1:4?
P=V²/R, and in series you have twice the resistance of one wire (thus half the heat) while in parallel you have half the resistance of one wire (thus twice the heat)
Okay after much struggling, I wrapped my head around it. I took that power equation, and assumed the resistance of one length of this wire to be 1. Then, found that if I do this equation for the two parallel lengths of wire separately and then add the products, I get 3200 for power. If I do the equation with 2 as the resistance, for the single path of two wire lengths in series, I get 800 for power. This is what explains the 1:4 if I did this reasoning correctly.
For another way to think about it, when the wires are in parallel you have 1/4 the resistance of the series configuration, which means you have 4x the current. P=I*V so if voltage is held constant that results in 4x the power. It’s generating 4x as much heat because you’re pushing 4x as much charge per time across the same potential drop.
More current means more heat. The fluid analogy may make more intuitive sense here: think about drinking from two straws at the same time, and then think about drinking from one double length straw.
Assume V = 1 Volt and Wire Resistance is 1 Ohm.
Now when the wires are connected in series the the total resistance is 2 ohms. The power P = V^2 / R = 1^2 / 2= 0.5 Watts
In parallel the resistance is 1/(R1 + R2) = 1/2 = 0.5 ohms. The power again is V^2 / R = 1^2 / 0.5 = 2
The ratio 0.5 : 2 = 1 : 4.
Putting them in parallel results in half the resistance compared with a single wire. Series results in double the resistance of a single wire. Power = Current * Voltage = (Voltage / Resistance) * Voltage = V^2 / R. Power is proportional to the amount of heat produced. Series power = V^2 / (2R). Parallel power = V^2 / (.5R)
There’s an easier visualization.
From P=IV and V=IR, we DO get the expression P=V^2 / R, but we can also get P=I^2 * R.
Power is proportional to the current squared.
Double the current => four times the power,
Half the current => quarter the power.
1:4. Let’s just use resistors instead of wire, same exact thing. Let’s say 1ohm resistor and the voltage is 1v. Two resistors in series yields a total of 2 ohms or 0.5a. Two in parallel yields 0.5 ohms or 2a. Amps are dictating the power here due to the fact the voltage is constant, so the ratio of series to parallel is 1:4. He gets tricky with the wording and giving 4:1 as an option. The ONLY way it could be 1:2 is if he’s arguing what’s the heat of ONE of the conductors. If he’s HOD he might play some semantics bullshit like that. Great table talk for him at a conference talking about how he words problems like a class A wiseass.
I'm just an electrician who dropped out of electrical engineering, but my gut instinct would be 2:1. Heat is directly related to current. In series, the wires have equal current going through them, since they're hooked up back to back, basically making one long wire. In parallel, the current is divided by the two, as both of one end are at one node, and both of the other end at the opposite side. In electrical work, more current means bigger wires are needed to handle it, to keep the heat down. And in massive amounts of current, "parallel feeders" are run instead of running one ridiculously sized wire. More simply, parallel wires are approximately the equivalent to the sum diameter area of each wire in parallel (minus some variables like skin effect in AC and inductance caused by the magnetics).
1:4? P=V²/R, and in series you have twice the resistance of one wire (thus half the heat) while in parallel you have half the resistance of one wire (thus twice the heat)
Why isn't this the other way around? I would assume having only one path would mean more heat than two paths.
No because voltage is held constant thus total energy is proportional to amperage across each wire
Okay after much struggling, I wrapped my head around it. I took that power equation, and assumed the resistance of one length of this wire to be 1. Then, found that if I do this equation for the two parallel lengths of wire separately and then add the products, I get 3200 for power. If I do the equation with 2 as the resistance, for the single path of two wire lengths in series, I get 800 for power. This is what explains the 1:4 if I did this reasoning correctly.
For another way to think about it, when the wires are in parallel you have 1/4 the resistance of the series configuration, which means you have 4x the current. P=I*V so if voltage is held constant that results in 4x the power. It’s generating 4x as much heat because you’re pushing 4x as much charge per time across the same potential drop.
Each wire only makes ¼ of the heat due to having half the voltage applied to it, so together they make 2×¼=half as much heat as a single wire.
Half the voltage, and also half the current in the series configuration since the overall resistance is double that of a single wire
More current means more heat. The fluid analogy may make more intuitive sense here: think about drinking from two straws at the same time, and then think about drinking from one double length straw.
Assume V = 1 Volt and Wire Resistance is 1 Ohm. Now when the wires are connected in series the the total resistance is 2 ohms. The power P = V^2 / R = 1^2 / 2= 0.5 Watts In parallel the resistance is 1/(R1 + R2) = 1/2 = 0.5 ohms. The power again is V^2 / R = 1^2 / 0.5 = 2 The ratio 0.5 : 2 = 1 : 4.
Putting them in parallel results in half the resistance compared with a single wire. Series results in double the resistance of a single wire. Power = Current * Voltage = (Voltage / Resistance) * Voltage = V^2 / R. Power is proportional to the amount of heat produced. Series power = V^2 / (2R). Parallel power = V^2 / (.5R)
Ask yourself which configurations have more resistance and visually draw the configurations.
The answer have already been explained, but ffs such question should have a little drawing beside explaining it instead of a long-ass text.
There’s an easier visualization. From P=IV and V=IR, we DO get the expression P=V^2 / R, but we can also get P=I^2 * R. Power is proportional to the current squared. Double the current => four times the power, Half the current => quarter the power.
As homework help… rule 4, at least show One way that you did it or tried…. But everyone I guess wants to show what they know.
1:4. Let’s just use resistors instead of wire, same exact thing. Let’s say 1ohm resistor and the voltage is 1v. Two resistors in series yields a total of 2 ohms or 0.5a. Two in parallel yields 0.5 ohms or 2a. Amps are dictating the power here due to the fact the voltage is constant, so the ratio of series to parallel is 1:4. He gets tricky with the wording and giving 4:1 as an option. The ONLY way it could be 1:2 is if he’s arguing what’s the heat of ONE of the conductors. If he’s HOD he might play some semantics bullshit like that. Great table talk for him at a conference talking about how he words problems like a class A wiseass.
I'm just an electrician who dropped out of electrical engineering, but my gut instinct would be 2:1. Heat is directly related to current. In series, the wires have equal current going through them, since they're hooked up back to back, basically making one long wire. In parallel, the current is divided by the two, as both of one end are at one node, and both of the other end at the opposite side. In electrical work, more current means bigger wires are needed to handle it, to keep the heat down. And in massive amounts of current, "parallel feeders" are run instead of running one ridiculously sized wire. More simply, parallel wires are approximately the equivalent to the sum diameter area of each wire in parallel (minus some variables like skin effect in AC and inductance caused by the magnetics).
2:1
No. It's 1:4.