Is the slope stands for resistance in a current vs voltage graph?
Yes.If Voltage is on X axis, Current on Y axis, Resistance is the slope.In the graph above, the 5 Volts point at V in x-axis when intercepted and then taken to the corresponding point in y-axis, gives 0.5 amps.Therefore, resistance of 5/0.5 = 10 ohms is indicated by this graph.If you draw a vertical line up from 3 volts in x to the slope, the corresponding y point will be 0.3, hence R = 3/.3 = 10 ohms.Same at any other point.The linearity (straight line) of the slope tells us the circuit obeys ohms law. At any point, the ration of 10 will be maintained.The angle of the slope to the x-axis is decided by the resistance, as well as by the scales used in the axes. Note the progression of voltage from 1, 2, 3 etc, compare to the y axis progressions of current from 0.1, 0.2, 0.3 etc.In this situation, a resistance of 5 ohms will have a slope that is twice s steep, and a resistance of 20 ohms will be more close to the horizontal axis half as steep as the current one.
Why does an current voltage graph have a negative gradient?
For a single resistor, increasing the voltage across it always increases current through it and the gradient of a V-I graph is always positive. But if you are considering a different circuit, e.g a power supply with internal resistance connected to a variable load, it is more complicated. If you measure the current and voltage and plot a graph you find a negative gradient. However you are NOT measuring the voltage across the fixed (internal) resistance; you are measuring (emf - voltage across internal resistor). Watch the video which might help. If that doesn't answer the question, repost with details of the circuit.
Tricky question with a capacitor + resistor..?
Am in a basic course of electrophys and physiology. Instructor started talking about time constants and capacitors. S he gave us homework which I solved all except this tricky one. A circuit with a switch, one R which is 1M ohms, one capacitor which is 1u F, and a 9V battery. He said once we close the switch plot: Vc (voltahe change in Capacitor), Vr (voltage change in R), and Ic (current of capacitor). He has three places with x (time) and y axis. Axes has Vc, Vr, or Ic. I plotted them all and want to see if they are correct. Here is what I have. For Vc: exponential growth until it reaches 9V. With Vr exponential drop from 9V to close to zero, and the same for Ic. Am not sure what units should time be.
Does the gradient of the graph in a non-linear curve, in current vs voltage, represent resistance?
Well, begin with a function for current. Is it a function of just voltage, or is it a function of spatial coordinates as well? If were taking the gradient of something, we imply a function of more than one variable. Otherwise, it's just a derivative.The derivative of voltage with respect to current will be resistance (derivation is division when it comes to dimensional analysis) because V/I = R. But note that this resistance, unless it is constant, is a function of current like our voltage. So the result is a function for how resistance changes with current.If we assume the current is a function of voltage, taking the derivative gives us conductance in terms of voltage.Neither of these seem quite right. This is because though we can perform these operations on simple functions of current and voltage functions, the reality is that voltage and current depend on each other as well as resistance meaning one is a function of the other two variables. Here, dimensional analysis helps less. Since the gradient of a function of n variables is a vector field in n dimensions, and resistance is a scalar not a vectir, we can rule out the idea of the resistance being a gradient of current.Even still, we don't have to throw out the results above for derivatives. In the case of semiconductors, which I might assume is the root of this question, the voltage and current relate to many physical variables relating to the modelling of semiconductor behavior. In the end, we can have a result in terms of a relationship between a voltage and current. Assuming one is a function of the other, we find the emitter resistance re' or equivalently the (trans)conductance gm. While these do not represent a true resistance (they are derived from the physical model) they are dimensionally equivalent to a resistance that changes with bias current Ic, and the thermal voltage Vt.I hope this shows how tricky analysis and calculus can be, depending on how you arrange things! But also I hope, in some way, I answered your question.Sam
What is the answer to this physics question, with some sort of explanation?
Please take the time to type out the question in the title and details. Provide high-quality images for graphs and other pictorial information. Otherwise, Quora will mark your question as needing improvement. It’s also a common courtesy to people who will be taking time to answer your question.Since the voltage across the resistor is growing linearly before [math]t_0[/math], the current through this section of the circuit must be a linear function by Ohm’s Law. And current is the time derivative of charge. So, integrating the line gives a quadratic function. So the correct answer is B.If you’re unsure of the differences between lines, parabolas, exponential curves, and logarithmic curves, then you should review those concepts before proceeding further.A good resource is https://www.khanacademy.org/math.
What are some technical questions asked during an interview in a core electrical engineering company?
First of all, there are not any fixed questions which are asked in interview. All depends on the Interviewer and how you respond to the questions asked by him.Some tips which you should take care of:Never underestimate yourself. You are much capable of performing in an interview than you think. So just be calm and composed.Keep all basic concepts of Electrical Engineering clear. This is the only key to crack the Interview. No one is going to ask you anything in detail. So your basics is weak, start preparing for it. Here is the book which you should buy for brushing up your basic concepts of Electrical Engineering.Buy Objective Electrical Technology Book Online at Low Prices in IndiaBefore entering the room for interview, take deep breath and just be calm. This will boost your confidence.Be honest in writing Resume. Because 2–3 questions will be directly from it.Here comes the main part,What are the questions asked?Introduce YourselfQuestions from Resume i.e. about your training and final year project.What is difference between Isolator and Circuit Breaker?Difference between MCB and MCCB.What is D.C Generator?What are your favorite subjects?What is Electric Traction?What is Current Transformer?In transformers, what remains constant?Draw waveform of D.C Frequency. This will check your presence of mind.What will happen if 40 Hz is provided to Transformer?Applications of D.C Series Motor.So you can see all the questions asked are not that tough. Just be confident and don’t be Nervous. Good Luck!!
What will be the capacitance vs voltage graph?
This is a good question. It is bit tricky and we need to understand the how a capacitor works to get the real answer.Normally people will say that capacitance of a capacitor depends on the area of the plates, the gap between the plates and what type of dielectric we are placing between the plates. There is no mention of voltage here and truly speaking voltage is not necessary to mention here.When we connect the plates with + & - voltages, opposite charges accumulate on the corresponding plate surface. Capacitance by definition is the amount of charge one capacitor can accumulate per unit applied voltage. Now, this capacity is fixed for a capacitor- by area, gap and dielectric. If we increase the voltage more charge will accumulate but the ratio of total charge and voltage will remain the same.So, the graph will be a flat line in Y axis if voltage is varied in X axis. This means, whatever the voltage value may be the capacitance of a capacitor will remain the same. Thanks for asking.
[Physics] Momentum and Collisions?
1. The momentum before Pb was equal to zero because momentum is a vector quantity and the carts both had identical momentum BUT their momenta were in opposite directions, exactly canceling. Thus the momentum after also = 0 2 The momentum before Pb = Momentum after Pa The vector sum of the momentum before the collision will equal the vector sum of the momentum after the collision. Since the collision is elastic, kinetic energy is also conserved meaning the total kinetic energy before is equal to the total kinetic energy after the collision. 3 Momentum is conserved in both elastic and inelastic collisions. The vector sum of momentum before the collsion is equal to the vector sum of momentum after the collision. Elastic collisions means the total kinetic energy before is equal to the total kinetic energy after the collision. KE is NOT a vector quantity. 4 Don't know