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How do voltage and voltage drops over a circuit relate to work done? The Volt unit is energy normalized to unit charge; Joule per Coulomb. Since the Amp unit is Coulomb per second, the product of the voltage across and current through a circuit element is the power associated with the circuit element. For a DC circuit, voltage and current are constant ...

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But then that means that the electron in the 5 ohm circuit would have done 5x the amount of work (or work done on it) of the 1 ohm circuit over 5x the duration. You're confusing work with power here. Work has nothing to do with duration. If an electron crosses a potential difference of $V$ with any resistance in between, the work done is the same, ...

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Typically the field strength is proportional to the voltage, so to get a higher field strength you need to increase the voltage. To see why this is you start from the basic formula for the field strength: $$B = k N I$$ where $B$ is the field strength, $N$ is the number of turns and $I$ is the current in the coil. $k$ is a constant that we'll ignore for ...

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A few preliminary ideas which might help: It doesn't really matter what the speed of the electrons is - a current of 1 C/s (=1 A) just means that a coulomb worth of charge (equal to $6.2 \times 10^{18}$ electrons) passes each point in the circuit each second. Perhaps there is one electron travelling so fast that it does $6.2 \times 10^{18}$ laps of the ...

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Voltage is similar to height. It plays the same role for electric charge as height*gravity does for a ball on a hill. So high voltage means high potential energy the same way a ball being high up on a hill means high potential energy. Voltage is not potential energy, the same way height is not energy. However, if you have a certain amount of charge $q$, you ...

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Why does a capacitor charge only upto the voltage of the source? Step by step: (1) When the capacitor voltage equals the source voltage, the voltage across the resistor in the series RC circuit is zero (2) By Ohm's Law, the current through the resistor must be zero too. (3) Because it is a series circuit, if there is zero current through the ...

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Here's how I would convince myself of the correct answer. Draw a circuit diagram showing the voltage source, resistor, and capacitor. (I assume it's in a simple series circuit?) Next, write out Kirchhoff's loop rule. You should find something like $V_\text{source}-V_\text{resistor}-V_\text{cap}=0$. Note that this equation is true at any time, not only for ...

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What kills you is the current not the voltage, as you read on your books. Of course that you to have a voltage difference so the current can flow, but it does not determine how strong the current will be. I do not know if I would die if I touch something with 1 kV. That's because the current will depend on the sum of the resistance between me and the ground ...

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I always thought this rule of thumb was a bit silly - current kills because it was driven by a voltage, otherwise there would have been no current. The rule arises because of the variability of skin resistance. Little voltage applied internally across your heart will kill you, but the skin's variability means that it is impossible to say what external ...

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The current you are going to get through your body depends on the voltage and on the resistance. You can touch a 110 V exposed cable using a piece of metal or a piece of plastic - in both cases the voltage is the same, but the resulting current - and hence the danger - is greater in the first (metal) case.

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Wires are in fact resistors, but with VERY VERY tiny amounts of energy being thermally dissipated by the current due to EXTREMELY low drops in voltage over large lenths of the wire. Thus, current DOES flow through the neutral wire, but the drop in potential along a length is literally far too small for your voltmeter to detect. Review Kirchoff's Voltage Law ...

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steps: emf= change in magnetic flux/ time Therefore, emf=(BA)/time Magnetic field strength B is constant. So, we just have to find change in area by time The length of AC will increase by 0.6m per second. As six second passed, AC=0.6*6. The BC can be calculated, BC= ACtan19 At the beginning, area is zero. After 6 seconds, area=0.5AC*BC. Thus, change is ...

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Draw the circuit using ideal circuit elements: Now, the series current is: $$I = \dfrac{\mathcal{E}}{R_{internal}+ R_{load}}$$ The voltage across the internal resistance is: $$V = \mathcal{E} \dfrac{R_{internal}}{R_{internal}+ R_{load}}$$ The power dissipated by the internal resistance can by found three equivalent ways: P = VI = ...

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The analogy is wrong. A voltage source can only shock us if it is able to pass a considerable amount of current through our body ( ~ 250 mA or so, I dont know the exact value but you can Google it ). The circuit that you are trying to discuss, does indeed have 36 Amps of current flowing through it, but once you connect yourself to the circuit, you are in ...

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well basically your missing some additional equations. The current coming out of the $10 \Omega$ resistor is the same as the one going into it. Hence $I_3$ in your equation would be just $0.5A$. Similar thing for $I_1$: The current going through the $20 \Omega$ resistor is $0.2A$. Note that the power supply also doesn't alter the current only the voltage. ...

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