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QUESTION: A 15v Battery is connected to parallel resistors R1 = 15 ohms, R2 = 25 ohms the battery has a internal resistance of 3 ohms.

what is the voltage dropped across the internal resistor?

ANSWER: Oh, you are learning ohms law.  Well, it never fails.  The voltage across a resistor will always give the current by E = IxR or I= E/R.

In a series circuit the sum of the voltage drops will equal the applied voltage.

Therefore, your current drops will add up to 15v.  There are two resistances in your series circuit; the 3 ohms internal to the battery and the 15||25 ohm parallel equivalent.

So, the current in the circuit is I = 15/(3 + 15||25).

The voltage across the 3 ohm internal resistor can now be computed using ohms law.

I will let you do the rest of it.  Let me know if you get stuck.





---------- FOLLOW-UP ----------

QUESTION: i have worked out that I=V/R I=1.6A and that the Rt is 9.375 using (R1*R2)/(R1+R2)

do the two resistance totals get added together eg, Rt+r as if they are in series?

if so then using V=IR then

I=15/12.375
I=1.212A
V=IR
V=1.212A*3ohms
V=3.636
15V - 3.636V
= voltage drop of 11.363V across the internal resistor

is it right?

Answer
You went one step too far!

The voltage drop across the internal resistance, 3 ohms, is just the current passing through it times the resistance which you determined to be 3.6 volts.

Got it?

What it really means is that with the battery open circuit you would get the 15 volts but when the load is added the voltage appearing across the battery terminals would be the 11.3 volts due to the internal voltage drop across the internal resistance being subtracted from the terminal voltage.  

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Electronics questions about AC, DC and digital theory.

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Graduate electrical engineer with over 40 years in electronic design, manufacturing, project organization and patent review. Experience in fields of industrial and consumer electronics (audio, video, acoustics, etc.)

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