Till the time none of the diodes is conducting , voltage across both of them will be 12 V but after germanium starts conducting voltage across it will be fixed at 0.3V .Assuming ideal components, you will never have 12V across the diodes in the conducting direction.
No, that violates circuit theory. That is not how voltages develop in a circuit.voltage across both of them will be 12 V
There is no specific answer to this question. How voltages develop across components in circuits is a very interesting topic, but it is out of the scope of circuit theory. In circuit theory, all that matters is Kirchhoff's laws, Ohm's law and v-i relationships of the components.Till the time the current is not flowing through any of the diodes , what is the potential difference across the two diodes ?
In the original diagram where both the diodes are in forward connection, you have two probable scenarios:How do we know what is the potential applied across the two diodes so as to decide which one is conducting ?
I don't understand . When none of the diodes is conducting , there is no current in the circuit which means no potential difference across the resistor . This means complete potential difference is applied across the two diodes .No, that violates circuit theory. That is not how voltages develop in a circuit.
Not at all.Isn't this contradictory to your objection that while the two diodes are not conducting , complete battery voltage 12 V is not applied across them ?
You are rightNot at all.
If the diodes have 12V across them, they should conduct, but the don't.
No, again, there is no "before" and "after" in circuit theory. Everything is instantaneous.Would it be correct to say that a potential drop occurs across a diode after it starts conducting ? But if that is correct , then why do we say that potential drop across the Si diode is also 0.3 V ?
Yes.Can we consider the potential drop across the Ge diode as a battery of voltage 0.3 V connected in series with opposite polarity ( such that it opposes the driving battery ) ?
Because the Silicon diode is in parallel with the Germanium one. And silicon diode conducts when its voltage drop is 0.7V. Anything below that, it will act as an open switch.then how is it that despite not conducting current through it , the drop across Si diode is 0.3 V ?
If two components are in parallel they are connected between the same two nodes so must have same voltage.But you do agree that once the Ge diode starts conducting , the potential drop across the Si diode is also 0.3 V ?
If you do agree , then how is it that despite not conducting current through it , the drop across Si diode is 0.3 V ?
If the input voltage is 0.2V (less than 0.3V), the diode voltage is also 0.2V and no diode will conduct. So, if you vary the input voltage from 0V to 0.3V, all of that voltage will appear across the diodes. I don't know if this is what you meant by "before any diode allows current to flow".what is the voltage across the two diodes before any one allows current to flow ?
It's not clear what you mean by "before anyone allows current to flow"?I agree .
But by that logic , what is the voltage across the two diodes before any one allows current to flow ?
If you directly apply 12V across a Ge (or any) diode, the drop across it will be 12V and not fixed 0.3V.When a 12 V battery is applied across a Ge diode , the drop across it is a fixed value . Why ?
No it's not (again, think of circuit theory). The Ge diode will draw an insanely high current at 12V and probably burn out.Sorry .This is incorrect
That is just an approximation. It is called the constant voltage drop model.Voltage drop across a diode ( conducting ) is a constant value .