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Byju's Answer

Standard VIII

Physics

Electric Power

A 1Ω resist...

Question

A $1 Ω$ resistance in series with an ammeter is balanced by $75 c m$ of potentiometer wire. A standard cell of emf $1.02 V$ is balanced by $50 c m$ . The ammeter shows a reading of $1.5 A$ . Then the error in ammeter reading is:

0.03 A

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3 A

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1.3 A

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0.3 A

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Solution

The correct option is A $0.03 A$
Since $50 c m$ of potentiometer wire balances $1.02 V$ of battery, the voltage per unit length of wire is $\frac{1.02}{50} V / c m$ .
Thus potential across $1 Ω$ resistance is $\frac{1.02}{50} \times 75 V = 1.53 V$
Thus current in the resistor= $1.53 A$
Reading in the ammeter= $1.5 A$
Thus error in reading= $0.03 A$

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A 1Ω resistance in series with an ammeter is balanced by 75 cm of potentiometer wire. A standard cell of emf 1.02 V is balanced by 50 cm. The ammeter shows a reading of 1.5 A. Then the error in ammeter reading is:

The correct option is A 0.03 ASince 50cm of potentiometer wire balances 1.02V of battery, the voltage per unit length of wire is 1.0250 V/cm.Thus potential across 1Ω resistance is 1.0250×75 V=1.53 VThus current in the resistor= 1.53 AReading in the ammeter= 1.5 AThus error in reading= 0.03 A

A $1 Ω$ resistance in series with an ammeter is balanced by $75 c m$ of potentiometer wire. A standard cell of emf $1.02 V$ is balanced by $50 c m$ . The ammeter shows a reading of $1.5 A$ . Then the error in ammeter reading is: