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 ammeter shows maximum deflection.

 Given that,

Resistance of the galvanometer is Rg = 100 Ω .

Maximum deflection current in the galvanometer is, Ig=100 μA .

Shunt resistance connected in parallel, S=0.1 Ω .

The minimum current I in the circuit must be determined in order for the ammeter to show maximum deflection. A shunt resistance is connected in parallel to the galvanometer resistance to convert a galvanometer to an ammeter. Both have equal potential differences because they are connected in parallel, and the majority of the current passes through shunt resistance. IgRg represents the potential difference across the galvanometer.

The Potential difference across the shunt resistance is (I−Ig)S .

As they are connected in parallel, IgRg  = (I−Ig)S

Substituting the values we get,

⇒ 100×10−6×100 = (I−100×10−6)×0.1

⇒ 0.01 = (I−100×10−6) × 0.1

Dividing both sides by 0.1

⇒ 0.01=(I−100×10−6) ×0 .1

⇒ I = 0.1 + 10−4

⇒ I = 0.1001A

Therefore, the minimum current required in the circuit for maximum ammeter deflection is, ∴ I=100.1 mA

Getting Info...
Cookie Consent
We serve cookies on this site to analyze traffic, remember your preferences, and optimize your experience.
It seems there is something wrong with your internet connection. Please connect to the internet and start browsing again.
AdBlock Detected!
We have detected that you are using adblocking plugin in your browser.
The revenue we earn by the advertisements is used to manage this website, we request you to whitelist our website in your adblocking plugin.
Site is Blocked
Sorry! This site is not available in your country.