I asked this question before and was told the answer is 6V butthe professor said

Question

I asked this question before and was told the answer is 6V butthe professor said the answer is 2V. My question really iswhy is 6 V wrong? A circuit consists of a 6 V battery connect to a 4ohm resistor that is in series with a 6 ohm resistor. Atpoint X a 3 ohm resistor is connected to the circuit between the 4ohm and 6 ohm resistors and at point Y the parallel circuitrejoins the main circuit after the 6 ohm resistor.                              4 ohmresistor            X                      battery  (6V)                                             3ohm resisor                           6 ohm resistor                                                                   Y Why is 6 V incorrect? I asked this question before and was told the answer is 6V butthe professor said the answer is 2V. My question really iswhy is 6 V wrong? A circuit consists of a 6 V battery connect to a 4ohm resistor that is in series with a 6 ohm resistor. Atpoint X a 3 ohm resistor is connected to the circuit between the 4ohm and 6 ohm resistors and at point Y the parallel circuitrejoins the main circuit after the 6 ohm resistor.                              4 ohmresistor            X                      battery  (6V)                                             3ohm resisor                           6 ohm resistor                                                                   Y Why is 6 V incorrect?

Explanation / Answer

Have you considered to analyze the network using the water flowanalogy? In the analogy, water flows through a canal. When branches divertfrom the main canal, their width determines whether the branch can carry the same amount of water,this is the analogy to current. The slope of the branch tells you whether the water is forcedfaster or slower through the branch than the through the main canal, this is the analogy tovoltage difference at the ends of each branch. Here, the main canal has a certain slope, according in its value to6V. Then, first the canal narrows due to the 4 resistor and thenfurther due to the 6 resistor. Ohm's law claims, in this analogy, that the same water flow mustsomehow come out at the other end than what was fed in at the beginning. Since Ohm's law states that V=I*R it is clear that V cannot be thesame when R changes, unless there is a way for I changing accordingly. For each point where branches divert (ie one branch comes in butmore than one go out - a parallel network), the current that flows in must equal the total currentthat flows out. For your network this means: You have to first find the R_eqresistance of the whole network to then use it to find the TOTAL current due to the 6V. Then you can usethat now known I_tot to apply it to the points where the current splits up into thebranches. The ratio of the resistances of the branches tells you how much current is drawn into eachbranch. On the other hand, from what we found out above, we already knowthat the current that flows through a diversion like that at X in sum must be equal to what flows in,so I_tot does not change, it just splits up. But the resistance of the branch seen for itself isdifferent from the resistance through which the current was coming in. Ohm's law thus requiresthat this can only be true, if the 'water' (current) flows in through a steeper or shallowerslope (voltage). Hope this helps. Be careful not to take the water analogy beyondsimple ohmic resistor networks. Like every analogy its range of validity is limited.

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