In the following circuit, the reading of the ammeter will be: (Take Zener breakdown voltage = 4 V)
We are given:
Step 1: Behavior of the Zener diode
Since the applied voltage \( V_s = 12 \, \text{V} \) is greater than the breakdown voltage of the Zener diode (\( 4 \, \text{V} \)), the Zener diode enters the breakdown region and maintains a constant voltage of \( 4 \, \text{V} \) across itself.
Step 2: Voltage across the load resistor
The Zener diode and the load resistor are in parallel. Therefore, the voltage across the load resistor is also: \[ V_{R_L} = V_Z = 4 \, \text{V} \]
Step 3: Current through the load resistor (measured by the ammeter)
\[ I = \frac{V}{R} = \frac{4}{400} = 0.01 \, \text{A} = 10 \, \text{mA} \]
Step 4: Ammeter Reading
The ammeter is in series with the 400 \( \Omega \) load resistor and thus measures the current through it.
Therefore, the reading of the ammeter is: \[ 10 \, \text{mA} \]
The output voltage in the following circuit is (Consider ideal diode case):
Which of the following circuits represents a forward biased diode?
Consider the discrete-time systems $ T_1 $ and $ T_2 $ defined as follows:
$ [T_1x][n] = x[0] + x[1] + \dots + x[n], $
$ [T_2x][n] = x[0] + \frac{1}{2}x[1] + \dots + \frac{1}{2^n}x[n]. $
Which of the following statements is true?
If $\lim_{x \to 1} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda + \mu$ is equal to
Let $f: [0, \infty) \to \mathbb{R}$ be a differentiable function such that $f(x) = 1 - 2x + \int_0^x e^{x-t} f(t) \, dt$ for all $x \in [0, \infty)$. Then the area of the region bounded by $y = f(x)$ and the coordinate axes is