Two bulbs are working in parallel order. Bulb A is brighter than bulb B . If `R_(A)` and `R_(B)` are their resistance respectively then

Options

A. `R_(A)=R_(B)`
B. `R_(B)gt R_(A)`
C. `R_(A)gtR_(B)`
D. None of these

Answer

To solve the problem step by step, we will analyze the relationship between the power consumed by the bulbs and their resistances.

### Step 1: Understand the arrangement of the bulbs
Two bulbs, A and B, are connected in parallel. In a parallel circuit, the voltage across each component is the same.

**Hint:** Remember that in a parallel circuit, the voltage across each bulb is equal.

### Step 2: Recall the formula for power
The power (P) consumed by an electrical device can be expressed using the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the device and \( R \) is its resistance.

**Hint:** Power is directly related to voltage and inversely related to resistance.

### Step 3: Compare the power of the bulbs
According to the problem, bulb A is brighter than bulb B. Since brightness is directly proportional to the power consumed, we can write:
\[ P_A > P_B \]
This means that the power consumed by bulb A is greater than that consumed by bulb B.

**Hint:** Brightness indicates higher power consumption.

### Step 4: Relate power to resistance
Using the power formula, we can express the powers of the bulbs in terms of their resistances:
\[ P_A = \frac{V^2}{R_A} \]
\[ P_B = \frac{V^2}{R_B} \]
Since \( P_A > P_B \), we can substitute the power expressions:
\[ \frac{V^2}{R_A} > \frac{V^2}{R_B} \]

**Hint:** Since voltage is constant for both bulbs, you can simplify the inequality.

### Step 5: Simplify the inequality
By canceling \( V^2 \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{1}{R_A} > \frac{1}{R_B} \]
This implies:
\[ R_B > R_A \]

**Hint:** An inverse relationship indicates that if one quantity increases, the other must decrease.

### Step 6: Conclusion
Since \( R_B > R_A \), we conclude that bulb B has a greater resistance than bulb A. Therefore, the bulb that is brighter (bulb A) has a lower resistance.

**Final Answer:** \( R_B > R_A \)