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Voltage, otherwise known as electrical potential difference (denoted ∆V and measured in volts) is the difference in electric potential between two points (adapted from Wikipedia). Voltage can be constant (DC) or varying (AC).

Your assumption is wrong. A high inrush current does not mean there's a high voltage. In electronic circuits, inrush current is caused by capacitors charging up. In a light bulb, the filament's … resistance is lower when the bulb is cool, which allows more current at startup. The input voltage from the AC power line stays at 240 VAC. EDIT: Here's an example circuit. I'm using DC to make it simple …
answered May 21 '15 by Adam Haun
Voltage represents electric potential energy. It's relative in the same way that gravitational potential energy is. Consider the example of a ball on a hill: What's the gravitational potential … ball will always gain more energy if you push it to the right than if you push it to the left. Voltage is the same way. Instead of a gravitational field, there's an electric field. Instead of …
answered Dec 3 '16 by Adam Haun
This circuit is supposed to act like an OR gate. You need the resistor for the diodes to work correctly, especially in the off state. \begin{array}{|c|c|c|c|} \hline \rm A & \rm B & \rm D1 & \rm D2 …
answered Apr 3 '18 by Adam Haun
[I'm ignoring non-ideal behavior since that doesn't seem to be what you're interested in.] Your assumption is wrong. Measuring the voltage drop across the resistor does measure the battery voltage … . The battery voltage and the resistor voltage are the same in your circuit. The general rules are: Components in parallel share the same voltage Components in series share the same current Ideally …
answered Jun 17 '15 by Adam Haun
kilovolt voltage source, you can rip apart air molecules, creating lots of free electrons! A less extreme example would be a vacuum tube, which "conducts" through empty space $(R = \infty)$ using electrons … electrodynamics and physical chemistry, and should not be treated as such. To answer your question more directly, regardless of whether a tiny current flows through the air, there can definitely be a voltage
answered Apr 23 '15 by Adam Haun
spends more time being off. Eventually, the inductor current will rise enough to deliver more average current to the load, which increases the voltage. But that takes time. Right after you increase the …
answered Sep 2 '15 by Adam Haun
: simulate this circuit – Schematic created using CircuitLab It looks like you measured the voltage across the mystery resistor correctly, then mistakenly measured the voltage across both other … resistors. Then you measured the voltage across all three resistors. What you want to do is measure the voltage from the positive battery terminal to point (a), then from (a) to (b), then from (b) to (c). …
answered Mar 7 '16 by Adam Haun
Voltage and current are fundamentally different things. They are not just different aspects of power. Is it possible to have a voltage difference without a current, and (in a superconductor) a … current without a voltage difference. As others have said, electrical components constrain the voltage-current relationship, but that doesn't mean we can only care about power. For one thing, it's a lot …
answered Aug 11 '18 by Adam Haun
The emitter voltage is limited by the base voltage since ~Vbe must be greater than ~0.6V to turn on the transistor. You've made an emitter follower, not a switch. If you want to switch the full voltage, use a PNP transistor instead. …
answered Jan 2 '15 by Adam Haun
VCC is an odd choice for a CMOS chip. I think the C is supposed to stand for "core". As Lorenzo says, it's the core voltage. Personally, I'd have called it VDD and named the external supply pin …
answered Jan 6 '17 by Adam Haun
After it's fully charged or discharged (i.e. at DC), a capacitor acts like an open circuit. That means there's no current through R3, which means there's no voltage across R3. That's just Ohm's Law. Since $V_{R3} = 0$, $V_C(0^-) = V = 12 V$. …
answered Mar 28 '15 by Adam Haun
You're mixing up conventions here. If you assume the charge carriers are negative (electrons), then the negative terminal of the battery is the high-potential terminal. Electric charge has a sign, not …
answered Feb 21 '16 by Adam Haun
is highly resistive, so the capacitor voltage and charge stay roughly constant. For someone with no physics background whatsoever, the pressure analogy is probably the best you're going to get …
answered Mar 18 '15 by Adam Haun
We haven't yet covered the general case where R1 and R2 are different. You can use nodal analysis or a Thevenin equivalent, but you can also take a more mathematical approach and use plain KVL. Each p …
answered Jun 3 '15 by Adam Haun
: The current on the secondary side is $1\cdot\frac {N_P}{N_S}$ amps. The voltage on the secondary side is $1000 \cdot\frac {N_S}{N_P}$ volts. The apparent input impedance on the primary side has a … actual load impedance on the secondary does not -- it's the physical impedance connected across the wires, what you're calling "appliances". But just as with Ohm's Law, if you know the voltage and current …
answered Sep 9 '15 by Adam Haun

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