Question

A zener diode having a breakdown voltage of $5.6V$ is connected in reverse bias to a battery of emf $10V$ and a resistance of $100\Omega $ in series. The current flowing through the zener is-
(A).$88mA$
(B).$0.88mA$
(C).$4.4mA$
(D).$44mA$

Answer 1

Hint: A diode is a semiconductor device which lets current flow through it. A zener diode is also a diode connected in reverse bias, which lets current flow through it only when a threshold voltage is crossed known as the zener voltage. After crossing the zener voltage, the remaining voltage operates in the circuit. Using ohm’s law we can calculate current by substituting corresponding values.

Formula Used:
$R=\dfrac{V'}{I}$

Complete answer:
A zener diode is a silicon semiconductor. It has infinite resistance until a zener voltage is reached, that means no current flows through it unless a threshold voltage called zener voltage is applied across its ends.
After the zener voltage is reached, current flows through it in reverse bias.
In reverse bias, the negative terminal is connected to the positive side of a transistor and the positive terminal is connected to the negative side.
Given, zener voltage is $5.6V$, battery voltage is $10V$, the leftover potential will account for the flow of current. Therefore, reverse bias voltage acting in the circuit is,
$\begin{align}
  & V'=10-5.6 \\
 & V'=4.4V \\
\end{align}$
Therefore, voltage operating in the circuit is $4.4V$.
According to ohm’s law,
$R=\dfrac{V'}{I}$ - (1)
Here,$R$ is the resistance
$V$ is the voltage applied to the ends of a circuit
$I$ is the current flowing through a circuit
Substituting given values in the above equation, we get,
$\begin{align}
  & 100\Omega =\dfrac{4.4V}{I} \\
 & I=\dfrac{4.4}{100} \\
\end{align}$
$\begin{align}
  & \Rightarrow I=4.4\times {{10}^{-2}}A \\
 & \therefore I=44mA \\
\end{align}$
Therefore, the current flowing through zener diode is $44mA$ so the correct option is (D).

Note:
As we increase the potential in reverse bias, the potential barrier also increases which restricts the flow of current through the junction between p-type and n-type of the diode. A very small amount of current flows in reverse bias which is almost negligible while, a large current flows in forward bias