Question

Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is:

A. 0.75A
B. zero
C. 0.25A
D. 0.5A

Answer 1

Hint: As a first step, one could read the question well and hence note down the given values from it. The main step that is to be done in this question is the analysis of the given circuit and checking the biasing of the diodes. Then apply Ohm’s law to get the current in the given circuit.
Formula used:
Ohm’s law,
$I=\dfrac{V}{R}$
Complete step by step answer:
In the question, we have a circuit given where there are two diodes ${{D}_{1}}$and ${{D}_{2}}$connected in series with $10\Omega $and $20\Omega $resistors and we have a battery of 5V connected parallel to them. We are supposed to find the current that is being supplied by the battery.
As a very first step, we could analyse the given circuit. The first diode ${{D}_{1}}$has its p-side connected to the positive of the battery and the n-side is connected to the negative terminal, in other words, this diode is forward biased and hence there will be current flow in this branch.
Now for the other diode ${{D}_{2}}$, very opposite of the other diode, has its n-side connected to the positive terminal and its p-side to the negative terminal. This diode is reverse biased and hence no current flow in this branch.
Now, we could apply ohm's law including only the forward biased branch to get,
$I=\dfrac{V}{R}$
Substituting the values,
$I=\dfrac{5V}{10\Omega }$
$\therefore I=0.5A$
Therefore, we found the current supplied by the battery to be 0.5A.

So, the correct answer is “Option D”.

Note: We could define a diode as a semiconductor device that is used to act as a one way switch for current. That is, they simply allow current flow in one direction very easily but restrict the flow in the direction opposite to it. From its very name we could conclude that it is a two terminal device.