Home
Class 12
PHYSICS
A zener diode having breakdown voltage 5...

A zener diode having breakdown voltage 5.6 V is connected in reverse bias with a battery of emf 10 V and a resistance of 100` Omega` in series . The current flowing through the zener is.

A

88 mA

B

0.88 mA

C

4.4 mA

D

44 mA

Text Solution

Verified by Experts

The correct Answer is:
D

Here `I=(10-5.6)/(100)=0.044 "A"=44xx10^(-3) =44 `mA
Promotional Banner

Topper's Solved these Questions

  • SEMICONDUCTORS AND ELECTRONICS

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE with solutions (JEE Main)|4 Videos

  • SEMICONDUCTORS AND ELECTRONICS

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE with solutions (AIPMT)|2 Videos

  • SEMICONDUCTORS AND ELECTRONICS

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE with solutions (WBCHSE)|21 Videos

  • REFRACTION OF LIGHT

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|11 Videos

Similar Questions

Explore conceptually related problems

A battery of internal resistances zero is connected to a galvanometer of resistance 80 Omega and a resistance of 20 Omega in series . A current flows through the galvanometer If a shunt of 1 Omega resistance is connected to the galvanometer, show that the current that will now flow through the galvanometer becomes 1/17 of the previous current.

A parallel combination of three resistances of 6 Omega, 12 Omega and 12 Omega is connected in series with an electric cell of emf 1.5 V and internal resistance 2 Omega . Find out the currents through the three resistances.

A shunt of resistance 1 Omega is connected with a galvanometer of resistance 100 Omega . What part of the main current will flow through the galvanometer?

A shunt of resistance 1 Omega is connected with a galvanometer of resistance 100 Omega . What part of the main current will flow through the galvanometer?

An electric cell of emf 10 V and of internal resistance 1 Omega is connected in series with the parallel combination of three resistances of 3 Omega,5 Omega and 8 Omega .Calculate the currents through the three resistances.

Two cells one of emf 1.4 V and internal resistance 0.6 Omega the other of emf 2.5. V and internal resistance 0.3 Omega are connected in parallel and the combination is connected in series with an external resistance of 4 Omega . What is the current through this resistance?

Two identical cells each of emf 1.5V are connected in series. IF this combination of cells is connected to a resistance and a galvanometer in series, the current that flows through the circuit is 1A.If the cells are connected in parallel the current becomes 0.6 A, Determine the internal resistance of each cell.

In the circuit shown, what will be the current through the 6v zener ?

The maximum output voltage of an unregulted voltage source is 10V. A Zener diode of rating 1 W is connected in reverse bias with this voltage source through a resistance R. The Zener diode will be burnt if R is less than 24 Omega . Find the maximum value of Zener voltage (in volt ) of the diode.