When the forward bias voltage of a diode is changed from $0.6\;V$to $0.7\;V$, the current changes from $5\;mA$ to $15\;mA$. Then its forward bias resistance is:
(A) $0.01\Omega $
(B) $0.1\Omega $
(C) $10\Omega $
(D) $100\Omega $

Hint: First we will calculate the change in the voltage across the diode. This comes out to be $(0.7 - 0.6)V = 0.1V$. Now for this $0.1\;V$ change in the voltage across the diode, the current increases by $(15mA - 5mA) = 10mA$. So we have an increase in the current on account of the increase in the voltage. We will apply ohm’s law using these changed values of voltage and current to get resistance.

Complete step by step solution:
In an ideal diode, there is no voltage drop as the forward bias resistance is zero and the reverse bias resistance is infinity. However, the diodes in the everyday application have a voltage drop of around $0.7\;V$ and a small forward bias resistance.
In the given question we have the initial voltage drop as given to be $0.6\;V$ with a $5\;mA$ current flowing.
This voltage drop increases to become $0.7\;V$ with a $15\;mA$ current flowing.
We will assume that the internal forward bias resistance of the diode is $r$. Thus using ohm’s law of the form $V = IR$, we will get the required value of the internal resistance.
Here since the voltage and the current changes, we will calculate the changed voltage as the voltage difference and the difference of the current to compute the value of the forward bias resistance.
Therefore applying $(\Delta V) = (\Delta I)R$, we get
$(0.7 - 0.6)V = (15mA - 5mA)r$.
Upon rearranging the equation we get,
$r = \dfrac{{0.1V}}{{10mA}}$
$ \Rightarrow r = 10\Omega $

Thus the correct answer is option (C).

Note: In case of such diode problems where the current and the voltages are increased with time, there must be some internal resistance of the diode. This internal resistance should be assumed to be constant. The value of the constant internal resistance can be calculated by computing the change in the voltage and dividing it by the change in the current. Thus we get the required resistance of $10\Omega $.