Electric bulb `50 W - 100 V` glowing at full power are to be used in parallel with battery `120 V`, `10 Omega`. Maximum number of bulbs that can be connected so that they glow in full power is

Options

A. 2
B. 8
C. 4
D. 6

Answer

To solve the problem step by step, we need to determine the maximum number of bulbs that can be connected in parallel to a battery while ensuring that each bulb operates at its rated power.

### Step 1: Calculate the resistance of one bulb
Given:
- Power (P) = 50 W
- Voltage (V) = 100 V

Using the formula for power:
\[ P = \frac{V^2}{R} \]

We can rearrange this to find the resistance (R):
\[ R = \frac{V^2}{P} \]

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

### Step 2: Determine the current through each bulb
Using the power formula again, we can find the current (I) through each bulb:
\[ P = V \cdot I \]
Rearranging gives:
\[ I = \frac{P}{V} \]

Substituting the values:
\[ I = \frac{50}{100} = 0.5 \, A \]

### Step 3: Set up the circuit with n bulbs in parallel
When n bulbs are connected in parallel, the total current (I_total) supplied by the battery is:
\[ I_{total} = n \cdot I \]
Where I is the current through one bulb.

### Step 4: Apply Kirchhoff's Voltage Law (KVL)
The total voltage from the battery is given as 120 V, and the internal resistance of the battery is 10 Ω. The equivalent resistance (R_eq) of n bulbs in parallel is:
\[ R_{eq} = \frac{R}{n} = \frac{200}{n} \]

Using KVL:
\[ 120 = I_{total} \cdot (R_{eq} + R_{internal}) \]
Substituting the values:
\[ 120 = n \cdot 0.5 \cdot \left(\frac{200}{n} + 10\right) \]

### Step 5: Simplify the equation
Expanding the equation:
\[ 120 = 0.5n \left(\frac{200}{n} + 10\right) \]
\[ 120 = 0.5n \cdot \frac{200 + 10n}{n} \]
\[ 120 = 0.5(200 + 10n) \]
\[ 240 = 200 + 10n \]

### Step 6: Solve for n
Rearranging gives:
\[ 240 - 200 = 10n \]
\[ 40 = 10n \]
\[ n = 4 \]

### Conclusion
The maximum number of bulbs that can be connected in parallel to the battery while ensuring they glow at full power is **4**.