# Tag Info

1

It doesn't. For DC, the current flows in one direction. Charges will flow to the capacitor plate. In the very instant where the capacitor is connected to the battery, the current flows as if there was no capacitor. Only after a short while will a bunch of charges have been accumulated at the plate, starting to repel the incoming charges, lowering the ...

-1

Step 1: redraw the network so we can more clearly visualize the series/parallel relationships: *------*-------*-------* | | | | | 3ohm 6ohm | | | | | 6V *---*---* 3ohm | | | | 4ohm | | | | *----------*-----------* Step 2: note that there are ...

0

If the only load is the wire, such as a hot wire cutter, then supply voltage and wire resistance are all that matter, and both formulas are equivalent. But most of the time, wire size is chosen so that most of the voltage appears on the load. As an example, a toaster uses a power cord with wires which drop much less voltage than the heaters - if this were ...

-2

Note that 3 Ω on the left and 6Ω on the top are in parallel: 6 Ω (on the top) // 3 Ω (the one on the left) = 2 Ω Now, we replace 6 Ω//3 Ω with 2 Ω found. Then we have 2 Ω in series with 4 Ω (in the middle): 2 Ω + 4 Ω = 6 Ω Now, in the middle we have only 6 Ω and 3 Ω in parallel on the right: 6 Ω // 3 Ω (the one on the right) = 2 Ω 2 Ω is the equivalent ...

1

A Zener is not like a normal diode. A normal diode lets current flow in only one direction and needs to be installed in the correct direction. A Zener diode is placed in the opposite direction, against the flow of current. A Zener diode will prevent current from flowing until it reaches a certain voltage, depending on the diode rating. Once this critical ...

0

Even if the circuit has no resistance(which isn't possible as every wire has some finite non-negligible resistance $\rho L /A$), the source will have some internal resistance causing the circuit to have resistance, after which you can use Ohm's law to find the current. Though it isn't recommended to connect two ends of a battery together.(It reduces the ...

0

if you were to have say a 5V source connected directly to ground, no components in the circuit at all, you would need to add resistance for the current to flow? No, for sure current will flow. If nothing resists the electrons, they will keep speeding up. Current $I$ will keep rising. Power will keep increasing, $P=VI$. For really no resistance, ...

1

Two identical hoses with the same pressure difference between their ends will carry twice as much water as each does individually. So you get twice as much flow (current) for the same pressure difference (voltage), which is another intuitive way of thinking about Dave's Answer that it is more enlightening in this case to think in terms of conductances rather ...

2

This is more intuitive if you think in terms of conductance, which is the inverse of resistance (1/R). When you put two equal resistors in parallel, you double the overall conductance. Why? You are adding a second path for current to flow, so you double that flow. Unfortunately, we tend to speak mostly in terms of resistance, which makes the math a bit ...

3

Use kirchhoff's first law, so for two resistors in parallel: $$I_\text{total}=I_1+I_2$$ Then just use I=V/R $$\\\frac{V}{R_\text{total}}=\frac{V}{R_1}+\frac{V}{R_2}$$ The voltage across any component, whether it be across resistor 1, 2, or the whole parallel portion of the circuit, is the same. It just cancels out so you can divide both sides by V.

1

This is niether series nor parallel circuit, because of resistor 4. We have to use Kirchoff's laws. We will assume that both ends are connected to a battery to simplify our analysis. First, reformulating the loop rule, we get that since the potential difference between the top and bottom end is the same, all paths from the top to bottom end must end up with ...

1

What the voltmeter is reading in the top circuit would be, if you used a Kirchhoff's loop rule in the loop containing 2 resistors and the voltmeter, the difference in the potential difference across the top-right resistor and the bottom-right resistor. The loop rule calculation would look something like this ($A,B,C,\&D$ are the top-left, top-right, ...

1

$E=\frac{V^2}{r}t$ only holds if $I=\frac{V}{r}$, which only holds if the internal resistance $r$ is the only resistance in the circuit. Taking into account the load resistance $R$ we would get $I=\frac{V}{r+R}$, and hence $E=\frac{V^2}{r+R}t$ (assuming the battery's voltage is constant; otherwise it would become an integral). If you want to take the ...

2

Your analysis doesn't apply to what you are applying it to. $P=V^2/r$ is the power dissipated in the internal resistance of the battery. It is the power dissipated by the internal resistance if the battery is shorted. What you want is "the power supplied by the battery", which is the power dissipated in the external resistor: $P = V^2/R$, where care ...

