Who can solve this question and where is zener diode's resistance ?

 

Yes

 

How far have you gotten? Show us your work.

 

Yes

can u help me about this question?

 

How far have you gotten? Show us your work.

 

Yes if you make two assumptions: That the ammeter has zero resistance (a perfect ammeter); that the resistor shown as variable is set to the resistance written next to it, 2.2kΩ.

You are given the voltage across the zener. The voltage across the ammeter/220Ω/2.2kΩ is the same as the zener voltage so you can calculate the total resistance of this branch and therefore the current through it (IL).
You are given the voltage at both sides of the 100Ω resistor so you can calculate the current through it.

Now Kirchoff's first law will give you the current through the zener.

 

Yes if you make two assumptions: That the ammeter has zero resistance (a perfect ammeter); that the resistor shown as variable is set to the resistance written next to it, 2.2kΩ.

You are given the voltage across the zener. The voltage across the ammeter/220Ω/2.2kΩ is the same as the zener voltage so you can calculate the total resistance of this branch and therefore the current through it (IL).
You are given the voltage at both sides of the 100Ω resistor so you can calculate the current through it.

Now Kirchoff's first law will give you the current through the zener.

how can i calculate total resistance? that is can i calculate just 220 ohm there?

 

What about the 2.2kΩ?

 

What about the 2.2kΩ?

2200+220 ? namely

 

2200+220 ? namely

Does that seem right to you?

 

Does that seem right to you?

i did 10-5.6 / 100 = 0.044 A = Iz and then 5.6/220+2.2k= 2.3mA=IL and then already vo=vz namely vo=5.6V But this is wrong.

 

i did 10-5.6 / 100 = 0.044 A = Iz

Is the current in the 100Ω resistor equal to the current in the zener?

 

Is the current in the 100Ω resistor equal to the current in the zener?

if i sketch 100ohm's next to ix , namely, ix=iz+iL =2.3mA+56mA= 58.3mA= ix ?????

 

but i couldn't find iL

You found IL in post #11

 

if i draw 100ohm's next to Ix current, ix=iz+iL is this true ?

Yes.

 

Yes.

iz= 5.6/100= 56mA so that iz=56mA and then il=2.3mA
if i draw 100ohm's next to ix , namely, ix=iz+iL =2.3mA+56mA= 58.3mA= ix ?????

 

ix = iz + iL Correct.
Now ix is the current in the 100Ω. So what is the voltage across that resistor and therefore what is the current?

 

ix = iz + iL Correct.
Now ix is the current in the 100Ω. So what is the voltage across that resistor and therefore what is the current?

10/100= 0.1A = ix therefore, 0.1A= 2.3*10^-3+ iz if i solve this equation, iz= 97.7mA is it true?
Why 2.3 mA very small value? and What is ampermeter of value? Where is Vo ?
YOU ARE A GREAT MAN !!!!!!!!!!!!!!!!!

 

The

10/100= 0.1A = ix therefore, 0.1A= 2.3*10^-3+ iz if i solve this equation, iz= 97.7mA is it true?

The voltage across the 100Ω is not 10V. Use the right voltage and then do that sum above again.

 

The

The voltage across the 100Ω is not 10V. Use the right voltage and then do that sum above again.

5.6/100= 56mA=ix
ix=iz+iL
56mA=2.3mA+iz
iz=53.7mA