Using Ohm's law, the current in the circuit is:
\(\ \ I = \frac{V - V_b}{R} = \frac{5.7 - 0.7}{5 \times 10^3} = 1 \text{mA}\)
Given below are two statements ; one is labelled as Assertion $A$ and the other is labelled as Reason $R$
Assertion A: Photodiodes are used in forward bias usually for measuring the light intensity
Reason R: For a p-n junction diode at applied voltage $V$ the current in the forward bias is more than the current in the reverse bias for $\left|V_2\right|>\pm V \geq\left|V_0\right|$ where $V_0$ is the threshold voltage and $V _2$ is the breakdown voltage
In the light of the above statements, choose the correct answer from the options given below