The heat generated in a circuit is given by `Q = I^(2) Rt` , where `I` is current , `R` is resistance , and `t` is time . If the percentage errors in measuring `I , R , and t are 2% , 1% , and 1%` , respectively , then the maximum error in measuring heat will be

Options

A. `2 %`
B. `3%`
C. `4%`
D. `6%`

Explanation

To find the maximum error in measuring heat generated in a circuit, we start with the formula for heat:

\[ Q = I^2 R t \]

Where:
- \( I \) is the current,
- \( R \) is the resistance,
- \( t \) is the time.

### Step 1: Identify the powers of each variable in the formula
In the equation \( Q = I^2 R t \):
- The power of \( I \) is 2,
- The power of \( R \) is 1,
- The power of \( t \) is 1.

### Step 2: Write the formula for maximum relative error
The formula for the maximum relative error in a product or power of variables is given by:

\[ \frac{\Delta Q}{Q} = a \frac{\Delta I}{I} + b \frac{\Delta R}{R} + c \frac{\Delta t}{t} \]

Where:
- \( a \), \( b \), and \( c \) are the powers of \( I \), \( R \), and \( t \) respectively.

### Step 3: Substitute the powers and the percentage errors
From the problem, we know:
- \( \Delta I/I = 2\% \) (current error),
- \( \Delta R/R = 1\% \) (resistance error),
- \( \Delta t/t = 1\% \) (time error).

Substituting the values into the formula:

\[ \frac{\Delta Q}{Q} = 2 \cdot \frac{\Delta I}{I} + 1 \cdot \frac{\Delta R}{R} + 1 \cdot \frac{\Delta t}{t} \]

This becomes:

\[ \frac{\Delta Q}{Q} = 2 \cdot 0.02 + 1 \cdot 0.01 + 1 \cdot 0.01 \]

### Step 4: Calculate the maximum relative error
Calculating each term:

\[ \frac{\Delta Q}{Q} = 0.04 + 0.01 + 0.01 = 0.06 \]

### Step 5: Convert relative error to percentage
To convert the relative error to a percentage, multiply by 100:

\[ \text{Percentage Error in } Q = 0.06 \times 100\% = 6\% \]

### Final Answer
The maximum error in measuring heat \( Q \) is **6%**.

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