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2

How do I identify which ones are parallel or series? If all of the current leaving one resistor enters another resistor, the two resistors are in series. The resistances of series connected resistors can be added together to find the equivalent resistance of a single resistor, e.g., $$R_{eq} = R_1 + R_2 $$ If all of the voltage across one resistor is ...


0

Since this is a homework type question, I can only offer you a tip and not solve it for you. Tip - Look at this series combination for example Notice that there is some current $I$ flowing through each of these three, and the value of potential at points 1, 2, 3 and 4 (across $R_1$, $R_2$ and $R_3$) aren't the same. On the other hand, in a parallel ...


0

You want to work this out in stages. Start at the right hand side where there are three resistors clearly in series. "In series" means that the current that flows through one resistor is equal to the current flowing through the other resistor. The simplest way to identify this is to look for nodes (points where components are connected) with just two wires: ...


2

A couple of suggestions: (1) the EE stackexchange site a better home for this question (2) simply solve for the voltage across the capacitor and the current through the inductor. Once you have those, the energies stored, as a function of time are just $$W_L(t) = \frac{L}{2}i^2_L$$ and $$W_C(t) = \frac{C}{2}v^2_C$$ Since this is evidently a DC circuit ...


0

This is a potential difference. In power transmission, it's the RMS amplitude of a alternating signal.


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In Diagram 4, I don't understand how the voltmeter is connected in parallel. Parallel vs. Series - needs a reference. Either Producer (Source of electricity) or consumer (the Resistor in your diagrams). Diagram 4: Vmeter is in Parallel with the Resistor because the current is split. The Ameter is in Series with the Resistor (and Vmeter, but you can ...


1

The inherent idea is, from that equation $$I = \exp (\frac{eU}{k_B T})$$ if you plot $(\ln I)$ versus $U$, that would be a straight line (of the form $y=mx$), with a slope $$\alpha = \frac{e}{k_B T}$$ That's all you have to do in the experiment, use least square fitting to find an accurate value of $\alpha$ and then find the Boltzmann constant using the ...


1

Isn't it just a convention you have to choose once and for all. It's all about how to pass from Maxwell to lumped element circuits. Especially, I can choose the two different conventions (all quantities are vectorial in the following) $$E=\pm\nabla V$$ since I didn't choose my field-to-potential rules yet. Usually one chooses to conform to the classical ...


0

Partial answer here about what the author means by flux: We now introduce the flux $\phi_n$ of a node $n$ which is defined by the time integral of the voltage measured along the path connecting the node to the ground on the spanning tree. (Defined a few pages prior to 363.) It then seems $\dot\phi$ is voltage as we would normally define it.


2

A motor-generator can be used to convert electric power from one voltage x current combination to another voltage x current combination. Such systems have been used, and are sometimes still used, for exactly this purpose. One advantage is that a large flywheel can be added to the shaft, effectively low pass filtering the average power. This can be useful ...


0

If your source has both a sinusoidal component and a constant (DC) component, simply perform two separate analyses - a DC analysis and an AC analysis. For the DC analysis, the sinusoidal component is zeroed and the (constant) voltages and currents are solved for (recall that capacitors are replaced with open circuit and inductors are replaced with wires). ...


1

DC bias is $A e^{i\theta} e^{i\omega t} = A e^{i\cdot 0} e^{i\cdot 0 t} = A$, however, quoting this, Phasors and Complex impedances are only relevant to sinusoidal sources.


4

know that the two conversion coefficients are close but I simply cannot see why the RMS value is the one that conforms to reality. It's true that we can ask for the average of the magnitude of a sinusoid over time and calculate it. This is a useful quantity if, for example, we want the time average voltage at the output of a full wave rectifier ...


2

Waffle's answer shows you exactly why the RMS isn't the average: Here, the average value of $V(t)$ clearly isn't $2V_p/\pi$ that you've obtained, it's zero. The half-cycle business doesn't make much sense given that if you chose $\pi/2$ to $\pi$, instead of 0 to $\pi/2$, you'd have a negative average, so which would be the true "average" $+2V_p/\pi$ or ...


5

The instantaneous power expended in a resistor is proportional to $V^2$ (i.e. independent of its direction), so the average power expended is given by the mean square voltage! The square of the average absolute voltage will not yield the power expended in the resistor. Edit: And actually this close-to-duplicate question does have more extensive answers ...


2

The definition of RMS is Root Mean Square and means the square root of the avarage of the square of some quantity, that is $a_{RMS}=\sqrt{\langle a^2 \rangle}$. So for the case of a sinusoidal current, we would get $$V_{RMS}^2=\frac{1}{T}\int_0^T\left(V_p\sin\frac{2\pi t}{T}\right)^2dt=\frac{V_p^2}{2}\implies\\ V_{RMS}=\frac{V_p}{\sqrt2}$$


0

What is the easiest way to solve this? The easiest way is unlikely to be the most illuminating way. What I propose is that you first work the problem from basic circuit principles and then explore more powerful solution methods such as, e.g., node voltage analysis, Thevenin equivalent circuit etc. Start from the right hand side with Ohm's law: $$I_6 ...


