Solution Electrical Numerical 11

Numerical 11: A moving coil ammeter, a thermal ammeter and a rectifier are connected in series with a resistor across a 110 V sinusoidal a.c. supply. The circuit has a resistance of 50Ω to current in one direction and, due to the rectifier, an infinite resistance to current in the reverse direction. Calculate: (i) The readings on the ammeters; (ii) The form and peak factors of the current wave.

Solution: (i) Max value of voltage: $\displaystyle \displaystyle \mathrm{V_m\ =\ \frac{V}{0.707}\ =\ \frac{110}{0.707}\ =\ 155.5\, V }$ 

Max Value of current: $\displaystyle \displaystyle \mathrm{I_m\ =\ \frac{V_m}{R}\ =\ \frac{115.5}{50}\ =\ 3.11\, A }$ 

During the +ve half cycle the current is proportional to the voltage and is sinusoidal (see figure). Thus average value of current over the +ve half cycle.

.

$\displaystyle \displaystyle \mathrm{I_{av}\ =\ 0.637\times I_m =\ 0.637\times 3.11 =\ 1.98\ A}$  

During -ve half cycle, the current is zero. But because of inertia of the moving system the moving coil ammeter reads the average value of the current over the whole cycle.

Therefore reading on the Moving coil ammeter: $\displaystyle \displaystyle \mathrm{\frac{1.98}{2} =\ 0.99\ A}$ 

The variation of the heating effect in the thermal ammeter is shown in the figure, the max power being $\displaystyle \mathrm{I_{m}^{2}R}$ where R is the resistance of the instrument.

The average heating effect over the +ve half cycle is $\displaystyle \mathrm{\frac{1}{2}I_{m}^{2}R}$  and since no heat is generated during the 2nd half cycle, it follows that the average heating effect over a complete cycle is $\displaystyle \mathrm{\frac{1}{4}I_{m}^{2}R}$

If I is the direct current which produce the same heating effect then,

$\displaystyle \mathrm{I^2R\ =\ \frac{1}{4}I_{m}^{2}R}$ 

$\displaystyle \mathrm{I\ =\ \frac{1}{2}I_{m}\ =\ \frac{3.11}{2} =\ 1.555\ A }$ 

i.e Reading on thermal ammeter =1.555 A

[ don't use $\displaystyle \mathrm{I\ =\ \frac{0.707\times 3.11}{2} =\ 1.1\ A }$ ]

(ii) Foam factor: $\displaystyle \mathrm{K_f\ =\ \frac{I}{I_{av}} =\ \frac{1.555}{0.99} =\ 1.57 }$ 

Peak factor: $\displaystyle \mathrm{K_P\ =\ \frac{I_m}{I} =\ \frac{3.11}{1.555} =\ 2.0 }$