A moving coil galvanometer of resistance 100 - is used as an ammeter using a resistance 0-1 - The maximum deflection current in the galvanometer is 100 - A- Find the minimum current in the circuit- so that the ammeter shows maximum deflection A- 100-1 mAB- 1000-1 mAC- 10-01 mAD- 1-01 mA

The correct option is A 100.1 mA nbsp; V_ab=i_gG=(i i_g)S ∴ i= ( 1+GS )i_g Substituting the values, we get i= ( 1+1000.1 )× 100× 10^ 6 or i=100.1 mA ...

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Standard XII

Physics

Galvanometer

A moving coil...

Question

A moving coil galvanometer of resistance $100 Ω$ is used as an ammeter using a resistance $0.1 Ω$ . The maximum deflection current in the galvanometer is $100 μ A$ . Find the minimum current in the circuit, so that the ammeter shows maximum deflection

100.1 mA

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1000.1 mA

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10.01 mA

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1.01 mA

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Solution

The correct option is A $100.1 mA$

$V_{a b} = i_{g} G = (i - i_{g}) S$
$∴ i = (1 + \frac{G}{S}) i_{g}$

Substituting the values, we get
$i = (1 + \frac{100}{0.1}) \times 100 \times 10^{- 6}$
or $i = 100.1 mA$

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A moving coil galvanometer of resistance 100 Ω is used as an ammeter using a resistance 0.1 Ω. The maximum deflection current in the galvanometer is 100 μA. Find the minimum current in the circuit, so that the ammeter shows maximum deflection

The correct option is A 100.1 mA Vab=igG=(i−ig)S ∴i=(1+GS)ig Substituting the values, we get i=(1+1000.1)×100×10−6 or i=100.1 mA

A moving coil galvanometer of resistance $100 Ω$ is used as an ammeter using a resistance $0.1 Ω$ . The maximum deflection current in the galvanometer is $100 μ A$ . Find the minimum current in the circuit, so that the ammeter shows maximum deflection

The correct option is A $100.1 mA$

$V_{a b} = i_{g} G = (i - i_{g}) S$
$∴ i = (1 + \frac{G}{S}) i_{g}$

Substituting the values, we get
$i = (1 + \frac{100}{0.1}) \times 100 \times 10^{- 6}$
or $i = 100.1 mA$