1

You were given the current through $R_1$ as 1mA. You should know that the 1mA current is shared between $R_2$ and $R_3$. You could determine the voltage drop across $R_1$, and then $R_{eq}$ (the equivalence of $R_2$ and $R_3$) so you can calculate the current through each resistor. Added: Yes, the voltage drop across $R_1$ is 6V. Remember that this circuit ...


1

What is the smallest resistance that can be used in a 5 V circuit if the resistors available are rated for 1/4 W? This is a trick question of sorts. It's true that one could work from the relationship $$p_R = \frac{v^2_R}{R}$$ and the range of $R$ for which $p_R \le 0.25W$. However, this completely ignores the issue of derating for reliability. ...


0

You are right. The second expression would work for the following diagram:


1

Why do we use this formula to find the total resistance? Because parallel connected conductances sum just as series connected resistances. Ohm's law is $$V_R = RI_R $$ The dual of Ohm's law is $$I_G = GV_G $$ Since, for parallel connected circuit elements, the voltage across is identical, it follows that, for parallel connected conductances ...


1

Think about current flow. If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ We now need ...


1

This is derived from the the formula for calculating total resistance in a parallel circuit $$\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ taking its inverse gives: $$R_t = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}$$


0

You have the right equation - now just solve for $R$. This means you multiply both sides by $R$, and divide both sides by $P$. In steps, you started with $$P = \frac{V^2}{R}$$ Multiply both sides by $R$: $$P\ R = \frac{V^2}{R}R = V^2$$ Now divide both sides by $P$, to get $$\frac{P}{P}R = \frac{V^2}{P}\\ R = \frac{V^2}{P}$$ Physics equations manipulate ...


1

Yes I also think that's what they did. And yes, you'd also change the voltage proportionally to the current. But all of this is as long as the resistance is constant (which is what they mean when they say the circuit is the same). A less-prone-to-mistake equation to use in these cases where the circuit is kept the same is gotten by substituting Ohms law, $V ...


1

Consider the current as the flow of positive charges. Positive chrges flow from higher potential to lower potential naturally if they are free to do so. But you can force them to move in the opposite direction too with the help of an extenal agency. Actually this is the work of the battery, it will move the charges from lower potential to higher potential ...


0

Idealizations are just that, approximately true in many situations. If you connect a wire across a battery, the non-ideal nature of the wire and battery will show-the wire will have some resistance and the battery will not be a perfect voltage source. In most circuits, the resistance of the other circuit elements are much higher than the resistance of the ...


1

What you are missing is called conservation of charge. The fundamental fact is that electrons do not simply 'disappear' or vanish into thin air inside a resistor. This means that, if there is a certain number of electrons flowing into the resistor, these electrons must also flow out on the other side. In particular, in a steady state (i.e. the flow of ...


1

Why does current need a closed loop to flow? Not in every case you need a closed circuit. A good example that current in an open "circuit" can flow is the antenna rod. A generator generates an alternating current and the electrons in the rod moves and there is an electric flow.


0

I think you have a problem with Kirchoff's laws.Now it states that the potential drop across a closed loop is 0. Your txt seems to have taken the direction of I as the direction of electron flow.Now suppose you move along the wire anticlockwise(The txt has moved clockwise.) Across the resistor ,potential drops but you travel along the original direction of ...


1

Your question has two parts. First addressing why electron needs a closed loop to flow- Consider a wire which is not closed for e.g in the diagram. Here Due to the electric field(marked in green) the electrons move to the right.But after some time the electrons accumulate in the right forms their own electric field with the positive charge(marked in ...


0

Electrical components are said to be in parallel if they are across the SAME potential. All the three capacitors are thus in parallel. Electrical components are said to be in series if the same current passes through them. If you decide to treat C2 and C3 together as one component, then C1 is in series with the combination of C2 and C3. but individually, ...


0

When the current is turned on for 10 seconds the capacitors have gradually charged according to the equation Q(t)=CV(1-e^(-RC)) Now after 10 seconds when the switch is open the two capacitors act as two voltage sources in parallel and the current could be found out by superposition principle which is used when we have more than one source


1

After the switch is opened, we have a series circuit - the current through each circuit element is identical and equal to $i(t)$. Choosing a clockwise reference direction for the series current, KVL clockwise yields $$\frac{1}{1\mu F}\int_{-\infty}^t i(\tau) d\tau + i(t) \cdot 1k\Omega + i(t)\cdot10k\Omega + \frac{1}{1000\mu F}\int_{-\infty}^t i(\tau) ...


