The meter element of a permanent-magnet moving coil instrument has a resistance of
5 Ω and requires 15 mA for full-scale deflection.
Calculate the resistance to be connected in parallel to enable the instrument
to read up to 1 A.
Solution
Given:
Meter resistance, Rm = 5 Ω
Full-scale meter current, Im = 15 mA = 0.015 A
Required full-scale current, I = 1 A
Let S be the value of shunt resistance in parallel with the meter.
For a shunted ammeter:
Im Rm = (I − Im) S
Therefore,
S = (Im Rm) / (I − Im)
= (0.015 × 5) / (1 − 0.015)
= 0.075 / 0.985 ≈ 0.076 Ω
Answer: Required shunt resistance ≈ 0.076 Ω.
A moving coil milliammeter with a resistance of 1.6 Ω is connected with a shunt of 0.228 Ω. What will be the current flowing through the instrument if it is connected in a circuit in which a current of 200 mA is flowing?
Solution
Given:
Rm = 1.6 Ω
S = 0.228 Ω
Total current, I = 200 mA = 0.2 A
For a shunted ammeter, current division is given by:
Im = I × S / (Rm + S)
So,
Im = 0.2 × 0.228 / (1.6 + 0.228)
= 0.0456 / 1.828 ≈ 0.02495 A ≈ 25 mA
Answer: Current through milliammeter ≈ 25 mA.
What should be the resistance of the moving coil of an ammeter which requires 2.5 mA for full-scale deflection so that it may be used with a shunt having a resistance of 0.0025 Ω for a range of 0–10 A?
Solution
Given:
Im = 2.5 mA = 0.0025 A
I = 10 A
S = 0.0025 Ω
For shunted ammeter:
Im Rm = (I − Im) S
Therefore,
Rm = (I − Im) S / Im
= (10 − 0.0025) × 0.0025 / 0.0025
≈ 10 × 0.0025 / 0.0025 = 10 Ω
Answer: Moving coil resistance Rm = 10 Ω.
A moving iron ammeter with a range of 0–1 A has an internal resistance of 50 mΩ and an inductance of 0.1 mH. To increase the range to 0–10 A for all operational frequencies, a shunt coil is connected. Find:
(i) The resistance of the shunt coil in mΩ.
(ii) The time constant of the shunt coil (in milliseconds) so that the range extension is valid
for all frequencies.
Solution
Given:
Original range: 0–1 A (meter current Im = 1 A)
New range: 0–10 A (total current I = 10 A)
Meter resistance, Rm = 50 mΩ = 0.05 Ω
Meter inductance, Lm = 0.1 mH = 0.0001 H
(i) Shunt resistance
At 10 A range: Im = 1 A, so shunt current Is = 10 − 1 = 9 A.
For DC distribution:
Im Rm = Is S
S = Im Rm / Is
= 1 × 0.05 / 9 ≈ 0.00556 Ω = 5.56 mΩ
(ii) Time constant for shunt
For frequency-independent ratio, the time constants must be equal:
Lm / Rm = Ls / S = Ï„
Ï„ = Lm / Rm = 0.0001 / 0.05 = 0.002 s = 2 ms
If required, shunt inductance:
Ls = Ï„ × S = 0.002 × 0.00556 ≈ 1.11 × 10−5 H ≈ 11.1 μH
Answers:
(i) Shunt resistance S ≈ 5.56 mΩ
(ii) Time constant Ï„ = 2 ms
In a moving coil galvanometer, the deflection becomes half of its original value when the galvanometer is shunted by a resistance of 20 Ω. Find the resistance of the galvanometer (G).
Solution
Let the original deflection be proportional to current I1, and the deflection with shunt
be proportional to I2.
Given that deflection becomes half:
I1 / I2 = 2 ⇒ m = 2
For a galvanometer converted to ammeter:
m = 1 + G / S
So,
2 = 1 + G / 20
G / 20 = 1 ⇒ G = 20 Ω
Answer: Galvanometer resistance G = 20 Ω.
A galvanometer has a full-scale current of 15 mA and a full-scale voltage of 750 mV. Find the value of shunt resistance required to convert this galvanometer into an ammeter of range 0–2.5 A.
Solution
Given:
Ig = 15 mA = 0.015 A
Vg = 750 mV = 0.75 V
Required range, I = 2.5 A
First, find galvanometer resistance G:
G = Vg / Ig = 0.75 / 0.015 = 50 Ω
In ammeter form, with shunt S:
Ig G = (I − Ig) S
S = Ig G / (I − Ig)
= (0.015 × 50) / (2.5 − 0.015)
= 0.75 / 2.485 ≈ 0.302 Ω
Answer: Required shunt resistance S ≈ 0.3 Ω.
The deflection of a galvanometer falls from 50 divisions to 20 divisions when a shunt of 12 Ω is connected across it. Find the resistance of the galvanometer.
Solution
Deflection ∝ current through galvanometer.
Without shunt: deflection = 50 divisions (current I1)
With shunt: deflection = 20 divisions (current I2)
m = I1 / I2 = 50 / 20 = 2.5
For shunted galvanometer:
m = 1 + G / S
So,
2.5 = 1 + G / 12
G / 12 = 1.5 ⇒ G = 12 × 1.5 = 18 Ω
Answer: Galvanometer resistance G = 18 Ω.
In an ammeter, 5% of the main current passes through the galvanometer. If the galvanometer resistance is G and the shunt resistance is S, find the relation between G and S.
Solution
Let total current be I.
Given: Ig = 5% of I = 0.05 I
Therefore,
m = I / Ig = 1 / 0.05 = 20
For a shunted galvanometer:
m = 1 + G / S
20 = 1 + G / S ⇒ G / S = 19 ⇒ S = G / 19
Answer: Relation between shunt and galvanometer resistance:
S = G / 19 (or G = 19 S).