A moving coil galvanometer (resistance 100 ohms) plays the role of ammeter via using resistance 0.1 ohms. If maximum deflection current is 100 mu-A, show minimum current in the circuit so that ammeter offers maximum deflection.

As per the given data, the following can be known:

• Resistance of galvanometer, Rg = 100 Ω
• Maximum deflection of the given current in galvanometer, Ig = 100 μA (Symbol: mu-A)
• Shunt resistance which is parallelly joined, S = 0.1 Ω

In the circuit, the minimum current (I) can be used to determine so that the ammeter shows maximum deflection. Shunt resistance has been attached parallelly to the galvanometer resistance in order to convert a galvanometer to an ammeter.

Simply, both of them have equal potential differences because they are parallelly connected, and the majority of the current passes through shunt resistance. I_gR_g represents the potential difference across the galvanometer.

Potential difference across shunt resistance = (I − I_g)S.

As they are connected in parallel, I_gRg  = (I−I_g)S

Replacing the values we get,

⇒ 100 × 10⁻⁶ × 100 = (I − 100 × 10⁻⁶) × 0.1

⇒ 0.01 = (I − 100 × 10⁻⁶) × 0.1

Now, after dividing both sides by 0.1

⇒ 0.01 = (I − 100 × 10⁻⁶) × 0.1

⇒ I = 0.1 + 10⁻⁴

⇒ I = 0.1001 A

Therefore, the minimum current required in the circuit for maximum ammeter deflection is, 

\(\therefore\) I = 100.1 mA

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