Cant understand this terminal voltage question

[nishantve1]

Homework Statement

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 don't understand what is terminal voltage + how it changes It would be great if someone explained it .

Homework Equations

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

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 ?

 

[azizlwl]

It is internal resistance.
You can picture it as this

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

 

[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.

 

[ehild]

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 ?

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