Question:
The voltage across a lamp is \((6.0\pm 0.3)\) volt and the current passing through it is \((4.0\pm 0.1)\) ampere. The power consumed in watt will be (using percentage error)
The voltage across a lamp is \((6.0\pm 0.3)\) volt and the current passing through it is \((4.0\pm 0.1)\) ampere. The power consumed in watt will be (using percentage error)
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Multiply V and I; percentage errors add for a product.
Updated On: Oct 1, 2026
- \((24.0\pm 1.8)\)
- \((24.0\pm 0.75)\)
- \((22.0\pm 0.4)\)
- \((18.0\pm 0.4)\)
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The Correct Option is A
Solution and Explanation
Step 1: Power:
\(P=VI=6.0\times4.0=24.0\) W.
Step 2: Percentage Errors:
\(\dfrac{\Delta V}V\times100=\dfrac{0.3}{6.0}\times100=5\%\). \(\dfrac{\Delta I}I\times100=\dfrac{0.1}{4.0}\times100=2.5\%\).
Step 3: Add for a Product:
For a product, \(\dfrac{\Delta P}P=\dfrac{\Delta V}V+\dfrac{\Delta I}I=5\%+2.5\%=7.5\%\).
Step 4: Absolute Error:
\(\Delta P=0.075\times24.0=1.8\) W. So \(P=(24.0\pm1.8)\) W, which is option (A).
Final Answer:
The power is \((24.0\pm1.8)\) W, option (A). \[ \boxed{\text{(A) } (24.0\pm1.8)} \]
\(P=VI=6.0\times4.0=24.0\) W.
Step 2: Percentage Errors:
\(\dfrac{\Delta V}V\times100=\dfrac{0.3}{6.0}\times100=5\%\). \(\dfrac{\Delta I}I\times100=\dfrac{0.1}{4.0}\times100=2.5\%\).
Step 3: Add for a Product:
For a product, \(\dfrac{\Delta P}P=\dfrac{\Delta V}V+\dfrac{\Delta I}I=5\%+2.5\%=7.5\%\).
Step 4: Absolute Error:
\(\Delta P=0.075\times24.0=1.8\) W. So \(P=(24.0\pm1.8)\) W, which is option (A).
Final Answer:
The power is \((24.0\pm1.8)\) W, option (A). \[ \boxed{\text{(A) } (24.0\pm1.8)} \]
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