Find emf and internal resistance of a battery

[Blu3eyes]

Homework Statement

When a 60 \Omega resistor and a 120 \Omega are connected in parallel across a battery there is a current of 700 mA flowing through the battery. When the 120 Ohms resistor is disconnected from the circuit the current through the battery drops to 500 mA. What is the emf and the internal resistance of the battery?

Homework Equations

emf= I(R load + ri)
V terminal = emf - I*ri

The Attempt at a Solution

 

[Zryn]

What is the equivalent resistance combination of the 60R||120R load?

Lets call it x and note that it must be less than 60R. Let's call the internal resistance in series with this parallel combo y (and I use R for Ohms).

So we know that yR + xR with a constant battery voltage generates a current of 70mA.

When we remove the 120R we are left with only 60R as the circuit load resistance. So now, the internal resistance yR + 60R generates a current of 500mA? But we know xR must be less than 60R, so the total circuit resistance must have increased, which should cause a decrease in current for a constant voltage.

Are the two current values (70mA and 500mA) correct or is my logic wrong?

Anyhow, to help solve this problem, use a drawing and simultaneous equations!

 

[gneill]

Something's fishy with the currents. If you increase the load resistance (by removing a parallel path -- the 120 Ohm resistor), the current through the circuit should decrease, not jump from 70 mA to 500 mA.

 

[Blu3eyes]

What is the equivalent resistance combination of the 60R||120R load?

Lets call it x and note that it must be less than 60R. Let's call the internal resistance in series with this parallel combo y (and I use R for Ohms).

So we know that yR + xR with a constant battery voltage generates a current of 70mA.

When we remove the 120R we are left with only 60R as the circuit load resistance. So now, the internal resistance yR + 60R generates a current of 500mA? But we know xR must be less than 60R, so the total circuit resistance must have increased, which should cause a decrease in current for a constant voltage.

Are the two current values (70mA and 500mA) correct or is my logic wrong?

Anyhow, to help solve this problem, use a drawing and simultaneous equations!

Something's fishy with the currents. If you increase the load resistance (by removing a parallel path -- the 120 Ohm resistor), the current through the circuit should decrease, not jump from 70 mA to 500 mA.

Sorry folks. My bad. It actually is 700 mA and then drops to 500 mA.
Thanks for helping me.

 

[gneill]

Sorry folks. My bad. It actually is 700 mA and then drops to 500 mA.
Thanks for helping me.

Ah. That change makes sense. So, are you good? Have you worked through the problem?

 

[Blu3eyes]

Ah. That change makes sense. So, are you good? Have you worked through the problem?

I understand the problem now, but still something is not clear to me.
I figured out x=40 R then,
I(y+40)=I(y+60)
Iy+40I =Iy +60I
0 =20Iy
this is where I got stuck, I am not sure I'm heading the right way

 

[SammyS]

It's also true that V_terminal=I·R_load .

So, EMF = I·R_load + I·r_internal

 

[gneill]

Okay. If y is your internal resistance, and if 40 Ω is the equivalent of the 60 Ω and 120 Ω resistors in parallel, and if we let E represent the internal battery voltage, and further, if I1 and I2 are the two currents noted for the two cases, then you've got the two equations:

E = I1(y + 40 Ω)
E = I2(y + 60 Ω)

Can you run from there?

 

[Blu3eyes]

Okay. If y is your internal resistance, and if 40 Ω is the equivalent of the 60 Ω and 120 Ω resistors in parallel, and if we let E represent the internal battery voltage, and further, if I1 and I2 are the two currents noted for the two cases, then you've got the two equations:

E = I1(y + 40 Ω)
E = I2(y + 60 Ω)

Can you run from there?

Thanks, I've got it. I totally forgot that I was given the two currents, so I was stuck.
After solving everything, ri=10 \Omega.
emf=35 V
Thank you folks!