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How Many Nodes Are There In The Following Circuit And Where Are They Located

How do I solve a circuit by using nodal analysis?

let in this circuit, there be four nodes A,B,C,D.We have taken Node A as reference node which depicts the node from which other node voltages are compared (or GND Node).writing a Nodal Equation consists of KCL, which tells us that sum of the current coming into a node is equal to the sum of the current going out of the node.the current is measured using Ohm’s law considering a single node at a time….In this example, Node B and Node D are not considered for any analysis as because there is no resistance between the nodes and the Battery(or Cell).the following expression shows the KCL eq, where LHS is the summation of the all outgoing current and RHS which is 0, is the summation of all incoming current. (Vc is the Node Voltage taken at the Node C).Always the Node Voltage is considered to be higher than the other voltage source by using which ,current is found out.Thank You

When we measure the EMF of a circuit using a voltmeter, why does the circuit need to be open?

Please provide additional information to help answer your question more specifically. In general the circuit does not have to be open. What circuit are you looking at? What nodes do you want to measure?Voltage is an across variable so it is simply a matter of finding an instrument that will properly read your desired Voltage. Then simply apply the input of the meter to the nodes in the circuit that you wish to measure. There are many special types of Voltmeters designed for measurement problems, they are not all the same.Generally Voltages are measured relative to some type of circuit reference such as power supply ground. This is not always true however. One key item about measuring Voltage is that touching some nodes in signal circuits may disturb the node enough to damage the reading. A common feature of high performance meters is a null balanced input which draws near zero current.Todays world of circuit simulation is tremendously useful in solving difficult measurement problems. Simulators are free of measurement artifacts caused by loading nodes with measurement input current. Further, problems which are related to undesired noise injection into sensitive circuits may disturb the circuit and are not present in simulation. Have you tried simulation on your circuit. The Linear Technology LTSim is a tremendous tool for this.Final note to consider, for some difficult measurements it is useful to build special isolation circuitry into the assembly to convert your desired measurement to a ground referenced signal which is more easily measured. An output directional rectifier built into a radio transmitter to measure standing wave ratio (SWR) is an example of this.

How do I find the construct equation of this circuit?

First notice there is a simple relationship between Vo and ix. Because the reference node is on the bottom node, the relationship is simply Ohm's law. So Vo = ix * 50k(ohms) This solves your problem because this relationship generates your "missing" equation

The easiest technique here would be nodal analysis. Sounds like you know the rest! Hope this helps, let me know if you need more!

How do I calculate voltages, currents and resistances in a circuit with resistors in both series and parallel circuits?

In the general case, you need Kirchhoff's circuit laws - Wikipedia.There are two laws: Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). I almost always use KCL which is based on the observation that charge does not accumulate in nodes (the mathematical point where two or more components are attached together). This allows you to write an equation like: The sum of all currents entering a node =0. If a node has three components, then you will have three currents and their sum must be zero. Yo can also state this law as the sum of the currents leaving a node must equal the sum of the currents entering a node. However, you don’t always know which currents are entering and which are leaving, so you just say that they are all entering and it will turn out that some of them are negative. You write the equation for every node. If you have three nodes, then you have three simultaneous equations. If the equations are linear, you can solve them by the techniques of linear algebra.Before you apply KCL, you eliminate as many nodes as possible. For example, if you have two components in series, they would have a node between them. You can replace those two components by a single component with impedance equal to the sum of the impedances of the two individual components. In this way you eliminate a node.The currents into a node are almost always defined by the voltage different between the nodes to which the component is attached, which means your system of simultaneous equations usually have node voltages as variables. Thus KCL allows to to directly determine the node voltages.

What is difference between a node and junction in electrical circuits?

A point at which two or more elements are joints together is called node.While a point where three or more branches meet together is called a junction. see below image.in the above figure we can say that point a,b,c,d,e,f.are nodes and point c & f are called junction. while point c & f itself a node.Thanks……..Image source: Google.

Hamilton Circuits Question?

You'll use counting for this:
a)
Pick a node to start at, you'll then have n-1 nodes to choose from for your next move (because you can't move to the node you're already at). Each move after (till you reach a length of n) will have n-1 choices.
Which looks like:
n(n-1)^(n-1).
b)
Again pick a node, you'll have n-1 nodes to choose from for your next move, but this time every move after that (till length of n) will have n-2 choices, because you can't go "retrace" the path you just took.
Which looks like:
n(n-1)(n-2)^(n-2)
c)
Finally, pick a node to start, then you'll have n-1 nodes to choose for a next move, then n-2, then n-3 and so on. Each move will remove a choice because you can't repeat nodes or arcs.
Which looks like:
n(n-1)(n-2)(n-3)(n-4).... which is the same as n!

Thanks, I was working on this myself and this helped put my thoughts together. Hope that works, and is correct.

-House

Three bulbs in a circuit?

a) With the switch open, indicate the approximate surface charge on the circuit diagram.(Do this on paper. Your instructor may ask you to turn in this work.) Refer to your diagram to decide which of the following statements about the circuit (with the switch open) are true:

The surface charge on the wire at location B is positive. is TRUE but the charge is bound, you cannot transfer it to electroscope

There is a large gradient of surface charge between locations M and L. FALSE
The electric field in the filament of bulb 3 is zero. TRUE 3.6/(distance in m between B and C) V/m

There is no excess charge on the surface of the wire at location C TRUE (negative charge), but the charge is bound, you cannot transfer it to electroscope

The electric field in the air between locations B and C is zero. FALSE

(b) With the switch open, find these potential differences:
VB - VC = 3.6 V
VD - VK = 0 V

(c) After the switch is closed and the steady state is established, the currents through bulbs 1, 2, and 3 are I1, I2, and I3 respectively. Which of the following equations are correct loop or node equations for this steady state circuit?

I2 = I3 FALSE as r2 and r3 are not equal
-I1*(10 )-I2*(38 ) + I3*(33 ) = 0 FALSE
+3.2V + -I1*(10 ) + I3*(10 ) = 0 FALSE
I1 = I2 + I3 TRUE
+3.2V + -I1*(10 ) + -I2*(38 ) = 0 FALSE
-I2*(38 ) + I3*(33 ) = 0 TRUE



(d) In the steady state (switch closed), which of these are correct?

VC - VF = +I2*(38 ) TRUE
VC - VF = 0 FALSE
VL - VA = -3.2V + I1*(10 ) TRUE
VC - VF = +I1*(10 ) FALSE
VC - VF = +I3*(33 ) TRUE


(f) Now find the unknown currents, to the nearest milliampere.
I1 = 115 mA
I2 = 53 mA
I3 = 62 mA
(g) How many electrons leave the battery at location N every second?
723014256619144603 electrons/s
(i) What is the numerical value of the power delivered by the batteries?
P = 0.3702 W
(j) The tungsten filament in the 38 ohm bulb is 9 mm long and has a cross-sectional area of 2 10-10 m2. What is the magnitude of the electric field inside this metal filament?
|| = 227 V/m

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