"""
A battery has an internal resistance of 0.5 ohms . The battery has an emf of 1.25 v .
When it's connected to a resistor ,the terminal voltage falls to 1.45 v.
what is the current flowing?and what is the value of the resistor?
"""
can you guys tell me the steps of solving this q ?!

 

You are stating a battery has an open circuit voltage (no load) of 1.25 Volt and then when a load is placed on the battery the voltage increases to 1.45 Volt?

A battery has an internal resistance of 0.5 ohms . The battery has an emf of 1.25 v .
When it's connected to a resistor ,the terminal voltage falls to 1.45 v.
what is the current flowing?and what is the value of the resistor?
"""
can you guys tell me the steps of solving this q ?!

Something isn't right here. How does the open circuit voltage increase under a load?

You may want to give this a read as it explains things well and again when a load is applied affording a current flow the battery voltage should decrease and not increase. When we apply Kirchoff's laws we just manipulate the formulas, including Ohm's law.

Ron

 

You have a negative resistor connected. The no load volts is less than the load volts.
I wish I had some of those!

 

Welcome to AAC!

can you guys tell me the steps of solving this q ?

Is this homework?

 

A battery has an internal resistance of 0.5 ohms .

The values of resistance is too big, the real values should be with mΩ.

 

Welcome to AAC!
Is this homework?

yes it is ,
Does it not allowed here ?!!

 

The values of resistance is too big, the real values should be with mΩ.

This is the prof. Q

 

Did you reread your original post?
Examine the numbers you posted more carefully.
How can the battery voltage increase from 1.25V to 1.45V?

 

You are stating a battery has an open circuit voltage (no load) of 1.25 Volt and then when a load is placed on the battery the voltage increases to 1.45 Volt?



Something isn't right here. How does the open circuit voltage increase under a load?

You may want to give this a read as it explains things well and again when a load is applied affording a current flow the battery voltage should decrease and not increase. When we apply Kirchoff's laws we just manipulate the formulas, including Ohm's law.

Ron

thanks bro that was helpful

 

This site will be a great help to you by guiding you as to how to solve the problem. Folks here DO NOT do your homework for you. You wouldn't learn much that way. Now, assuming you have the voltages reversed in your original post, what if you had an ideal voltage source (one that will hold a constant voltage no matter what load is placed upon it). What if you placed a 0.5 ohm resistor is series with it. Can you take it from here?

 

Did you reread your original post?
Examine the numbers you posted more carefully.
How can the battery voltage increase from 1.25V to 1.45V?

Sorry it was 1.52v not 1.25v

 

Please type the complete word, don't use the abbreviation or just type the first character.

 

Sorry it was 1.52v not 1.25v

That makes a whole world of difference, doesn't it?

Now show us how you would go about solving this.

 

Did you reread your original post?
Examine the numbers you posted more carefully.
How can the battery voltage increase from 1.25V to 1.45V?

The current warmed up the battery...

 

yes it is ,
Does it not allowed here ?!!

Homework questions are definitely allowed, but they belong in the Homework Forum. In that forum, members know not to simply provide answers.

I always ask if it seems like a homework question because you'll learn more from doing the work yourself or being helped versus someone thinking they're being helpful by just supplying an answer.

 

Dyslexia causes numbers and letters to be mixed up.
At first I thought a battery that measured only 1.25V could easily be a cheap Chinese "Sooper Heavy Dooty" one and almost dead with an internal resistance of 0.5 ohms.
Then I also wondered about the magic (heating?) that causes its voltage to rise when loaded.

 

Now show us how you would go about solving this.

Emf = V + Ir >>>>> 1.52 = 1.45 + I(0.5) >>>>>> then I = 0.14 A
so the R of the outer resistance is :
R=V/I >>>> R = 10.36 ohm
Is that true :} ?

 

Emf = V + Ir >>>>> 1.52 = 1.45 + I(0.5) >>>>>> then I = 0.14 A
so the R of the outer resistance is :
R=V/I >>>> R = 10.36 ohm
Is that true :} ?

How can you check your answer?

 

How can you check your answer?

I can't .

 

I can't .

Sure you can.

Draw the complete circuit. Use Ohm's Laws formulas and calculate the current through the total loop.
Does it agree with your initial estimate?