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Standard XII

Physics

Ammeter

A 1Ω resistan...

Question

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

0.002 A

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0.03 A

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1.01 A

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no error

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Solution

The correct option is B 0.03 A
$i p = \frac{V}{l} = \frac{1.02}{50}$
Voltage across ammeter = $\frac{V}{l} l_{1}$
$= \frac{1.02}{50} \times 75 = \frac{1.02}{2} \times 3 = 0.51 \times 3$
The true current $I_{A}^{1} = \frac{V_{A}}{R} = \frac{1.53}{1} I_{A}^{1} = 1.53 A E r r o r = I_{A}^{1} - I_{A} = 1.53 - 1.5 = 0.03 A$

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

The correct option is B 0.03 Aip=Vl=1.0250 Voltage across ammeter = Vll1 =1.0250×75=1.022×3=0.51×3 The true current I1A=VAR=1.531I1A=1.53AError=I1A−IA=1.53−1.5=0.03A

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