10-07-2015, 12:06 PM
(This post was last modified: 10-07-2015, 12:08 PM by SurreyNick.)
If someone can spare a few seconds I’d like to just check ’ve done my calculations correctly
I could only get a 6v 60mA miniature filament bulb for the Ratio Meter, which means the 120 ohm resistor specified in the circuit is too high. The transformer drops the supply voltage to 9v and so I calculate I need a 51 ohm 1.4W resistor…
Voltage drop: 9v minus 6v = 3v
Resistor value: 3v divided by 0.06a = 50 ohm (round up to nearest standard value = 51 ohm)
Resistor power: 0.06a times 0.06a times 50 ohm = 0.18 watts
I haven’t done this enough times for it to become second nature, so just want to make sure I have done this right? I am assuming I don’t need to compensate for the fact the bulb is receiving an a.c. supply either. Is that right too?
Thanks
Nick
I could only get a 6v 60mA miniature filament bulb for the Ratio Meter, which means the 120 ohm resistor specified in the circuit is too high. The transformer drops the supply voltage to 9v and so I calculate I need a 51 ohm 1.4W resistor…
Voltage drop: 9v minus 6v = 3v
Resistor value: 3v divided by 0.06a = 50 ohm (round up to nearest standard value = 51 ohm)
Resistor power: 0.06a times 0.06a times 50 ohm = 0.18 watts
I haven’t done this enough times for it to become second nature, so just want to make sure I have done this right? I am assuming I don’t need to compensate for the fact the bulb is receiving an a.c. supply either. Is that right too?
Thanks
Nick
