`40` electric bulb are connected in series across a `220 V` supply. After one bulb is fused the remaining `39` are connected again in series across the same supply. The illumination will be

Options

A. more with `40` bulbs than with `39`
B. more with `39` bulbs than with `40`
C. equal in both the cases
D. in the ratio `40^(2) : 39^(2)`

Answer

To solve the problem, we need to analyze the situation step by step:

### Step 1: Understand the Initial Setup
We have 40 electric bulbs connected in series across a 220 V supply. When connected in series, the total resistance \( R_t \) of the circuit is the sum of the individual resistances of the bulbs.

### Step 2: Calculate the Total Resistance with 40 Bulbs
Let the resistance of each bulb be \( R \). Therefore, the total resistance with 40 bulbs connected in series is:
\[ R_t = 40R \]

### Step 3: Calculate the Current with 40 Bulbs
Using Ohm's Law, the current \( I \) through the circuit can be calculated as:
\[ I = \frac{V}{R_t} = \frac{220}{40R} = \frac{5.5}{R} \]

### Step 4: Calculate the Power (Illumination) with 40 Bulbs
The power \( P \) consumed by the circuit, which is related to illumination, can be calculated using the formula:
\[ P = I^2 R_t = I^2 \times 40R \]
Substituting the value of \( I \):
\[ P = \left(\frac{5.5}{R}\right)^2 \times 40R = \frac{30.25}{R^2} \times 40R = \frac{1210}{R} \]

### Step 5: Analyze the Situation After One Bulb Fuses
When one bulb fuses, we have 39 bulbs left. The new total resistance \( R_t' \) is:
\[ R_t' = 39R \]

### Step 6: Calculate the Current with 39 Bulbs
The new current \( I' \) through the circuit is:
\[ I' = \frac{V}{R_t'} = \frac{220}{39R} = \frac{220}{39R} \]

### Step 7: Calculate the Power (Illumination) with 39 Bulbs
The power consumed by the circuit with 39 bulbs is:
\[ P' = I'^2 R_t' = \left(\frac{220}{39R}\right)^2 \times 39R = \frac{48400}{1521R^2} \times 39R = \frac{1885200}{1521R} \]

### Step 8: Compare the Illumination
To compare the illumination with 40 bulbs and 39 bulbs, we can find the ratio of the powers:
\[ \text{Ratio} = \frac{P'}{P} = \frac{\frac{1885200}{1521R}}{\frac{1210}{R}} = \frac{1885200}{1521} \times \frac{R}{1210} \]
This simplifies to:
\[ \text{Ratio} = \frac{1885200}{1521 \times 1210} \]

### Conclusion
From the calculations, we can conclude that the illumination with 39 bulbs will be less than with 40 bulbs. The illumination will decrease when one bulb is fused.