Hi i am having some obstacles in building up my concept in dc circuit First of all, as we know, voltmeter is made up of a series circuit with both ends connected to two points of other circuits, while inside the series circuit, a galvanometer and resistor is connected in series. How the voltmeter measure the potential difference?? is it that the reading appears on the voltmeter directly equal to the potential difference across the built-in resistor with very very high resistance inside the voltmeter? Now for the circuit B, why will the reading of voltmeter be equal to emf of the cell?? supposedly there should be no potential difference across the built-in resistor in the voltmeter right? But for the circuit A, what should be the potential difference across the diode?? in this case there should be potential difference across the diode right? since the emf is having tendency to do work against the built-up opposing voltage in the diode. Now, what should the voltmeter read ? and the potential difference across the internal resistor should be what? Hope you can guide me.