Quote:
Originally Posted by cdm1christopher
So I did not need the heater volts as they where always going to be high. So not sure why I was being asked for them? Ha.
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Some time ago at the start of the thread you asked about running a set marked 220V on UK mains. By measuring the heater voltages you established (i) that the set had not been modified for modern UK operation and (ii)that the heaters were being significantly overrun -more than double the accepted +/-10pc for voltage.
A valid point was also made that in series heater strings like this it is the current which is the primary parameter and instructions were given several times by me and others on how to measure this without needing to break the heater chain.
Instructions were also given on how to add a series resistor of appropriate value. Herald and Lawrence have both gone through the calculations with actual figures so that you know the value of the component necessary.
One (post 108) provided an option to deal only with the overrun heaters, (and,as an option, the HT circuit if necessary) the other (post 121) provided an option to deal with dropping the voltage for the whole current requirements of the set.
It may be worth re -reading both of those posts as they both provide solutions with resistor values necessary to make the heater voltages (and HT voltages if necessary ) correct for this set without adding a transformer, and without guessing at component vales.
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Throughout this we have been trying to explain
1. ohms law ( see my post 109) V=IR
2. Power (P) (measured in Watts) P=IV
These two equations form the basis of electricity and I'm going to have another go at trying to show how you can use them as I am determined that we should not have gone through 120 -odd posts without you having the chance to master ohms law.
The datasheet says the radio consumes nominally 40W at 220V. We know the house mains are 245V. Without doing any measuring at all can we work out the value of the series resistor necessary to operate the radio on the house mains?
We need to choose a resistor which will drop 25V across it. To work out R (from V=IR ) we need both the voltage drop across it V (25V) and the current flowing through it (I)
The resistor is will be in series with the radio and the current (I) is the same at any point in a series circuit so if we can find the current drawn by the radio we can do the s necessary to find R (because we will have two of the three terms in the equation)
For the radio, we don't have I but we do have P (40w)
we can re-arrange P=IV as I=P/V
so I (the current drawn by the radio)= 40/220 = 0.181A
we know that we want to drop 25V (245-220) across the reisistor and the current flowing through the dropping resistor should be 0.181A
So from ohms law rearrange V=IR for R : R=V/I
R= 25/0.181=138 ohms - pretty close to Lawrence's result.
Note that the above calculation is a gross simplification and 'quick and dirty' ;eg. the 40W figure quoted by the manufacturer is nominal only- for actual component values, use the sums and methdology which have been done in previous posts. These two sums were intended to illustrate ohms law, to show how calculations can be done so that guessing is unnecessary -and how by tackling the same problem in different ways you can 'prove' your answer is in the right ballpark.