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Byju's Answer

Standard XII

Physics

Galvanometer

In the follow...

Question

In the following circuit a $10 m$ long potentiometer wire with resistance $1.2 o h m / m$ , a resistance $R_{1}$ and an accumulator of emf $2$ $V$ are connected in series. When the emf of thermocouple is $2.4$ $m V$ then the deflection in galvanometer is zero. The current supplied by the accumulator will be :

4 \times 10^{- 4} A

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8 \times 10^{- 4} A

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4 \times 10^{- 3} A

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8 \times 10^{- 3} A

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Solution

The correct option is D $4 \times 10^{- 4} A$
WE know that $E = x l = i ρ l$
where symbols have their usual meaning.
According to question:
$\Rightarrow i = \frac{E}{ρ l} = \frac{E}{ρ l} = \frac{2.4 \times 10^{- 3}}{1.2 \times 5} = 4 \times 10^{- 4} A$ .

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In the following circuit a 10 m long potentiometer wire with resistance 1.2ohm/m, a resistance R1 and an accumulator of emf 2 V are connected in series. When the emf of thermocouple is 2.4 mVthen the deflection in galvanometer is zero. The current supplied by the accumulator will be :

The correct option is D 4×10−4AWE know that E=xl=iρlwhere symbols have their usual meaning.According to question:⇒i=Eρl=Eρl=2.4×10−31.2×5=4×10−4 A.

The correct option is D $4 \times 10^{- 4} A$
WE know that $E = x l = i ρ l$
where symbols have their usual meaning.
According to question:
$\Rightarrow i = \frac{E}{ρ l} = \frac{E}{ρ l} = \frac{2.4 \times 10^{- 3}}{1.2 \times 5} = 4 \times 10^{- 4} A$ .