - Power (P1) = 100 watts - Voltage (V1) = 200 volts Using the formula for resistance: \[ R = \frac{V^2}{P} \] Substituting the values: \[ R_1 = \frac{200^2}{100} = \frac{40000}{100} = 400 \, \text{ohms} \] ### Step 2: Calculate the current for the 100-watt bulb (Bulb 1) Using Ohm's law, we can find the current (I): \[ I = \frac{V}{R} \] Substituting the values: \[ I_1 = \frac{200}{400} = \frac{1}{2} \, \text{amperes} \] ### Step 3: Calculate the resistance of the 200-watt bulb (Bulb 2) Now, we know the power and voltage for Bulb 2: - Power (P2) = 200 watts - Voltage (V2) = 100 volts Using the same formula for resistance: \[ R_2 = \frac{V^2}{P} \] Substituting the values: \[ R_2 = \frac{100^2}{200} = \frac{10000}{200} = 50 \, \text{ohms} \] ### Step 4: Calculate the current for the 200-watt bulb (Bulb 2) Using Ohm's law again to find the current: \[ I_2 = \frac{V}{R} \] Substituting the values: \[ I_2 = \frac{100}{50} = 2 \, \text{amperes} \] ### Step 5: Determine the ratio of resistances Now we can find the ratio of the resistances of the two bulbs: \[ \text{Resistance ratio} = \frac{R_1}{R_2} = \frac{400}{50} = 8:1 \] ### Step 6: Determine the ratio of maximum current ratings Now, we find the ratio of the maximum current ratings: \[ \text{Current ratio} = \frac{I_1}{I_2} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \] ### Conclusion From the calculations: - The resistance of the first bulb is 400 ohms and the second bulb is 50 ohms, giving a resistance ratio of 8:1. - The current ratings are in the ratio of 1:4. Thus, the correct answer is that the maximum current ratings are in the ratio of 1:4.
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