Batteries / Charging + Resistance.

Thread Starter

stewi3

Joined Feb 24, 2013
2
Hey,

Got this question on my tutorial sheet and 2 different teachers have given different answers. No one else in the class really knows for certain which way is right, and it's doing my head in to be honest.

I've attached a picture of the question, and a solution that one teacher gave us.

Could someone please tell me if this is the correct way of working this stuff out?

 

WBahn

Joined Mar 31, 2012
33,187
Hey,

Got this question on my tutorial sheet and 2 different teachers have given different answers. No one else in the class really knows for certain which way is right, and it's doing my head in to be honest.

I've attached a picture of the question, and a solution that one teacher gave us.

Could someone please tell me if this is the correct way of working this stuff out?

As The_RB says, the method looks reasonable, but check your calculations.

I wouldn't actually recommend this approach to actually charge the batteries, but for a paper exercise in circuits that's not the point.

What was the other method that was suggested?
 

Thread Starter

stewi3

Joined Feb 24, 2013
2
What was the other method that was suggested?
Find Rt for the circuit by doing Total Circuit voltage, Vt =[110 - (V=E-Ir)]

Divided by Total Amps (10). = Total Circuit resistance, Rt

Rt = Rs + Ir
Therefore Rs = x

However, you get a different answer.
 

WBahn

Joined Mar 31, 2012
33,187
Find Rt for the circuit by doing Total Circuit voltage, Vt =[110 - (V=E-Ir)]

Divided by Total Amps (10). = Total Circuit resistance, Rt

Rt = Rs + Ir
You KNOW this is wrong because the units don't work out.

Rt and Rs have units of resistance. Ir has units of voltage. The two can't be added together.

Always, always, always check the units!

Therefore Rs = x

However, you get a different answer.
The basic approach is valid, it just goes off the tracks midway. The total resistance is the resistance of the batteries plus the resistance you are going to put into it, call that Rb and Rx respectively.

The total voltage pushing current through that total resistance is 110V-Vb, where Vb is the voltage of the battery stack.

Your current is then

I = (110V-Vb)/(Rb + Rx)

Solve for Rx.

You should get the same result as before.
 
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