The voltage across a lamp is `(6.0pm0.1)V` and the current flowing through it is `(4.0pm0.2)A`. Find the power consumed with maximum permissible error in it.

Answer

To find the power consumed by the lamp and the maximum permissible error in it, we can follow these steps:

### Step 1: Understand the formula for power
The power \( P \) consumed by an electrical device is given by the formula:
\[ P = V \times I \]
where \( V \) is the voltage and \( I \) is the current.

### Step 2: Identify the values and their uncertainties
From the problem, we have:
- Voltage \( V = 6.0 \, \text{V} \) with an uncertainty of \( \pm 0.1 \, \text{V} \)
- Current \( I = 4.0 \, \text{A} \) with an uncertainty of \( \pm 0.2 \, \text{A} \)

### Step 3: Calculate the nominal power
Using the values of voltage and current, we calculate the nominal power:
\[ P = 6.0 \, \text{V} \times 4.0 \, \text{A} = 24.0 \, \text{W} \]

### Step 4: Calculate the relative errors
To find the maximum permissible error in power, we need to calculate the relative errors in voltage and current:
- Relative error in voltage \( \left( \frac{\Delta V}{V} \right) \):
\[ \frac{\Delta V}{V} = \frac{0.1}{6.0} \approx 0.01667 \]

- Relative error in current \( \left( \frac{\Delta I}{I} \right) \):
\[ \frac{\Delta I}{I} = \frac{0.2}{4.0} = 0.05 \]

### Step 5: Combine the relative errors
For multiplication, the relative errors add up:
\[ \frac{\Delta P}{P} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]
Substituting the values:
\[ \frac{\Delta P}{P} = 0.01667 + 0.05 = 0.06667 \]

### Step 6: Calculate the absolute error in power
Now, we can find the absolute error \( \Delta P \):
\[ \Delta P = P \times \frac{\Delta P}{P} = 24.0 \times 0.06667 \approx 1.6 \, \text{W} \]

### Step 7: State the final result
Thus, the power consumed by the lamp is:
\[ P = 24.0 \pm 1.6 \, \text{W} \]