I am stuck on a problem with a simple electrical circuit. (I am studying independently, and so I'm not trying to cheat on a homework). The problem is sort of shaped like this: [-]. It has a 2.0 volt battery at the top with an internal resistance of 1 ohm. It has a 1.0 volt battery at the bottom with an internal resistance of 2 ohms. And across the middle (parallel to both batteries and also in between them) is a resistor of 2 ohms. I need to find the current across the 2 ohm resistor. I know the answer to be 0.42 amps, but I have know idea how to get it. Please can you help?
2007-02-18 06:52:48 · 7 answers · asked by eazylee369 4 in Science & Mathematics ➔ Engineering
The positive of the top battery flows into the negative of the bottom battery and visa versa
2007-02-19 08:36:09 · update #1
I'm assuming that the positive poles of the two batteries are connected to each other, rather than the positive pole of one connected to the negative pole of the other. If this is not the case, then my answer is unfortunately not valid, but it should give you an indication of how to proceed.
If the current through the upper battery is i and that through the lower battery is j, then the current across the middle is i+j. (One of Kirchoff's Laws).
Applying Kirchoff's other law to the upper and middle limbs gives
2 = i + 2(i+j) = 3i + 2j
and to the lower and middle limbs gives
1 = 2j + 2(i+j) = 2i + 4j
Multiply the first equation by 2 and the second by 3. Then subtract.
-1 = -8j
j = -1/8 amp
Substitute this in the 2nd equation to get 1 = 2i - 1/2
i = 3/4 amp
Current across middle is i + j = 3/4 - 1/8 = 5/8 amp. Not quite the answer you have. Perhaps one of the batteries was the other way round after all.
Hope this has helped.
2007-02-18 08:17:53 · answer #1 · answered by Anonymous · 1⤊ 0⤋
.do u mean 0.42amps through the 2 ohm resister in the middle if so
Then voltage across the middle resistor is 0.84 volts
amps through top circuit = 0.84 towards negative terminal
amps through bottom circuit = 0.42 away from positive terminal
current through middle resistor flows left to right and of course 0.42amps.
Using Kirchhoff's first law then ignoring your information 0.42amps in middle resistor I make it 0.6 amps top circuit
0.2amps bottom circuit and 0.4 amps through Central 2ohm resistor.
this assumes of course that the +ve of 2v battery is connected to
-ve 1v battery.
2007-02-19 11:39:51 · answer #2 · answered by mad_jim 3 · 0⤊ 0⤋
You need to make sure we know the circuit. Are the cells in parallel or series? Current flows THROUGH a resistor NOT ACROSS IT. Ask the right question, and you will get a good answer.
2007-02-18 09:53:23 · answer #3 · answered by wrightcharliey 1 · 0⤊ 0⤋
the galaxies rotate at speeds inconsistent with their obvious mass is with the help of the fact we do see all of it. i'm relating it as being the theoretical dark count number. There are very reliable proofs that exhibits that dark count number exist. One the is the inconsistent speed of and obvious mass. dark count number makes up approximately seventy 5% to 80% of the situation in the Universe...
2016-09-29 07:06:31 · answer #4 · answered by ? 3 · 0⤊ 0⤋
Ohm i really wouldn't know m8 but is Ohms law involved?
2007-02-18 06:56:58 · answer #5 · answered by scrambulls 5 · 0⤊ 0⤋
could u email me the scheme of the circuit?
i believe i could solve it for u.
asal_barakchi@yahoo.co.uk
2007-02-18 18:29:23 · answer #6 · answered by asal 2 · 0⤊ 0⤋
yeah lee, i got ur battry and u got mine!
2007-02-18 06:56:36 · answer #7 · answered by slip_knott 1 · 0⤊ 1⤋