Finding the voltage of the circuit (resistor ladder)

In summary: It is the same problem, with different numbers. The hint is there, but the answer is not. You might be able to get the same answer as the given one by using ##V(x)=V_0e^{-\frac{x}{\sqrt{\gamma}}}## in the differential equation for I(x).In summary, the conversation discusses a problem involving equivalent resistance and Ohm's law, with a focus on the purpose of a specific variable and the resulting exponential drop in voltage. The conversation also touches on the background knowledge necessary to approach the problem, including Kirchhoff's laws and transmission line theory. The given answer uses a differential equation and a specific equation for voltage, but it may be possible to arrive at the same answer using a
  • #1
deskochan
6
1
Homework Statement
Finding the voltage of the circuit (resistor ladder)
Relevant Equations
Resistor Ladder?
The answer is B.
My question:
(a) What is the basic approach to start to think about this problem: Equivalent resistance? ohm's law?
(b) What is the purpose of gamma = R2/(R1+R3) because I have no idea how it uses?
(c) Why the answer is an exponential drop of voltage?
1.jpg
 
Last edited:
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  • #2
This reminisces me of transmission line theory. You basically have to choose an infinitesimal loop and use KCL and KVL to analyze it. If all go well you ll get one differential equation for Voltage ##V(x)## and one for the current ##I(x)##
 
  • #3
Delta2 said:
This reminisces me of transmission line theory. You basically have to choose an infinitesimal loop and use KCL and KVL to analyze it. If all go well you ll get one differential equation for Voltage ##V(x)## and one for the current ##I(x)##
I have only GCE A level physics background. How can I start to think about it? I do not expect to solve this question but at least how can I start to think from A level physics.
 
  • #4
Hello @deskochan,
:welcome:
​
Picture is rather unsharp. Nice to know that b (not: B) is the answer, but
1626196686870.png
is illegible for me.

deskochan said:
(a) What is the basic approach to start to think about this problem: Equivalent resistance? ohm's law?
That is rather obvious: they even give a hint ! What do you do with that hint ?

(b) What is the purpose of gamma = R1/(R1+R3) because I have no idea how it uses?
I don't see that definition. I do see ##\gamma = {R_2/R_1+R_3}##. It comes back in the answer, so it seems to be useful. And it reduces the amount of writing needed.

(c) Why the answer is an exponential drop of voltage?
Because that is what comes out ? :rolleyes:
Probably some infinite sum, but from the hint it can also be the solution of a differential equation ...

##\ ##
 
  • #5
deskochan said:
I have only GCE A level physics background. How can I start to think about it?
Not sure how good can be a GCE A level physics background. Are you familiar with Kirchhoff's laws, Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL)?
 
  • #6
BvU said:
Hello @deskochan,
:welcome:
​
Picture is rather unsharp. Nice to know that b (not: B) is the answer, but View attachment 285911is illegible for me.That is rather obvious: they even give a hint ! What do you do with that hint ?I don't see that definition. I do see ##\gamma = {R_2/R_1+R_3}##. It comes back in the answer, so it seems to be useful. And it reduces the amount of writing needed.Because that is what comes out ? :rolleyes:
Probably some infinite sum, but from the hint it can also be the solution of a differential equation ...

##\ ##
Sorry for the poor resolution. it is -2x/(sqrt (4*gamma+1)+1).
What is the hint? Sorry for this question but I really have an interest.
Would you tell me more and I just have GCE A level physics and I think this question is out of A level.
 
  • #7
Delta2 said:
Not sure how good can be a GCE A level physics background. Are you familiar with Kirchhoff's laws, Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL)?
A little bit for self-study but not familiar because GCE A level seldom use it.
 
  • #8
Ok well, if you have not seen a treatment on transmission line theory, and you are not familiar with Kirchhoff's laws then probably this problem is beyond your level of knowledge. The differential equations I get by properly applying KVL and KCL are :
$$\frac{dV(x)}{dx}=-I(x)(R_1+R_3)$$
$$\frac{dI(x)}{dx}=-\frac{V(x)}{R_2}$$
But these have a bit different solution for ##V(x)## than the B) of the answer. I get $$V(x)=V_0e^{-\frac{x}{\sqrt{\gamma}}}$$
 
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  • #9
Delta2 said:
Ok well, if you have not seen a treatment on transmission line theory, and you are not familiar with Kirchhoff's laws then probably this problem is beyond your level of knowledge. The differential equations I get by properly applying KVL and KCL are :
$$\frac{dV(x)}{dx}=-I(x)(R_1+R_3)$$
$$\frac{dI(x)}{dx}=-\frac{V(x)}{R_2}$$
But these have a bit different solution for ##V(x)## than the B) of the answer. I get $$V(x)=V_0e^{-\frac{x}{\sqrt{\gamma}}}$$
Yes, you are right. It is beyond my level. Thank you very much.
 
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  • #10
https://www.pravega.org/events/science_eve/decoherence/Decoherence_Prelims__Objective_.pdf

1626200116912.png
 
  • #11
BvU said:
https://www.pravega.org/events/science_eve/decoherence/Decoherence_Prelims__Objective_.pdf

View attachment 285914
Yep.
 

FAQ: Finding the voltage of the circuit (resistor ladder)

How do you find the voltage of a circuit with a resistor ladder?

The voltage of a circuit with a resistor ladder can be found by using Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by the resistance (R). In a resistor ladder, the total resistance can be calculated by adding up the individual resistances in series. Once the total resistance is known, the voltage can be found by multiplying it by the current flowing through the circuit.

What is a resistor ladder and how does it affect voltage?

A resistor ladder is a series of resistors connected in a circuit, with each resistor having a different value. The total resistance of the ladder affects the voltage in the circuit by limiting the flow of current and causing a voltage drop across each resistor. This voltage drop can be calculated using Ohm's Law and the individual resistances of each resistor in the ladder.

Can the voltage of a circuit with a resistor ladder be changed?

Yes, the voltage of a circuit with a resistor ladder can be changed by adjusting the individual resistances in the ladder. By increasing or decreasing the resistance, the total resistance of the ladder will change, causing a change in the voltage of the circuit according to Ohm's Law.

How does the placement of resistors in a ladder affect the voltage?

The placement of resistors in a ladder does not affect the voltage of the circuit. As long as the resistors are connected in series, the total resistance and voltage will be the same regardless of the order in which they are placed in the ladder.

What is the purpose of using a resistor ladder in a circuit?

A resistor ladder is commonly used in circuits to divide the voltage and reduce the amount of current flowing through the circuit. It can also be used to create a specific voltage output by adjusting the individual resistances in the ladder. Additionally, a resistor ladder can be used to create a smoother, more linear response in certain electronic components.

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