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I'm looking at this question in a design exercise.
By simple calculation the voltage is 9V, the impedance of the heater is 3ohms. So the current through R is 3A?
OR
Do I need to take into account the PWM out? Will that be producing a voltage of 3.3V in line with V out on the Mbed. So the voltage across the resistor is actually 6.7V?
The math is correct using the 9 V as a start, but understand that the current you calculate only represents instantaneous current through the heater coil. The average current is a more meaningful number, and to get that you do need to take the PWM into account and specifically the duty cycle. Consider that PWM ( meaning pulse width modulation) is controlling the coil heat by rapidly switching the current through the coil (and thus the voltage across it as well) on and off at a frequency of 1 kHz (according to your post). But the frequency is not as important as the duty cycle of the PWM waveform.
Consider the simplest cases... if the duty cycle of the PWM square wave is 100% (meaning its always on), then the PWM is effectively doing nothing, and you might as well just connect the heater to the 9 V supply. But lets say the square wave is an even 50% duty cycle. That means that the heater is on half the time and on half the time. Fortunately the averaging math for a square wave is pretty straightforward. The current 1 (as you calculated) 3 A for half the time and zero of the other half. So 50% of 3 A would be 1.5 A average current.
Unfortunately there is not enough information in the diagram or your description to determine the duty cycle of that PWM output. But at least you have a starting point.
