A moving coil ammeter, a thermal ammeter and a rectifier are connected in series with a resistor across a 110 V sinusoidal ac supply. The circuit has a resistance of 50 Ω to current in one direction and, due to rectifier , an infinite resistance to current in the reverse direction. Calculate
a) The readings on ammeter
b) The form and peak factor of current wave
If you are noticing some error in problems kindly comment below.Thanks
Given
V = 110V
R = 50 Ω
To find
a) The readings on ammeter
b) The form and peak factor of current wave
Solution
a) We know that
V = 0.707 x Vmax
So Maximum voltage,
Vmax = V / 0.707
= 110 / 0.707
= 155.5 V
As per Ohm's Law Maximum Current
Imax = Vmax / R
= 155.5 / 50
= 3.11 A
As we know that moving coil ammeter, a thermal ammeter and a rectifier connected in series with a resistor across an ac supply.
So during positive half cycle average value of current
Iavg = 0.637 x Imax
= 0.637 x 3.11
= 1.98 A
During negative half cycle the current value is zero. But due to inertia of moving parts the moving coil ammeter will show the average value of current through the whole cycle
Reading on the moving coil ammeter = Iavg / 2
= 1.98 / 2
= 0.99 A
Reading on ammeter is proportional to heating effect
So during positive half cycle average heating effect
Havg = (1/2) x I2max x R
During negative half cycle the heat dissipated is zero. But the thermal ammeter will show the average value of heat through the whole cycle
Reading on thermal ammeter = [(1/2) x I2max x R] /2
= (1/4) x I2max x R -------------------------------(1)
Total heating effect = I2 x R ------------------------------------(2)
Equating (1) & (2)
I2 x R = (1/4) x I2max x R
I = Imax / 2
= 3.11 / 2
= 1.55 A
Reading on thermal ammeter = 1.55 A
b) Form factor kf = I / Iavg
= 1.55 / 0.99
= 1.57
Peak factor kp = Imax / I
= 3.11 / 1.55
a) The readings on ammeter
b) The form and peak factor of current wave
If you are noticing some error in problems kindly comment below.Thanks
V = 110V
R = 50 Ω
To find
a) The readings on ammeter
b) The form and peak factor of current wave
Solution
a) We know that
V = 0.707 x Vmax
So Maximum voltage,
Vmax = V / 0.707
= 110 / 0.707
= 155.5 V
As per Ohm's Law Maximum Current
Imax = Vmax / R
= 155.5 / 50
= 3.11 A
As we know that moving coil ammeter, a thermal ammeter and a rectifier connected in series with a resistor across an ac supply.
So during positive half cycle average value of current
Iavg = 0.637 x Imax
= 0.637 x 3.11
= 1.98 A
During negative half cycle the current value is zero. But due to inertia of moving parts the moving coil ammeter will show the average value of current through the whole cycle
Reading on the moving coil ammeter = Iavg / 2
= 1.98 / 2
= 0.99 A
Reading on ammeter is proportional to heating effect
So during positive half cycle average heating effect
Havg = (1/2) x I2max x R
During negative half cycle the heat dissipated is zero. But the thermal ammeter will show the average value of heat through the whole cycle
Reading on thermal ammeter = [(1/2) x I2max x R] /2
= (1/4) x I2max x R -------------------------------(1)
Total heating effect = I2 x R ------------------------------------(2)
Equating (1) & (2)
I2 x R = (1/4) x I2max x R
I = Imax / 2
= 3.11 / 2
= 1.55 A
Reading on thermal ammeter = 1.55 A
b) Form factor kf = I / Iavg
= 1.55 / 0.99
= 1.57
Peak factor kp = Imax / I
= 3.11 / 1.55
= 2.0
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