Read about voltage | 295 Discussions | Page 1

Having more difficulty understanding the concept, thus im not showing values. What is causing me confusion is the line in the middle. The first aR and bR are obviously in parallel, but the second aR and bR confuse me. I tried calculating the equivalent resistance from the first aR and bR and...

I used the potential at the surface of the sphere for my reference point for computing the potential at a point r < R in the sphere. The potential at the surface of the sphere is ## V(R) = k \frac {Q} {R} ##. To find the potential inside the sphere, I used the Electric field inside of an...

The problem is for a solid sphere uniformly charged with Q and radii R. First I calculated taked ##V(\infty)=0##, giving me for : $$ \begin{align*} V(r)=&\frac{3Q}{8\pi\varepsilon_0 R}-\frac{Q}{8\pi\varepsilon_0 R^3}r^2\qquad\text{if $r<R$}\\ V(r)=&\frac{Q}{4\pi\varepsilon_0 r}\quad\text{if...

Here is a circuit diagram: . We have three capacitors, with capacitances ##C_1##, ##C_2## and ##C_3##. Plates are labelled as ##A_1, A_2, A_3 ... A_6##. Point P is connected to the positive terminal of the battery and point N is connected to the negative terminal of the...

Well i don't you to solve the question for me but I want you to clarify the concepts pertaining to this question. My question is how do I write a equation for the circuit since the there is same charge on one of the capacitors. While writing the equation should i put the voltage across the...

In my opinion, the voltage across the C1 should be 9V as the potential on the side of the positive plate of the capacitor should be (15-6)V and on the other be 0V. Similarly the potential across C2 should be (7-0)V. Here I'm basically assuming that the voltage at the negative terminals of the...

This is in python: #ELECTRIC POTENTIAL from mpl_toolkits.mplot3d import Axes3D from matplotlib import cm import numpy as np import matplotlib.pyplot as plt dx = 0.1 dy = 0.1 xrange=np.arange(-1,1,dx) yrange=np.arange(-1,1,dy) X,Y = np.meshgrid(xrange, yrange) max_dV = 10e-5 blockRadius = 3...

Hi everyone! I ask some help in understanding better the concept of voltage. The voltage is a difference in electric potential between two points ##a## and ##b##. It is defined as However, I'm a bit confused with the use of notation: - Is ##V_{ab}## the same as ##\Delta{V}##, or rather...

I honestly don't know how to quite even begin this problem. Looking at Fig 3-2, the slopes of the graphs are 1/R, and hence where the slopes are 0, we have infinite resistance, in which case current wouldn't flow through that resistor and hence simplify the circuit. So I was trying to find...

From the circuit I have: ##-v_b + v_a + V = 0## ##v_b - V = v_a## ##i_1 = (v_b - V)/R_1## ##I + i_2 = i_1## ##(v_b - V)/R_1 = I + v_b/R_2## From this last equation I get ##v_b = 10.8## and hence ##v_a = 5.8##. However, apparently that is wrong. (And hence my answers to #2 were all wrong as...

I don't get this. Since we have analyzed the circuit separately for each source, adding them should give me the final values of I1, V2, I2, V1 etc. However, that's not quite true—from cases 1 and 2, I should have I1 = 4 + 5 = 9 mA, but it's 8mA. Hence, I thought that the black box consumes 1mA...

I am having a hard time solving this. Letting \$i\$ be the current flowing into ##R_3##, ##i_1## the current flowing through ##R_1## and ##i_2## the current through ##R_1##—and the node between R3 and R1 be e_1; Using KVL and KCL, I've managed to find that ##i + I = i_1 + i_2## Hence...

I have no problem getting the ##R_{TH}## since from the special element's POV, the resistors are in parallel, and that's the answer. However, I don't really understand how to get ##V_{TH}##. Ignoring the special element, it seems that I have the resistors in series this time. But I'm not too...

I don't really understand or see the correct way to approach this. Letting the current in question be ##i_x## (as shown in Fig. 1), and the unknown (changing) resistance be $R_x$, I can write: ##-V_s + R_s i_s + i_x R_x = 0##, and ##R_p i_p = i_x R_x##. Hence we can also write ##-V_s + R_s i_s...

