# A circuit question involving internal resistances

## Homework Statement

A battery has an emf of 6v. The battery is connected in series with an ammeter and a voltmeter.If a certain resistor is connected in parallel with the voltmeter,the voltmeter reading decreases by a factor of 3,and the ammeter reading increases by the sme factor. What is the initial reading V of the voltmeter? All elements except battery in the circuit have unknown internal resistances

2. The attempt at a solution
R3 is the resistor that is connected in parallel with the voltmeter.
I also use R1,R2 to represent internal resistances of voltmeter and ammeter which i assume are resistors connected in series to the voltmeter and ammeter.(ammeter then measures current flowing through R2 while voltmeter measures pd across R1 or R3)

I tried to construct equations using basic rules about resistors connected in series or parallel,after solving some (overly complex) equations,i had V=4.8v while the answer is v=4.5v

I also tried to use Kirchhoff's rules to solve,but it did not work out.

Can you post your equations? I was able to get 4.5V using voltage divider, KCL, and ohms law (x2).

I really just started writing equations until something popped out at me - then I had an idea where to go!

Can you post your equations? I was able to get 4.5V using voltage divider, KCL, and ohms law (x2).

I really just started writing equations until something popped out at me - then I had an idea where to go!

I simplified my equations and now i can get the correct answer of 4.5v
my equations are as follows
assume R1//R3 is r

IR1+IR2=6V
3IR2+3Ir=6V
solve these to get R1-2R2=3r

one more equation:
IR1=3(3I)r → R1=9r

so R1=3R1-6R2
R1=3R2

therefore IR1=4.5V