Cant understand this terminal voltage question !

1. Jun 12, 2012

nishantve1

1. The problem statement, all variables and given/known data
So the Problem says . A 8V battery of internal resistance 0.5 Ω is being charged using another storage battery of 120 V using a resistance of 15.5 Ω calculate the Terminal Voltage of the battery .
I really dont understand what is terminal voltage + how it changes It would be great if someone explained it .

2. Relevant equations

V =IR
For Batteries in series
ε(total) = ε1 - ε2 (as positive is connected to the positive terminal)

3. The attempt at a solution
Heres how I interpreted the circuit

I did the solution and i got the right answer but I have no idea how it worked
First I calculated the Equivalent ε
= 120 - 8 = 112V
also r equivalent = 15.5 Ω + 0.5 Ω = 16Ω
So the Current flowing is 7 Amps
Now what I did is
Potenital Drop across the 120 V = potential drop across the 15.5Ω resistor + Terminal Voltage drop across the battery
that gives Terminal Voltage = 11.5 V
Which is correct now What I am not able to understand is what is Terminal Voltage and the way of calculating it and how it gets affected . Will there be terminal voltage change across 120 V battery too ? how ? also keeping the internal resistance of 8V battery in mind I can say that if 7A of current is flowing through it then the potential difference should be
V = ε - I* internal resistance
= 8 - 3.5
= 4.5V
Why is this wrong ? How is it 11.5 V instead ?

2. Jun 12, 2012

azizlwl

It is internal resistance.
You can picture it as this

(+)---R---battery......(-)
(+) and (-) battery terminals.
R internal resistor.

3. Jun 12, 2012

Infinitum

Terminal voltage is the voltage at the terminals of the power source(here, equivalent battery). Basically, $V_t = V \pm IR$, where V is the emf of the cell, and Vt is the terminal voltage. Plus or minus depends on flow of current as this equation is from Kirchoff's Voltage Law.

4. Jun 12, 2012

ehild

V=ε-IR(internal) is the terminal voltage if the current flows out from the positive terminal of the battery. That happens if the battery is the only source in the circuit. There is a more powerful voltage source here, with 120 V emf, charging the 8 V battery. So the = 7A current flows into the battery, making the terminal voltage higher than the emf.

ehild