Diode problem - Find Current I

I am unable the solve this problem.
Could someone help to find out which diode is ON and which is OFF and tell me how to find out ?
I am unable to write the KVL.

And help to find the current "I"

Thanks in advance!

Both ON and I=0.3 A

CataM said:

Both ON and I=0.3 A

Click to expand...

Can u please tell me how the problem was solved ?
I am finding difficult to write the KVL for the circuit ?
Which loop must be considered first and how the KVL was written to solve this problem ?

Hi,

Calculate the right path and the left path independently.

Then imagine what happens when you connect/join both.
What is "fix" and what will change.

Btw: CataM's solution is correct.

Klaus

You can apply Superposition theorem. Consider only one source at a time and apply superposition theorem twice and then merge the results.

https://everycircuit.com/circuit/5422887705509888

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There are only two loops. it doesn't matter which loop you consider first.

Note that a limitation of the Superposition Theorem is that it must involve all linear devices:

You should not use superposition theorem for this because diodes are non-linear devices. However, since both diodes are on in this case and the diodes are ideal diodes with no voltage drop, it will work in this case since both diodes are acting as short circuits.

To solve, note the direction of both current lops given by the polarity of the two voltage sources. The direction of the currents tell us that both diodes are forward biased and therefore will be on.
Now you can replace both diodes with short circuits since they are assumed to be ideal diodes as per the problem.

Then examine the left current loop:
From Kirchoff's Voltage Law, the resistor voltage will be -3V.
The current I=-3V/10 ohm = -0.3A ( the -ve sign only indicates the direction of current )

Now we can analyze the right loop: since V2 is 5V, this will create 2V ( 5V - 3V using Kirchoff ) across the resistor next to D2 and 0.2A through D2.
We can also see from Kirchoff's Current Law that Id1=0.3A - 0.2A = 0.1A
Therefore currents are:
Id1=0.1A
Id2=0.2A
I=0.3A

Note that all current directions support forward bias of the diodes which confirms that they are on.
It is good to do this backwards check to make sure that our initial assumptions about the diodes being on are indeed correct.

Cheers