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Old 6th July 2007, 01:31 PM   #11
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Join Date: Mar 2001
Location: Sydney
The impedance of the cap is determined by 1/(2*pi*f*c), where pi=3.141, f=50Hz and c=capacitance in farads (you need to divide the 0.33uf by 1000000). For 0.33uF, this produces an impedance of about 9.6k. Calculate the current drawn by the relay coil (24v coil divided by whatever the coil resistance is), then use ohms law to calculate the voltage drop across the cap. By changing the value of the cap it changes the impedance and hence changes the voltage drop for a given current.
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Old 6th July 2007, 01:39 PM   #12
jarthel is offline jarthel  
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Quote:
Originally posted by smithy666
The impedance of the cap is determined by 1/(2*pi*f*c), where pi=3.141, f=50Hz and c=capacitance in farads (you need to divide the 0.33uf by 1000000). For 0.33uF, this produces an impedance of about 9.6k. Calculate the current drawn by the relay coil (24v coil divided by whatever the coil resistance is), then use ohms law to calculate the voltage drop across the cap. By changing the value of the cap it changes the impedance and hence changes the voltage drop for a given current.

I assume the 9.6K value is then paralleled to the 2x 470K to get the final resistance equivalent?

why not just use resistance to drop the voltage?

thanks again
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Old 7th July 2007, 01:39 PM   #13
mzzj is offline mzzj  
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Quote:
Originally posted by jarthel



I assume the 9.6K value is then paralleled to the 2x 470K to get the final resistance equivalent?

why not just use resistance to drop the voltage?

thanks again
Using only resistor would result pretty high power dissipation for small circuit like that (around 5W for 230v operation)
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