I assume that because there is a resistance, the polarity of the voltage must be the same as the charge flow, and thus the current, in order for energy to flow in the same direction. For instance, could I use the example of a light bulb (the resistor) plugged into AC lines; we know that if the...

I already found ##I(t)## using Kirchhoff's laws, I got the equation ##V-RI-L\frac{dI}{dt}=0\Rightarrow L\frac{dI}{dt}=V-RI## then I solved the differential equation getting ##I(t)=\frac{V}{R}\left[1-e^{-\frac{R}{L}t}\right]##. My problem is founding the voltage as a function of time ##V(t)##, I...

Given: U = 220 V f = 50 Hz r = 20 Ohm C = 100 μF Find: I Solution: 1) Xc = 1/(ω*C) = 1/(2*π*f*C) = 1/(2*π*50*10^-4) = 31.83 Ohm 2) R_eq - equivalent resistance R_eq = (r*Xc)/(r+Xc) = (20*31.83)/(20+31.83) = 12.28 Ohm 3) I = U/R_eq = 220/12.28 = 17.9 A True answer given in the textbook is 13 A...

Hi all. I'll get to the point. I've been interested in electricity since I was in college. Concepts such as current and resistance seemed easy to grasp for me but voltage remains a little bit obscure. It's thanks to this forum (specially forum members Jim Hardy r.i.p and SophieCentaur, sorry if...

Let ##Q## - charge of one of conductor, ##\phi_1## --- potential of charged conductor, ##\phi_2## --- potential of uncharged conductor. For the charged conductor: \begin{equation} \phi_1 = D_{11}Q , \end{equation} for uncharged conductor: \begin{equation} \phi_2 = D_{21}Q \end{equation}

The problem: My attempt:

I see that DC power supply have voltage between it's + & - and its 24V. However, there is no voltage with ground. I don't understand - if device's "point" has some potential, why doesn't it give some voltage with ground (which has ~0 potential) I tried this with phoenix contact...

I connected a PVC insulated alligator clip test lead to a 12V (give 20V) DC power supply positive terminal and a multimeter. After that I connected an other one to the multimeter ground, and a third one to the power supply negative terminal. I turned the multimeter to DC V measure mode. When I...

Hi all Trying to carry out an experiment and need a high voltage supply Looking for say upto 40kv dc adjustable generator been looking on the net and only able to find low cost voltage boosters on eBay Or if anyone know of a good website to try Any help would be much apreciated Thanks....Niki

Hi! I've been struggling with this. Original exercise here: Find the value of Iy when R=0. And the value of Vx when R is infinite. For the first part of the question I did this since R=0: I've tried to solve this circuit and I get that Iy and I5 are 0 A, and this can't be possible since Iy...

Homework Statement Moderators note: link removed. All images should be uploaded to PhysicsForums. Homework Equations V=IR Series in parallel Series in current The Attempt at a Solution I managed to get the current and voltage of R1 and R2. I've been trying to get the next one R3, but...

Homework Statement L = 20mH = 20 x 10-3 H i = 40 mA for t≤0 i = A1e-10,000t + A2e-40,000t A for t≥0 The voltage at t=0 is 28 V. I have to find the equation for the voltage for t>0. Then I have to find the time when power is zero. Homework Equations v(t) = L* di/dt p(t) = L*i* di/dt The...

Let's assume a 2:1 transformer which has a 100V Source connected on the primary circuit and has no/negligible resistance, on the secondary circuit a 5 Ohms resistor is connected. Using the 'Impedance Transfer/Reflection' method, the primary circuit would act as if there was a 25 Ohms resistor...

Hi. The derivation of the capacity of an ideal parallel-plate capacitor is inconsistent: On the one hand, the plates are assumed to be infinitely large to exploit symmetries to compute an expression for the electric field, on the other the area is finite to get a finite expression for the...

<Moderator's note: Moved from a homework forum.> Calculate the required voltage to produce a electric arc between 2 iron nails (distance: 3cm). I´ve read in the internet that you need 1 kV per mm. But how can I calculate this value, that I need 1kV per 1mm?

If this is better for the HW section please tell me, sorry about that. So Voltage is the change in potential. I'm puzzled by how potential in circuits is the same along different points until we reach a resistor - at which point the potential changes. As in, if we take two points before a...