In a given circuit diagram, a 5 V zener diode along with a series resistance is connected across a 50 power supply. The minimum value of the resistance required, if the maximum zener current is 90 mA will be __________ $\Omega$.
To determine the minimum value of the resistance R required, we start by understanding the circuit. The zener diode maintains a constant voltage Vz of 5 V across it. The supply voltage Vi is 50 V.
Step 1: Calculate the Voltage Drop
The voltage across the series resistance R is the difference between Vi and Vz:
ΔV = Vi - Vz = 50 V - 5 V = 45 V
Step 2: Use Ohm's Law
Ohm’s Law states that V = IR. Rearranging for R, we have:
R = ΔV/Izmax = 45 V / 0.09 A
Substituting the values gives:
R = 500 Ω
Step 3: Verify the Solution
The computed resistance value is 500 Ω, which falls within the given range of 500,500. Thus, the minimum required resistance to ensure the maximum zener current does not exceed 90 mA is indeed 500 Ω.
Similar Questions
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Important Questions from Modern Physics
- A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ______ $\Omega$.
- A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of $1.21\text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is __________. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J.s}$)
- The average energy released per fission for the nucleus of ${}^{235}_{92}\text{U}$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure ${}^{235}_{92}\text{U}$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is ________.
(Avogadro Number = $6 \times 10^{23} \text{ per mole}$) - The energy released when $\frac{7}{17.13} \text{ kg}$ of ${}^7_3\text{Li}$ is converted into ${}^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32} \text{ eV}$. The value of $\alpha$ is _________. (Nearest integer)
(Mass of ${}^7_3\text{Li} = 7.0183 \text{ u}$, mass of ${}^4_2\text{He} = 4.004 \text{ u}$, mass of proton = $1.008 \text{ u}$ and $1 \text{ u} = 931 \text{ MeV/}c^2$ and Avogadro number $= 6.0 \times 10^{23}$) - A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy $3.2$ Mev. The estimation of the radius of the nucleus of gold element $(Z = 79)$ by him is $\times 10^{-16}$ m
(Take $\frac{1}{4\pi \epsilon_0} = 9\times10^9 \frac{Nm^2}{C^2}$)
