A 5V battery is connected across the points X and Y. Assume \[{D_1}\] and \[{D_2}\] to be normal silicon diodes. Find the current supplied by the battery if the +ve terminal of the battery is connected to point X.

A. \[ \sim 0.86A\]
B. \[ \sim 0.5A\]
C. \[ \sim 0.43A\]
D. \[ \sim 1.5A\]

Hint:Before we start addressing the problem, we need to know about the diodes. A diode is a semiconductor device that has a two-terminal that conducts current in one direction.

Formula Used:
The formula to find the current using ohm's law is given by,
\[V = RI\]
Where, V is voltage in the circuit, R is the resistor used and I is current.

Complete step by step solution:

Image: A circuit consists of 2 diodes and 2 resistors

Consider a battery of 5V connected across points X and Y as shown in the figure. Assume that \[{D_1}\] and \[{D_2}\] are normal silicon diodes. Now, we need to find the current supplied by the battery when the +ve terminal of the battery is connected to point X.

The diode \[{D_1}\] is in forward bias therefore the current passes through it, but since the diode \[{D_2}\] is in negative bias condition, there is no flow of current. So, we are not considering the \[5\Omega \] resistor.

Now, we need to find the current only through the \[10\Omega \] resistor. In order to do that we consider ohm’s law that is,
\[V = RI\]
\[\Rightarrow I = \dfrac{V}{R}\]
They have given that the diode is made up of silicon so the forward voltage of silicon is 0.7V so, the silicon diode drops a potential of 0.7 volts across it, and only \[{D_1}\] will conduct.

So, \[V = 5 - 0.7\] and \[R = 5\Omega \]. Substitute the values in the above equation we obtain,
\[I = \dfrac{{5 - 0.7}}{{10}}\]
\[\therefore I = 0.43A\]
Therefore, the current supplied is 0.43A.

Hence, Option C is the correct answer

Note:Silicon Diode is a semiconductor that has both positive and negative charge polarity and allows the current to flow in one direction while restricting another. These are used in radios, computers, AC and DC power supplies, temperature and radiation sensors etc.