Two electric bulbs rated 50 W and 100 V are glowing at full power, when used in parallel with a battery of emf 120 V and internal resistance 10 `Omega`. The maximum number of bulbs that can be connected in the circuit when glowing at full power, is

Options

A. a) 6
B. b) 4
C. c) 2
D. d) 8

Answer

To solve the problem of determining the maximum number of bulbs that can be connected in parallel while glowing at full power, we will follow these steps:

### Step 1: Calculate the resistance of each bulb
The resistance \( R \) of each bulb can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
where \( V \) is the voltage rating of the bulb and \( P \) is the power rating.

Given:
- Voltage rating \( V = 100 \, \text{V} \)
- Power rating \( P = 50 \, \text{W} \)

Substituting the values:
\[ R = \frac{100^2}{50} = \frac{10000}{50} = 200 \, \Omega \]

### Step 2: Calculate the equivalent resistance of the two bulbs in parallel
When two resistors \( R_1 \) and \( R_2 \) are in parallel, the equivalent resistance \( R_{eq} \) can be calculated using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
For two identical bulbs:
\[ \frac{1}{R_{eq}} = \frac{1}{200} + \frac{1}{200} = \frac{2}{200} = \frac{1}{100} \]
Thus,
\[ R_{eq} = 100 \, \Omega \]

### Step 3: Calculate the total resistance in the circuit
The total resistance \( R_{total} \) in the circuit includes the internal resistance of the battery \( r = 10 \, \Omega \) and the equivalent resistance of the bulbs:
\[ R_{total} = R_{eq} + r = 100 + 10 = 110 \, \Omega \]

### Step 4: Calculate the power of the circuit
The power \( P_{circuit} \) supplied by the battery can be calculated using the formula:
\[ P_{circuit} = \frac{V^2}{R_{total}} \]
where \( V \) is the EMF of the battery.

Given:
- EMF \( V = 120 \, \text{V} \)

Substituting the values:
\[ P_{circuit} = \frac{120^2}{110} = \frac{14400}{110} \approx 130.91 \, \text{W} \]

### Step 5: Calculate the maximum number of bulbs
To find the maximum number of bulbs \( n \) that can glow at full power, we divide the total power of the circuit by the power of one bulb:
\[ n = \frac{P_{circuit}}{P_{bulb}} = \frac{130.91}{50} \approx 2.618 \]
Since \( n \) must be a whole number, we take the integer part:
\[ n = 2 \]

### Conclusion
The maximum number of bulbs that can be connected in the circuit while glowing at full power is **2**.

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