Question

Discuss the conversion of galvanometer into an ammeter and also a voltmeter.

Answer 1

1. The ammeter must offer low resistance, no change in current.
2. Galvanometer is converted to an ammeter by connecting low resistance in parallel.
3. Low resistance – Shunt resistance.
4. I – current in the circuit.
5. I_g – Galvanometer current.
6. R_g – Galvanometer resistance.
7. Current through the shunt, I_s = I – I_g
8. `"V"_"galvanometer" = "V"_"shunt"`

I_g R_g = (I – I_g)S

S = `("I"_"g" "R"_"g")/("I - I"_"g")`    .....(1)

`"I"_"g" = "S"/("S" + "R"_"g")"I"`

I_g α I

Ammeter

The deflection in the galvanometer is proportional to current θ α I_g

`1/"R"_"eff" = 1/"R"_"g" + 1/"s"`

`1/"R"_"eff" = ("R"_"g""S")/("R"_"g" + "S") = "R"_"a"`

R_a is very low.

An ideal ammeter has zero resistance.

Galvanometer to a voltmeter:

1. The voltmeter must have high resistance.
2. A galvanometer is converted to a voltmeter by connecting high resistance in series.
3. R_g – Galvanometer resistance.
4. I_g – Galvanometer current.
5. R_h – High resistance
6. Since it is connected in series, the current is the same.
   
   Voltmeter
   i.e. I = I_g
   `"I"_"g" = "Potential difference"/"total resistance"`
   R_v = R_g + R_h
   `"I"_"g" = "V"/("R"_"g" + "R"_"h")`
   `"R"_"h" = "V"/"I"_"g" - "R"_"g"`
   ∴I_g α V
7. An ideal voltmeter has infinite resistance.