Question:
The heat generated in a circuit is given by \(Q=I^{2}Rt\). If the percentage error in the measurement of \(Q\) is \(1\%\), \(R\) is \(2\%\), and \(t\) is \(1\%\), then the percentage error in the measurement of current \(I\) is
The heat generated in a circuit is given by \(Q=I^{2}Rt\). If the percentage error in the measurement of \(Q\) is \(1\%\), \(R\) is \(2\%\), and \(t\) is \(1\%\), then the percentage error in the measurement of current \(I\) is
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Write the current in terms of Q, R and t, then add the fractional errors multiplied by the powers. The power of Q in the expression for current is one half.
Updated On: Nov 12, 2026
- \(3\%\)
- \(2\%\)
- \(4\%\)
- \(5\%\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Rearranging \(Q=I^{2}Rt\) gives \(I=\left(\dfrac{Q}{Rt}\right)^{1/2}\). In a product or quotient the maximum fractional errors add, each multiplied by its power.
Step 2: Write the error relation
\(\dfrac{\Delta I}{I}=\dfrac{1}{2}\left(\dfrac{\Delta Q}{Q}+\dfrac{\Delta R}{R}+\dfrac{\Delta t}{t}\right)\)
Step 3: Calculation
\(\dfrac{\Delta I}{I}\times100=\dfrac{1}{2}(1+2+1)\%=2\%\)
Final Answer:
The percentage error in the current is 2 percent. \[ \boxed{\text{(B) }2\%} \]
Rearranging \(Q=I^{2}Rt\) gives \(I=\left(\dfrac{Q}{Rt}\right)^{1/2}\). In a product or quotient the maximum fractional errors add, each multiplied by its power.
Step 2: Write the error relation
\(\dfrac{\Delta I}{I}=\dfrac{1}{2}\left(\dfrac{\Delta Q}{Q}+\dfrac{\Delta R}{R}+\dfrac{\Delta t}{t}\right)\)
Step 3: Calculation
\(\dfrac{\Delta I}{I}\times100=\dfrac{1}{2}(1+2+1)\%=2\%\)
Final Answer:
The percentage error in the current is 2 percent. \[ \boxed{\text{(B) }2\%} \]
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