Two electric bulbs A and B are rated as 60 W and 100 W . They are connected in parallel to the same source. Then,

Options

A. B draws more current than A
B. Currents drawn are in the ratio of their resistances
C. both draw the same current
D. A draws more current than B

Answer

To solve the problem of two electric bulbs A and B rated at 60 W and 100 W respectively, connected in parallel to the same voltage source, we can follow these steps:

### Step 1: Understand the Power Ratings
The power ratings of the bulbs are given as:
- Bulb A: \( P_A = 60 \, W \)
- Bulb B: \( P_B = 100 \, W \)

### Step 2: Use the Power Formula
The power consumed by an electrical device is given by the formula:
\[ P = \frac{V^2}{R} \]
Where:
- \( P \) is the power,
- \( V \) is the voltage across the device,
- \( R \) is the resistance of the device.

### Step 3: Calculate the Resistance of Each Bulb
Since both bulbs are connected in parallel, they experience the same voltage \( V \). We can rearrange the power formula to find the resistance:
\[ R = \frac{V^2}{P} \]

For Bulb A:
\[ R_A = \frac{V^2}{P_A} = \frac{V^2}{60} \]

For Bulb B:
\[ R_B = \frac{V^2}{P_B} = \frac{V^2}{100} \]

### Step 4: Find the Ratio of Resistances
To find the ratio of the resistances \( R_A \) and \( R_B \):
\[ \frac{R_A}{R_B} = \frac{V^2/60}{V^2/100} = \frac{100}{60} = \frac{5}{3} \]

### Step 5: Determine the Current Through Each Bulb
The current through each bulb can be calculated using Ohm's law:
\[ I = \frac{V}{R} \]

The current through Bulb A:
\[ I_A = \frac{V}{R_A} = \frac{V}{V^2/60} = \frac{60}{V} \]

The current through Bulb B:
\[ I_B = \frac{V}{R_B} = \frac{V}{V^2/100} = \frac{100}{V} \]

### Step 6: Find the Ratio of Currents
To find the ratio of the currents:
\[ \frac{I_A}{I_B} = \frac{60/V}{100/V} = \frac{60}{100} = \frac{3}{5} \]

### Step 7: Conclusion
Since the current is inversely proportional to the resistance, we can conclude:
- Bulb B, which has a lower resistance, draws more current than Bulb A.

### Final Answer
Thus, the correct option is that Bulb B draws more current than Bulb A.

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