# Tag Info

18

The identity $$V = K \frac{dV}{dt}$$ is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, ...

6

This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between. In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between: $$\dot V=-\frac1\tau V,$$ where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, ...

5

Generally, no. You've written two equations. The first relates voltage and current for an isolated resistor. The second relates voltage and current for an isolated capacitor. Given only those two expressions, there's no reason at all that they can or should be combined. That is, you have no circuit, only isolated components. There are principles for ...

3

It is quite easy to use a step-up converter to generate almost any voltage. The question is "voltage at what current". The power a battery can deliver is finite (power = voltage times current), but you can convert voltages in many different ways. The most obvious is an oscillator (inverter) followed by a transformer and a rectifier, but there are more ...

3

The $KV$ in $30KV$ refers to the motor velocity constant $K_v$. It does not refer to kilovolt $kV$, cf. comments by John Rennie.

3

The simple reason is that you have only passive, linear components in your circuit; when you push with a certain voltage, you get a certain current response. If the voltage cycles a certain number of times per second, that's how often the current must respond. Think about it the other way - if you _didn't_have the same frequency, what response would make ...

2

First, the reactive power is not dissipated, but which corresponds to power delivered by the power stored in the reactive component (inductor or capacitor) during a semi-cycle; in the next half cycle, the component returns the stored energy to the source. For this to occur, the component must have the ability to store energy. In the case of a capacitor, the ...

2

You are massively overthinking the problem. The collector current is given (by the diagram) to be 150x the base current. The sum of base and collector current has to flow through the emitter... That's all you need to solve this. In particular, a current source will look to a circuit like "whatever resistance" it needs to be in order for the correct current ...

2

Current flows in a closed loop, so $$I_c = I_r\tag1$$ There is a voltage across the capacitor due to the integrated charge, across the resistor due to the instantaneous current, and across the whole loop due to the motion of the resistor. The net voltage around the loop is zero, so $$V_c - I_r R + \mathcal{E}=0\tag2$$ The force on the wire is given by ...

2

You're quite correct that there will be some current flowing, so there must be a voltage drop due to the internal resistance of the battery. The EMF measured by any voltmeter will always be less than the true EMF. If the internal resistance of the battery is $R_b$ and the resistance of your voltmeter is $R_m$ then the voltage you measure will be: $$V = ... 2 The 3 A that leaves the battery depends on two things: the amount of supply voltage, and the impedance (here resistance) that draws this current. So the second part simply says: The 3 A current that leaves the voltage source = The voltage source divided by the total resistance (which is the internal resistance, plus the equivalent of the parallel ... 2 AC has each wire positive half the cycle and negative the other half. It will charge the capacitor on one half cycle and discharge it on the other half. The net charge will be zero. 2 Of course you can charge a capacitor with AC. The problem is that you keep changing how it is charged. While you apply a positive voltage to one plate, it will get a positive charge; half a cycle later, it will attempt to get a negative charge; and so it continues. The capacitor is always a little bit behind - as your AC voltage is changing, the capacitor ... 1 Actually V=0! Notice that the left branch shorted the right one, because the points A & C are the same, thus Va = Vc. 1 In your diagram, the devices appear to be connected in parallel and V appears to be voltage across the left-most wire which must be zero. Since you've designated the devices with the letter D, I assume they are diodes but diodes aren't drawn as rectangles so I honestly can't write much more about this. As I'm typing this, I see a comment to the effect ... 1 Like any set of equations in physics, this is true when the conditions stated apply. Here, you are using two equations:$$ V(t) = RI(t) $$when the current passes through an element that is a perfect resistor, and$$ C \frac{dV}{dt} = I(t) $$when the current passes through an element that is a perfect capacitor. So, in a simple RC circuit consisting of a ... 1 Well I imagine you know how to add resistors in parallel:$$R = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n}}$$Of course if you try to put a resistor with R=0 in there you run into \frac{1}{0}. You can probably take a limit and show that the equivalent resistance is 0 if there's a zero resistor in parallel. This makes sense. If current is ... 1 One approach to gaining insight to the answer is to note that, for two parallel resistors, the equivalent resistance is$$R_{EQ} = \frac{R_1R_2}{R_1 + R_2} = \frac{R_2}{1 + \frac{R_2}{R_1}}$$as long as either resistance is not equal to zero. Now, let R_2 go to almost zero while holding R_1 constant. When R_2 \ll R_1, the equivalent resistance is ... 1 The answer-as requested by HDE. So my foolish misconstruct of the electric field is that since it accelerates charged particles more or less depending on the distance from the source, it should create less motion in electrons a further distance from the source. I'm not sure why I thought resistance had much to do with it-on further analysis resistance ... 1 There is a "capacitor" of some kind holding the charge at one end of the wire - with capacitance C. It has initial charge Qo, and initial voltage Vo = Qo/C. In this relatively simple case, a discharging capacitor has an exponential rate of discharge. This website describes how to derive the exponential equation: ... 1 One equation is for resistive circuit and the other is for capacitive circuit. Two can not be merged together. 1 Yes. It depends on how you define the direction of positive voltage in your equivalent voltage source. In other words, if you switch the positions of the "+" and "-" on a typical diagram, the sign of the voltage will switch. 1 The answer depends on what you're trying to do. If you're trying to find the voltage between A and B, then note that the circuit can be redrawn as From this redrawn circuit, it's clear that the resistors need to be treated as being in series in the process of calculating the voltage between A and B. But if you're trying to find the Thévenin equivalent ... 1 "1) What does it mean for a voltage to be reflected bacK?" It means that the wave that was traveling in one direction on the transmission line is now traveling in the opposite direction. "What is physically happening?" Absolutely nothing, and that is exactly what is causing the reflection. The energy in the wave has nowhere to go at the end of an ... 1 Antimony is a donor in Germanium. That means that that 5th electron on the Antimony occupies a discrete level in the gap, not as part of the Germanium electronic bands. Since this level is close to the conduction band of the Germanium, it is highly likely that the electron will thermally get enough energy to jump into the conduction band. This leaves ... 1 In a perfectly ideal circuit without any resistive losses (even internal to the battery), energy will flow to fill capacitance with charge, and be stored as an electrical field but the transient flow will also create magnetic fields in any inductance. Without loss, the energy will bounce back and forth between the inductance and capacitance as magnetic and ... 1 The battery doesn't do any significant work if there's nothing connected to it. You can think of a battery with nothing connected to it as being a battery with an extremely large resistor connected to it. That's even essentially physically accurate, instead of just being an idealization, because although the resistivity of air is extremely high, it's not ... 1 You're just thinking 'upside down'. The 3 ohm resistor is just 1/4 the resistance of the 12 ohm resistor but, power is inversely proportional to the resistance. Thus, if you decrease the resistance by a factor of X, you increase the power by a factor of X. In this case, the resistance is decreased by a factor of 4 so the power is increased by a factor of ... 1 take only first 2 resistors and rest as x now the vertical resistor and your x will be in parallel, effective resistance would be Rx/R+x with series in horizontal resistor. Now equivalent resistance would be$$Req. = R+ (Rx/R+x).  take Req. as $x$ again Form quadratic equation and solve for $x$. This will be the answer.

1

You are increasing the number of electrons in the wire, but only by a very small amount. There's a somewhat clichéd but still excellent analogy for electrical circuits called the hydraulic analogy. In the hydraulic analogy the power supply is a pump, and the pressure is the voltage. The water represents the electrons, so the pressure generated by the pump ...

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