0

There's a really awesome trick for problems like this. This is going to be a long post but the method presented makes problems like this really easy. The idea is to turn the series branch $C_2$, $R_2$ into an effective parallel $R$ and $C$. See the diagram. The effective parallel values are denoted $C_{2,p}$ and $R_{2,p}$. Parallel capacitances just add, so ...


1

Here the capacitors look like in series but they are not actually. Capacitors can be said to be in series only if they carry same amount of charge which is not the case here.If you calculate charges will come out to be different.Look at the time constants(R*C) for the first branch and second branch, they are 1ms and 10sec respectively. As switch is on ...


5

When the switch is opened, the circuit is the equivalent of this, so I think you can clearly see that the resistors are in series and so are the capacitors. It seems that you already understand how to calculate series resistance, so I'll show how to understand series capacitance. First, you probably already know that capacitance is defined as the ratio ...


0

The best way to go about these kind of problems, is to use Kirchoff's loops. Whenever there is ambiguity about parallel/serial, Kirchoff is the way, since you disregard the question of serial vs. parallel altogether. Also remember that while resistors observe: $$ \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2} $$ for parallel and, $$ R_T=R_1+R_2 $$ for serial, ...


3

Let us crudely imagine the voltage source as a pump pumping water up to the top of a water slide, and the resistor as the slide itself with water flowing down through it. The height difference between the top of the slide and the bottom of the slide is the same as the height difference between the top of the pump and the bottom of the pump. The voltage ...


0

Consider this concrete scenario. Let's say that there's a voltage source connected to an inductor (maybe throw in a resistor in series to make it more applicable to a real-world scenario). The voltage source is increasing in time such that it drives an increasing current from point a to point b across the inductor, as you say. We can even do a step in ...


-1

If your inductor has a 1.0 amp current and you want to increase it to 1.1 amp, you need to increase the voltage to force more current through the inductor. Since there was already 1.0 amp flowing in the inductor, then it would resist the increase in voltage forcing more current (0.1 amp) to flow. So, yes there is an inverse voltage from B to A, but its ...


0

Update: After simulating a little more, it seems that Drew Noel had the right hunch after all: by inserting a 100pF capacitor in parallel to the 110 Ohm and 11kOhm resistors, we can shift the resonance frequencies up from 11.25kHz to 12.11kHz, which gets us into the right ballpark. 150pF will give 12.71kHz for the upper frequency and 200pF will result in ...


2

This is as much a chemistry question as it is an electricity question, because batteries are chemical devices. A battery is constructed such that there's a redox reaction split into two half-reactions which must move electrons through your circuit in order to complete. The reaction happens at a finite speed, and may also be limited by how fast various ions ...


2

The safety of a low voltage DC power supply is not established by the voltage on its output, but by the isolation between its input and output terminals. For example, a defective 12V power supply may have a short between the 120V AC input terminal and its negative output terminal. A user who would be connected to ground would then experience a 120V AC ...


2

The internal resistance of batteries is caused by a number of different mechanisms. A major contribution comes from the ionic conduction mechanisms in the electrolyte solution. Ions are large and can only move very slowly in electrolytes. Another source for the voltage drop is the concentration and current density dependent polarization of electrodes. To ...


1

Low voltage sources don't have enough potential to conduct through your skin or body so touching either the positive or negative doesn't make a difference. For you to feel it or get tissue damage, current must flow through you. This won't happen with very low voltage sources unless you're covered with something more conductive like wet salt water. But 12V ...


2

They are absolutely connected. In both cases, you have the following components: an "inertial term": this can be mass, or inductance. Something that resists change (in velocity, in current) a "linear force term": the spring ($F=-kx$) or the capacitor ($V=Q/C$). a "displacement term": this is $x$ for mechanical, and $Q$ for electrical. Their derivatives $v$ ...


2

The drift velocity is the average velocity due to an applied electric field. In a conductor, electrons scatter around at the Fermi velocity but have a net zero average (i.e., equal scattering in all directions). When the electric field is applied, the electrons are given a small velocity in one direction. Thus, we can say, $$ v_{drift}=\eta E $$ where $\eta$ ...


0

1) You're asking about non ideal capacitors (see e.g. here), i.e. answer to your question is 'yes'. The formula assumes that capacitor's plates are the perfect conductors. 2) Oblique - will not work since you connect the plates, the last one can be decomposed into three capacitors joined in parallel.


11

If you express power loss in a power line as $V^2/R$, the $V$ in that expression is the voltage difference between the two ends of the power line, not the voltage difference between the power line and ground. To supply a fixed amount of power $P_L$ to a load, if the voltage at the load $V_L$ is larger, the current $I=P_L/V_L$ can be smaller. If the power ...



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