In a Zener diode regulated power supply, unregulated DC input of \[10\,\,{\text{V}}\] is applied. If the resistance \[\left( {{R_S}} \right)\] connected in series with a Zener diode is \[200\,\,\Omega \] and the Zener voltage \[{V_s} = 5\,\,{\text{V}}\] , the current across the resistance \[{R_S}\] is:
A. \[15\,\,{\text{mA}}\]
B. \[10\,\,{\text{mA}}\]
C. \[20\,\,{\text{mA}}\]
D. \[5\,\,{\text{mA}}\]
E. \[25\,\,{\text{mA}}\]

Hint:First of all, we will calculate the voltage drop across the resistance from the given input voltage and the Zener voltage.As resistance is given so using the Ohm’s law \[V = IR\] , calculate the required current across \[{R_S}\] .

Complete step by step answer:
Given,
DC input voltage, \[{V_i} = 10\,\,{\text{V}}\]
Resistance, \[{R_S} = 200\,\,\Omega \]
Zener voltage, \[{V_s} = 5\,\,{\text{V}}\]

Therefore, voltage drop across \[{R_S}\] is,
\[V = {V_i} - {V_s}\]
$\Rightarrow V = \left( {10 - 5} \right)\,\,{\text{volt}} \\
\Rightarrow V = 5\,\,{\text{volt}}$

According to ohm’s law, the current between the two points through a conductor is directly proportional to the voltage between the two points. The formula is given by, \[V = IR\] .

Therefore, current across the resistance \[{R_S}\] is given by,
\[I = \dfrac{V}{{{R_S}}}\] …… (i)

Substitute the values of \[V = 5\] and \[{R_S} = 200\,\,\Omega \] in equation (i).
Therefore,
$I = \dfrac{5}{{200}}\,\,{\text{A}} \\
\Rightarrow I = 0.025\,\,{\text{A}} \\
\therefore I = 25\,\,{\text{mA}} \\$
Hence, the current across the resistor is \[25\,\,{\text{mA}}\] .The correct option is E.

Additional information:
A Zener diode is a particular type of diode that is engineered to allow current to move backwards predictably when a certain reverse voltage set is reached, defined as the Zener voltage. With a large range of Zener voltages, Zener diodes are produced and some are also variable.In reverse biased state, a zener diode still works. As such, the zener diode can be used to build a basic voltage regulator circuit to sustain a steady DC output voltage through the load, through fluctuations in the input voltage and changes throughout the load current.

Note:It is generally seen that while solving the problem, most of the students tend to make the mistake by adding the input voltage and the Zener voltage. This is absolutely wrong. It is important to note that we actually need to find the voltage drop. Again, after obtaining the answer in amperes we need to multiply with \[1000\] to convert it into milliamperes, but most of the students multiply with \[100\] .