# Terminal Voltage of unknown battery

1. Mar 23, 2005

### hkhandrika

The problem is:

The current I through the 4.0 k-ohm resistor in Fig. 26-42 is 6.50 mA. What is the terminal voltage Vba of the "unknown" battery? (There are two answers.)

The two answers depend on the direction of the current. Here is the picture:

http://home.san.rr.com/hkhandrika/26-42.gif

First I looked at the 4.0 k-ohm resistor and said that the 8.0 k-ohm resistor had 1/2 the current flowing thru it, then I added up the parallel resistors to get one that was (8/3)k-ohm with 9.75 mA of current going thru it.

I thought I could use this current and then say that the voltage at point B (if you go left) would be 9.75mA * the 8/3 resistor + 9.75 * the 5 k-ohm resistor. But then how would you figure out the voltage at point A? If you go right, could you then do 9.75mA * 8/3 resistor - 12V and set that equal to Va, and then add 9.75 mA * the 5 k-ohm resistor and say that's the voltage at B? I tried this, and my answer is incorrect. I am doing this on webassign so I can see whether my answer is wrong or not, but I dont have the solutions to this problem.

Am I doing the process correctly in the first place? Any help would be greatly appreciated.

Thank you for your time and effort.

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2. Mar 24, 2005

### stunner5000pt

for the 4Kohm and 8kohm resistors the CURRENT IS NOT THE SAME

they are in PARALLEL so the curretn will split up (but not equally!). However there is something that is equal between the two of them. What is it?
V = iR, must be on these variables right?? .
Now add up the resistors in parallel, and mkae one resitor. SO now you have two reistors in series with the 5 and the equivalent (4-8 kohm resistor). WHen ressistros are in series what is equal amongst them??

now around the circuit and form the equations which you need

3. Mar 24, 2005

### hkhandrika

If the current in the 4-kohm resistor is say 10 A, then the current in the 8kohm resistor should be 5 A shouldn't it? Also, how can u figure out how much current is going thru the sum of the parallel resistors if you simply add them up. Thanks for all your help. I will look into it shortly... need to grab some lunch first :tongue2:

4. Mar 26, 2005

### hkhandrika

Ok, sorry for seeming like someone new to the field, but maybe i guess i just made this problem harder than it should be.

Since the resistors are in parallel, the voltage at the point before the resistors and after the resistors is the same, so the voltage at all the resistors in parallel are the same. So I got that voltage from simply doing current through resistor A*resistance of A. Then I thought ok if you move right you have a battery which you traverse from plus to minus, so you have negative voltage difference so you would subtract 12V (correct?). Do you keep going like this?

The thing that got me was that terminal voltage is measured as

Vab = emf - Ir

But how do you find the emf, the current, and r? Is r just the 5.0 k-ohm resistor. If that's the case, then is the 1.0 ohm resistor = r for the 12 V on the bottom?

Last edited: Mar 26, 2005
5. Mar 26, 2005

### ramollari

hkhandrika - Is the direction of I given in the question? The value you find for $$V_{ab}$$ when the current is flowing left is far different from the value you find for it when the current is flowing right.

If you know the direction, then the solution is straightforward. Simply apply Kirchoff's rule for one loop only (first try to find the voltage across the resistors in parallell).

6. Mar 26, 2005

### hkhandrika

Ramollari,

Thank you for your advice. There are indeed 2 answers to the question. One is for the current traveling left and the other for the right. The voltage for the resistors in parallel is 4.0*6.5 = 26 V. When the parallel resistors are put together we get 1/4 + 1/8 = 3/8 so the final resistor is (8/3) k-ohm. The current in the final resistor is then 6.5 + (26/8 = 3.25) or 9.25 mA. Is this the same current through the 1 ohm resistor (if you travel right)?

7. Mar 28, 2005

### plusaf

piece of pie... easy as cake....

first, pick a direction for the current through the 4k resistor. let's start with "right-to-left", or from the positive side of the 12 v battery, towards the 5k resistor.

6.5mA through a 4k resistor, does, indeed, produce 26v. in this first case, the left side of the 4k resistor would be 26v lower than the right side, so if you take the negative side of the 12v battery as neutral, or ground reference, you go up 12v to the right side of the parallel resistors, then down 26v to the junction of the 5k and the 4k-8k parallel combo. so the right side of the 5k resistor is at -14volts.

since the 4k and 8k resistors are in parallel, the voltage across them are the same, and by Ohm's law, the current through the 8k is 1/2 that through the 4k. the total current is 6.5 + 3.25mA.

now, if the left side of the 12v battery is ground and the right side of the 5k resistor is at -14v, and there are 9.75mA flowing counter-clockwise through the other 6k worth of resistor (yes, is doesn't matter if the 6k comprises one, two, or a thousand resistors.....) what's the unknown voltage. hint: E=I*R

if the current through the 4k resistor is flowing in the opposite direction, the right side of the 5k resistor will be at what voltage, relative to the negative side of the 12v cell, and what will the unknown battery's voltage be then?

cheers!
+af

8. Mar 28, 2005

### tutor69

where I1 is current through 4k and I2 is current through 8 K resistor. Since I1 = I2/2 Therefore I = 9.75 mA
Now assume the current flowing in anticlockwise direction through the loop The parallel combination gives 8/3 K resistor. For this direction apply KVL and get Vba= 96.5 V
and for reverse direction the answer would be 72.5 V

9. Mar 29, 2005

### hkhandrika

Got the answer you guys! Woohoo! Thanks everyone for the help. This helped me understand circuits which is really good since we have a midterm coming up.

10. Apr 5, 2005