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: \[ 10 \, \text{mA} \]
Given the circuit with a Zener diode in series with two resistors, the voltage \(V_1\) across the 100Ω resistor can be calculated as: \[ V_1 = \frac{400}{100 + 400} \times 12V = \frac{4}{5} \times 12 = \frac{48}{5} \, \text{V} \] Here, \(V_1 > V_z\), where \(V_z\) is the Zener voltage. Thus, the Zener breakdown will take place, and the voltage across the 400Ω resistor will be \(4V\). Now, the current \(I\) across the 400Ω resistor is given by Ohm’s law: \[ I = \frac{4}{400} = 10 \, \text{mA} \] \[ \boxed{I = 10 \, \text{mA}} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,



Which of the following circuits represents a forward biased diode?
The output voltage in the following circuit is (Consider ideal diode case): 
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)