Multiplier & Shunt

Multiplier:

A small galvanometer which is used for measuring small currents 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 a voltmeter.

• V = V_g + V_m
• V = Voltage across the whole circuit
• V_g = voltage across the galvanometer
• V_m = voltage across the multiplier
• R_g = resistance across the galvanometer
• R_m = resistance across the multiplier

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

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

Example 8.1.1:

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

Solution 

Galvanometer current = 50 mA = 0.05 A

Voltage across galvanometer V_g = IR_g = 0.05 × 6 = 0.30 V

Voltage across the whole circuit, V = V_g + V_m

Voltage across multiplier, V_m = V – V_g

V_m  = 60 – 0.30 = 59.7 V

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

 R_m =\( \frac {59.7 \times 6}{0.30}\)

R_m = \( \frac {358.2}{0.30}\)

 R_m = 1194 Ω

Shunt:

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

• R_g = resistance of the galvanometer
• I_g = current in the galvanometer
• R_s = resistance of the shunt
• I_s = current in the shunt

From the diagram

I = I_g + I_s

Since the same p.d passes through them, being parallel:

V_g =  V_s

I_gR_g = I_sR_s

Example 8.1.2:

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

Solution

I_g = 100 mA = 0.1 A

R_g = 3 Ω

I = I_g + I_s

5 = 0.1 + I_s

5 – 0.1 = I_s 

I_s = 4.9 A

V_g =  V_s 

I_g × R_g = I_s × R_s

R_s  = \( \frac{0.1 \: \times \: 3}{4.9} \)

= 0.06 Ω