[amp_mcq option1=”11 mA” option2=”12 mA” option3=”10 mA” option4=”14 mA E. None of the above” correct=”option1″]
The correct answer is $\boxed{\text{E}}$.
The voltage drop across a diode is a function of the current through it. The higher the current, the higher the voltage drop. In this case, the first diode has a voltage drop of 0.75 V and the second diode has a voltage drop of 0.8 V. This means that the current through the second diode is less than the current through the first diode.
The current through the second diode can be calculated using Ohm’s law:
$$I = \frac{V}{R}$$
where $I$ is the current, $V$ is the voltage, and $R$ is the resistance. The resistance of a diode is a function of its temperature. The higher the temperature, the lower the resistance. In this case, the temperature of the diodes is not given, so we cannot calculate the resistance of the diodes. Therefore, we cannot calculate the current through the second diode.
The answer to this question is not given by any of the options.