Solving for Vemf and Internal Resistance

AI Thread Summary
To find the emf and internal resistance of a battery connected to different resistors, two currents are measured: 3.98 A with a 110-ohm resistor and 1.11 A with a 400-ohm resistor. The voltages calculated from these currents are 438.7 V and 444 V, respectively, indicating the influence of the battery's internal resistance. The internal resistance affects both measurements, and it is important to treat the resistors as separate circuits rather than in series or parallel. Understanding the concept of internal resistance is crucial for solving the problem accurately. The discussion emphasizes the need to check calculations and apply textbook knowledge for clarity.
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When a battery is connected to a 110.-ohm resistor, the current is 3.98 A. When the same battery is connected to a 400.-ohm resistor, the current is 1.11 A.

Find the emf supplied by the battery and the internal resistance of the battery.





Equations:

Vemf = i*R

I'm not sure how to approach this since you get two voltages (438.7 and 444) from the different resistors.

Should I treat these resistors like they are in the same circuit? In series or in parallel?

Please help!
 
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pleasehelpme6 said:
I'm not sure how to approach this since you get two voltages (438.7 and 444) from the different resistors.
Yes, you'll get two different voltages, by using different resistors, because there is another resistance involved that is always there: the internal resistance of the battery.

(By the way, check your math regarding the voltages you calculated.)
Should I treat these resistors like they are in the same circuit?
Not the 110 and 400 Ω resistors, no.

But the battery's internal resistance is always there, so yes, the internal resistance is within the circuit in both cases.
In series or in parallel?
What do you think? (Hint: look in your textbook for "internal resistance" of a battery, and the answer should be clear.)
 
Thanks! that explanation makes a lot of sense. I'm sure I can figure the rest out from there!
 
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Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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