1

In general the effective resistance of resistors in parallel can be calculated with, $$R_{eff} = \frac{1}{\sum{\frac{1}{R_i}}}.$$ When the ratio between any of the resistors, $R_i$, is an irrational number it will not be possible to use the method you mentioned. For methods, which can be used to calculate something, the most robust method is usually ...

3

It perhaps is not as rigorous as you want, but it is simple and intuitive: $$R = \frac{\rho L}{A}$$ Where $\rho$ is the resistivity, $L$ is the length, $A$ is the cross-section area. When you plug resistances in series, you "are" increasing $L$, and thus $R$ increases. If you put in parallel, you "are" increasing $A$, and thus $R$ decreases.

3

Resistance is Voltage per Current. $R = V/I$ or $V = RI$ if you like. So if you put two resistors in series, the voltage is that due to the first, plus that due to the second. You understand that perfectly. Now, what is the inverse of Resistance, Current per Voltage? Give it a name, call it Conductance, perhaps. $C = 1/R = I/V$ Then $I = CV$ if you like. ...

0

In the first one notice that the loop is being moved and thus the charged particles in the loop. Hence there are electrons moving in a magnetic field, which is basically Lorentz force. Griffiths argues that the Emf created in this scenario is due to, or better yet can be explained by the Lorentz force. In the second and the third one there is no charged ...

1

A rating of A filament lamp rated 12 V, 1.0 A means that it is designed to work at a voltage of 12V, and at this voltage, it consumes 1A current. a heating element rated 230V, 500W is intended similarly. It means that the element is designed to work at 230V, at which it consumes 500 W power. These sort of specifications made no sense to me ...

1

then shouldn't the variable resistor be connected to the negative terminal of the battery? It makes no difference at all where the variable resistor is connected. The current will be given by Ohms law. viz i = V/R. Regardless of where the resistor is located in the circuit shown the electrons pass through it in the same direction, and the same voltage ...

2

A negative charge attracts a positive charge and repels an also negative charge. If you have many (negative) electrons in one end of a wire and non in the other end, then these electrons will be pushed away from each other towards the end from which they are not pushed away. This principle is the key - the electron-dense end is called the negative end, the ...

1

If there were no other absolute voltage references in the problem, then you would be correct. But there is already a connection to ground on the upper trace of the diagram. The voltage of that trace does not change (it starts at $0$ and remains $0$ after closing $S$).

3

Your teacher's explanation makes sense - IFF there is a "missing component" in your circuit diagram. For example, if there is some series resistance (possibly inside the power supply). In that case, the current through $R_1$ will indeed depend on the current through $L$ - since the voltage source plus internal resistance behaves "a little bit" like a ...

0

imagine a small section of the conductor, say 1mm thick...indeed on the left side, there is a given amount of charge exiting this section, but from the right side, there is an equal inflow of charge. For that reason, the total amount of charge in the section of conductor does not change

3

You can think in terms of the current density vector. Its definition: $$\mathbf J = \sum_i n_i q_i \mathbf v_i$$ Where $n_i$ is the charge carrier density in the media, $q_i$ is the charge of the charge carrier, and $\mathbf v_i$ is the average velocity of the charge carrier. Assuming we have several charge carriers (electrons, ions, etc), you have to sum ...

3

... but what happens if you connect it in series? Consider a circuit which has a 3-V (ideal) battery connected in series with a 100-$\Omega$ and a 200-$\Omega$ resistor (series resistors). The voltage across the 100-$\Omega$ resistor will be 1 V, and across the 200-$\Omega$, 2 V. Next, connect a 1 M$\Omega$ resistor (a voltmeter) in parallel with the ...

1

I think it's due to the high resistance of a voltmeter but I don't really see why this would be a problem: It just means that the current going through the circuit is shifted down a level, but all things remain the same right? Ideally, a voltmeter should have such a high resistance that it's equivalent to an open circuit. This means that when ...

