If voltage across a bulb rated 220 V, 160 W drops by 3% of its rated value, the percentage of the rated value by which the power would decrease is

Options

A. 0.2
B. 0.025
C. 0.06
D. 0.1

Answer

To solve the problem step by step, we need to find out how much the power decreases when the voltage drops by 3%.

### Step 1: Understand the relationship between power and voltage
The power \( P \) consumed by a bulb is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the bulb and \( R \) is the resistance. Since the resistance remains constant, power is directly proportional to the square of the voltage.

### Step 2: Calculate the new voltage after the drop
The rated voltage of the bulb is \( V = 220 \, \text{V} \). A drop of 3% means the new voltage \( V' \) can be calculated as follows:
\[ V' = V - 0.03 \times V = V(1 - 0.03) = 220 \times 0.97 = 213.4 \, \text{V} \]

### Step 3: Calculate the initial power
The rated power of the bulb is \( P = 160 \, \text{W} \).

### Step 4: Calculate the new power with the reduced voltage
Using the formula for power:
\[ P' = \frac{(V')^2}{R} \]
We can express \( P' \) in terms of \( P \):
\[ P' = \frac{(213.4)^2}{R} \]
Since we know \( P = \frac{(220)^2}{R} \), we can relate \( P' \) to \( P \):
\[ P' = P \left( \frac{(213.4)^2}{(220)^2} \right) \]

### Step 5: Calculate the ratio of the new power to the original power
Now we calculate the ratio:
\[ \frac{P'}{P} = \frac{(213.4)^2}{(220)^2} \]
Calculating this gives:
\[ \frac{P'}{P} = \left( \frac{213.4}{220} \right)^2 \approx (0.97)^2 \approx 0.9409 \]

### Step 6: Calculate the decrease in power
The decrease in power can be found as:
\[ \Delta P = P - P' = P \left( 1 - \frac{P'}{P} \right) = P \left( 1 - 0.9409 \right) \approx P \times 0.0591 \]

### Step 7: Calculate the percentage decrease in power
To find the percentage decrease in power:
\[ \text{Percentage decrease} = \left( \frac{\Delta P}{P} \right) \times 100 = 0.0591 \times 100 \approx 5.91\% \]

### Final Answer
The percentage of the rated value by which the power would decrease is approximately **5.91%**.

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