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SS3: PHYSICS - 1ST TERM

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Multiplier

A small galvanometer which is used for measuring small current can be converted to a multiplier. This is achieved by connecting a high resistance (resistor) in series with the galvanometer in order to convert it to voltmeter

Screen Shot 2020 10 29 at 9.29.19 AM

V = Vg+ VR

V = Voltage across the whole circuit

Vg = voltage along the galvanometer

Vm = voltage across the multiplier

Rg = resistance across the galvanometer

Rm = resistance across the multiplier

Since galvanometer and multiplier are in the same series, the same amount of current flows through them

I = \( \frac {V_g}{R_g} = \frac {V_m}{R_m} = \frac {V_r}{R_m} \)

Example

A galvanometer of resistance 6Ω and full scale deflection (f.s.d) of 50MA is used to measure a P.d of 60V. Calculate the resistance of the multiplier.

Solution 

Galvanometer current = 50MA = 0.05A

Voltage across galvanometer Vg = IRg = 0.05 × 6 = 0.30V

Voltage across multipliers, V = Vg + Rm

    60 = 0.30 + Vm 

Vm  = 60 – 0.30 = 59.7V

The galvanometer and multiplier are in series, the same current passes through them;

\( \frac {V_g}{R_g} = \frac {V_m}{R_m}\)

\( \frac {0.3}{6} = \frac {59.7}{R_m}\)

 Rm =\( \frac {59.7 \times 6}{0.30}\)

Rm = \( \frac {35.82}{0.30}\)

 Rm = 1191Ω

Shunt

A low resistance (resistor) connected in parallel across a galvanometer to convert it to ammeter is called shunt.

Screen Shot 2020 10 29 at 9.37.45 AM

Rg = resistance of the galvanometer

Ig = current in the galvanometer

Rs = resistance of the shunt

Is = current in the shunt

From the diagram

I = Ig + Is

Since the same P.d passes through them, being parallel,

Vg =  Vs

IgRg = IsRs

Example:

A galvanometer of resistance 3Ω has an f.s.d for a current of 100MA. Calculate the resistance of the shunt required to convert it to a 5A ammeter

Solution

Ig = 100MA, Rg =3Ω

I = Ig + Is

5 = 0.1 + Is

5 – 0.1 = Is 

Is = 4.9A

Vg =  Vs 

Ig x Rg = Is x Rs

Rs  = \( \frac{0.1 \times 3 }{49} \)

=0.06Ω

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