3

First of all, notice that (for the initial 4 terminal resistor), the port potential $V_i$ is equal to the $v_i$ in you first image, but the port current $I_j$ is equal to $-i_j$ or $i_{(j+1) mod 4}$ and what you call resistance matrix $R$ is generally said impedence matrix $Z$ (because impedance parameters are calculated under open circuit conditions $Z_{nm} ... 1 Resistance is directly proportional to the length of the wire, and inversely proportional to the cross sectional area of the wire. R = pl/A, where R is resistance, p is the material's resistance in ohms, l is the length, and A is the cross sectional area in m^2. As a wire gets longer its resistance increases, and as it gets thinner its resistance also ... 0 Model the current through the resistor as $$i(t) = v(t)/R,$$ the current through the capacitor as you have above, that is $$i(t) = C \dfrac{dv}{dt},$$ and the turning on/off of the switch can be modelled using the Heaviside function which would pre-multiply the above expressions. The current has to be continuous throughout so we can equate the two ... -1 You can substitute the switch with a resistor$R_2$, that has a value of either$0$or$\infty$. 0 If the switch is open, then no current can pass through the circuit. Your ODE should then take the form $$\frac{di}{dt}=0\tag{1}$$ When the switch closes, then current can pass through, leading to $$R\frac{di}{dt}+\frac{i}{C}=0\tag{2}$$ When the switch opens again, we revert the ODE back to Equation (1). Thus, the time range$0\leq t\leq2$gives you ... 0 Your calculation is basically correct. If a battery is rated at 9.88WHrs it will produce a power of 1 Watt for 9.88 hours, or 8 Watts for 1.235 hours. However the power rating depends on what current the battery is producing. There will be an optimal current that gives the highest capacity, and the battery power rating is calculated for this optimum ... 0 The maximum of $$\left|\cfrac{V_2}{V_1}\right| = \cfrac{\sqrt{1+(\omega RC)^2}}{\sqrt{4 + (\omega RC)^2}}$$ is$1$at$\omega=\pm \infty$, and you find the half power frequency by solving: $$\frac{1+(\omega RC)^2}{4+(\omega RC)^2}=\frac{1}{2}$$ which gives$\omega=\pm \sqrt{2}/RC$1 IN ac frequency,the range of 50 to 60hz even though the connected led blinks it cannot be visible to human eye. in above mention that led can with stand in dual polarity but inrush current in ac will affect it. If you slow down the freq then blinking will visible 4 They are obviously talking about "conventional" current, not the flow of electrons. When you want to know the direction of magnetic force on a current you need to use "conventional" current direction. There is an interesting corollary to this relating to the Hall effect - if current in a semiconductor is carried by "holes" the polarity of the Hall effect ... 2 No. Step number two already mentions electrons from the battery flowing into the negative terminal of the capacitor, giving the negative terminal a negative charge. Step number three is talking about electrons flowing out of the positive terminal of the capacitor, giving the positive terminal a net positive charge. 1 I'm a little new here, so I can't comment yet. However, I do think you're supposed to indicate what you've already tried, so please try to give this problem you're best shot before looking to my answer below and next time, give some indication that you've put some effort into the problem. ... 1 Consider the left branch of the bridge. The total resistance is$P + Q$, so the current is: $$I_{left} = \frac{V}{P+Q}$$ The voltage drop across$Q$is just$V = IR$, so the voltage at the$PQ$midpoint is: $$V_{PQ} = V_{in}\frac{Q}{P+Q}$$ We argue in the same way for the right hand branch to get the corresponding equation: $$V_{RS} = ... 1 ...and V is zero so there is no energy ? The other answers all touch on this part of your question, but none of them explicitly says, that there is no energy dissipated in the superconductor itself. Ohm's Law applies to a conductor. The I in Ohm's Law refers to the current flowing in the conductor, the V refers to the voltage difference between ... 17 In a superconductor, the current can keep flowing "forever" since there is no resistance. But since conductors have inductance (in fact, superconductors are used most often to create magnets like for an MRI scanner), applying a voltage would not (immediately) cause an infinite current to flow. It is instructive to see how an MRI magnet is "ramped" (turned ... 6 If they have 0 resistance then I (V/R) should be infinite? According to Ohm's law, the voltage and current associated with a conductor are proportional:$$V = R \cdot I$$where the resistance R is the constant of proportionality. This equation holds for an (ideal) ohmic material. We can rearrange this equation to be$$I = \frac{V}{R}$\$ except ...

5

The voltage is zero. That's the point. The main way current gets started, like in an NMR magnet, is by inductive coupling.

0

@Jerry Schirmer is right, the best way to think of a superconductor is like a skier being able to ski at a constant speed without a slope. The power output is zero since no power is dissipated through any heating affects of the conductor so no energy is transferred except when resistance is met at the other end (like a bulb).

0

In a normal electric circuit, with copper wires at ambient temperature, electric potential does decrease around the circuit, as there will be some resistance and so there has to be a voltage/potential gradient for the current to flow. Copper wires DO have some resistance, so your assumption is invalid in that case. However, your assumption would be valid ...

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