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In general, if the power consumed would depend on the circuit structure. But for a simple case, such as two resistors connected in series versus the same resistors connected in parallel (with identical voltage sources in both), the power dissipated in the parallel combination will be greater.
5. The trick is to look at the nodes in the circuit. A node is a junction in the circuit. Two resistor are in parallel if the nodes at both ends of the resistors are the same. If only one node is the same, they are in series. So, R1 and R2 are in parallel and R3 is in series with R1||R2. Share.
Two resistors on the same path are in series. There must be nothing else in between. Two resistors that follow parallel paths are in parallel. If the wire splits in to paths and gatheres into one again alter, this would be parallel paths. In your circuit, you must piece by piece collect resistors into one equivalent resistor.
Current = amount of water flowing through pipe. Voltage = pressure of water. Power = water pressure x water flow (voltage x current) Resistors = constrictions in pipe. Pressure (voltage) drops occur across them. Any point in a (series circuit) pipe has the same amount of water flowing past that point as any other.
Which can be solved for Rtot R t o t: Rtot = ⋯ = 1 1 R1+ 1 R2+ 1 R3 R t o t = ⋯ = 1 1 R 1 + 1 R 2 + 1 R 3. Finally, we get the same total resistance we know we would get if R1 R 1, R2 R 2 and R3 R 3 were in parallel. Therefore, the two circuits are equivalent. Well, I guess this post turned itself into a reminder why you shouldn't try this ...
0. Say we have a series circuit (containing a battery) with charges flowing through it. We add a resistor to the circuit. The current should then decrease. If we add a 2 ohm resistor, the current would become lower than if we added a 1 ohm resistor. Then say we add another resistor of 2 ohms. Based on what I have read, current is the same ...
Manipulating this to get V/I in terms of R 1 and R 2 will get you the formula for parallel resistors, Req = V I = 1 R e q = V I = (1 R + 1 R) − 1. When two resistors are in series, the current through them is the same, but the voltage across them might be different. The voltages add to give the total voltage: V = IR + IR V = I R 1 + I R 2.
$\begingroup$ @john: (in series resistors) current is the charge passing through a certain area per unit time. let this certain area be the resistance R1, R2, R3 and R4. now if we see theoretically, all the resistors should offer some resistance so, shouldn't the current through a certain area (R1, r2, r3 or r4) be different as all of them ...
If the resistors were parallel connected such that they have identical voltage across, then the above is correct (if you change greater to less). But the resistors are not parallel connected. Instead, they are series connected which, by definition, means that they have identical currents through and, thus, the voltage across R1 is greater than ...
Current is not just the speed of the flow. It is the amount of charge passing a given point per unit time. If it is not the same amount all around the loop forming the circuit then that means the charge must be building up somewhere, and falling away somewhere else.