ADC and battery from camera

When i make measurment on my 1.5 battery ( battery from some camera ) on the scale " 10 ADC ( unfused )" i get 50mA. If i do the same thing but connected to the fused part of the voltmeter i get no result.

I dont understand why is this happening becuse im measuring the voltage from DC battery and not from some other DC source.

And one more thing,this unfused plug on the multimeter... how is this "unfused part" made (in multimeter) ?

---------------------------------------------
I have some problems with understanding the charging process in this camera.
If the DC current of the battery is 50mA how can capasitor fill himself so quickly ( under 5 seconds ) and be ready to give flash?

Guys, can someone give me an answer to my post, please ?

Beginner19,
Did your camera work with that battery after you made the measurement? If it worked your ammeter is faulty.
Testing a battery by short circuiting its terminals isn't exactly the best method I would vote for!

Giri

what is the fuse rating of the multimeter ? is it 200 mA ?

about the charging process .... it doesnt depend on how much current but on the time constant (R*C) ... because
v(t)=Vfinal (1-e^(-t/(R*C))) .... so if RC is small , the charging process will be quick

Beginner19 said:

When i make measurment on my 1.5 battery ( battery from some camera ) on the scale " 10 ADC ( unfused )" i get 50mA. If i do the same thing but connected to the fused part of the voltmeter i get no result.

I dont understand why is this happening becuse im measuring the voltage from DC battery and not from some other DC source.

And one more thing,this unfused plug on the multimeter... how is this "unfused part" made (in multimeter) ?

---------------------------------------------
I have some problems with understanding the charging process in this camera.
If the DC current of the battery is 50mA how can capasitor fill himself so quickly ( under 5 seconds ) and be ready to give flash?

Click to expand...

First take a good look what you actually measure. If you measure voltage, you should set your (digital) multimeter appropriately. If you want to measure how much courrent you can get by measuring short circuit current of the battery (because that is what happens when you put the probes to COMM(on) and 10A UNFUSED) - DON'T DO THAT!!! You will only cause damage to the battery.

Did your camera work with that battery after you made the measurement? If it worked your ammeter is faulty.

Click to expand...

--- The camera is working normal.

what is the fuse rating of the multimeter ? is it 200 mA ?

Click to expand...

--yes
-- If i connect COM and VΩmA then i get no measurment
-- If i connect the same battery to COM and 10ADC ( on the same scale ) i get 50mA
-- If i connect 9V on the COM - VΩmA way then i get 50mA

So, i dont understang why i have to measure (when i make direct connection to battery ) 1,5V camera battery on unfused part, and 9V block battery on other part ?

Testing a battery by short circuiting its terminals isn't exactly the best method I would vote for!

Click to expand...

--- what would you suggest ?

v(t)=Vfinal (1-e^(-t/(R*C)))

Click to expand...

-- Thank you very much for this formula! If i would make some calculations ( of i i connect other DC source ) this formula should do the trick ...

it doesnt depend on how much current but on the time constant (R*C)

Click to expand...

-- Let take a look into this situation. If you have 1,5V camera battery (50mA) and if you want to charge the capasitor within 6 seconds, how much "power" do i have to pull out from the battery ?

-- What are the standard values for such capasitors (V,F,C) ?

what is the fuse rating of the multimeter ? is it 200 mA ?

Click to expand...

--yes
-- If i connect COM and VΩmA then i get no measurment
-- If i connect the same battery to COM and 10ADC ( on the same scale ) i get 50mA
-- If i connect 9V on the COM - VΩmA way then i get 50mA

So, i dont understang why i have to measure (when i make direct connection to battery ) 1,5V camera battery on unfused part, and 9V block battery on other part ?

Testing a battery by short circuiting its terminals isn't exactly the best method I would vote for!

Click to expand...

--- what would you suggest ?

Click to expand...

actually i dont know exactly what u wanna do ....
if u want to measure the current output of the battery .... connect the battery to a small resistance (100 ohm for example) and measure the voltage across it --> I=V/100

i dont understand what is the 9V u r talking about ... ?

v(t)=Vfinal (1-e^(-t/(R*C)))

Click to expand...

-- Thank you very much for this formula! If i would make some calculations ( of i i connect other DC source ) this formula should do the trick ...

it doesnt depend on how much current but on the time constant (R*C)

Click to expand...

-- Let take a look into this situation. If you have 1,5V camera battery (50mA) and if you want to charge the capasitor within 6 seconds, how much "power" do i have to pull out from the battery ?

Click to expand...

if u want the capacitor to charge within 6 seconds .... then u set R*C <<6seconds




if u want to calculate power drawn from the battery , then we calculate the current :

i(t)=[Vfinal-V(t)]/R=(Vfinal/R)*exp(-t/(R*C)) --> it is clear it is not a DC current

P(t)=Vfinal*i(t)=((Vfinal)^2)/R)*exp(-t/(R*C)) --> power drawn from the battery

if u want the total energy drawn from the battery :

E=∫P(t)dt from 0 to infinity

=C*(Vfinal)^2)

-- What are the standard values for such capasitors (V,F,C) ?

Click to expand...

i have no idea

I have used standard 9V battery just as test. On this battery i can make measurments on VΩmA - COM connections but for 1.5 camera battery i have to switch to unfused plug. Why ?

Anyway, thank you for your replays !