
The given question involves analyzing the I-V characteristics of an electronic device to determine its type. The options provided are:
To solve this, we need to analyze the given I-V characteristics graph:
Now, let's reason through the options:
Therefore, the correct option is a Zener diode which can be used as a voltage regulator.
This choice is confirmed by the graph showing the constant current in reverse bias, indicative of a Zener diode's voltage regulating behavior.
The I-V characteristics of the given device show a typical breakdown region where the voltage remains fairly constant while the current increases. This characteristic is a signature of a Zener diode in its breakdown region, which is used as a voltage regulator.
Thus, the correct answer is Option (3).
Manufacturers supply a zener diode with zener voltage \( V_z=5.6\,\text{V} \) and maximum power dissipation \( P_{\max}=\frac14\,\text{W} \). This zener diode is used in the circuit shown. Calculate the minimum value of the resistance \( R_s \) so that the zener diode will not burn when the input voltage is \( V_{in}=10\,\text{V} \). 
The output voltage in the following circuit is (Consider ideal diode case): 
In the following circuit, the reading of the ammeter will be: (Take Zener breakdown voltage = 4 V)
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}