# Getting currents in ammeter, with some resistances.

• dhkdeoen
In summary, the conversation discusses the process of finding the current in an ammeter in a circuit with an ideal battery and an ammeter with zero resistance. The conversation includes the use of Kirchhoff's laws and the equations V=IR and 1/R=1/R1+1/R2 for parallel resistors. The attempt at a solution includes a mistake in the calculation of the equivalent top resistance and a mix-up of assumptions, leading to a wrong answer. The correct solution involves simplifying the equations and using the relation i=ε/Req, where Req is the equivalent resistance in the circuit. Thevenin equivalents are also mentioned, but it is noted that they have not yet been covered in the class.
dhkdeoen

## Homework Statement

In the figure R1 = 1.29R, the ammeter resistance is zero, and the battery is ideal. What multiple of ε/R gives the current in the ammeter?

## Homework Equations

V=IR, 1/R=1/R_1+1/R_2..(when resistances are parallel)
etc

## The Attempt at a Solution

since ammeter doesn't have resistance, two top resistances and two bottom resistances are basically parallel. First, I set the currents coming out from the battery is i, and I was getting R_eq

R_top: R_1+R=(119/229)R
R_bottom: R/2
R_eq=R_top+R_bottom=(467/458)R

set the voltage as ε

ε=(467/458)*iR
ε=V_top+V_bottom=V_top+iR
(since two bottom resistances are same, so V_bottom=iR)

(467/458)*iR=V_top+iR
V_top=(467/458)*iR-iR=(9/458)*iR

and

i_topleft=V_top/1.29R=(9/458)iR*1/(1.29R)=(0.015233...)*i
i_A=current through ammeter
i_A+i_topleft=i
i_A=i-(0.015233..)i=(0.9847..)i

so I got 0.9847.. for the answer, which is wrong. where did I go wrong? I spent like 3 hours on this problem, and still can't get it.

The equivalent top resistance is not 119/229 R.

It is also wrong that Vbottom=iR.

It is better to do the derivation symbolically, and substitute the numerical data at the end.

ehild

I couldn't managed to pull out the answer, so I did what you said.
R_t=(R*R1)/(R+R1)
R_b=R/2

R_eq=(R*R1)/(R+R1)+R/2

ε=R_eq*i

ε=V_t+V_b=V_t+0.5iR
V_t=R_eq*i-0.5iR
i_topleft=V_t/(1.29R)

so
i_A+i_topleft=i
i_A=i-i_topleft

i_A=i-(V_t/(1.29R))=i-((R_eq*i-0.5iR)/(1.29R))=i-(((R*R1)/(R+R1)+R/2*i-0.5iR)/(1.29R))
R1=R*1.29

i_A=i-((1/2.29+1/2)i-0.5i)/1.29
since I won't need i in the answer
A=1-((1/2.29+1/2)-0.5)/1.29)=0.66148
http://www.wolframalpha.com/input/?i=1-((1/2.29+1/2)-0.5)/1.29

which is still wrong. I'll appreciate your correction on my attempt.

dhkdeoen said:
I couldn't managed to pull out the answer, so I did what you said.
R_t=(R*R1)/(R+R1)
R_b=R/2

R_eq=(R*R1)/(R+R1)+R/2

ε=R_eq*i

ε=V_t+V_b=V_t+0.5iR
V_t=R_eq*i-0.5iR
i_topleft=V_t/(1.29R)

so
i_A+i_topleft=i

The equation in red is wrong if iAmeans the current through the ammeter.

ehild

I just realized I've mixed up my assumptions.

i_A+i_topleft=0.5i

is this right?
and, yes i_A is the current through the ammeter.

dhkdeoen said:
I just realized I've mixed up my assumptions.

i_A+i_topleft=0.5i

is this right?
and, yes i_A is the current through the ammeter.

Yes, it is correct.

ehild

then it should be

i_A=0.5i-(((R*R1)/(R+R1)+R/2)*i-0.5iR)/(1.29R))

substitute R1=1.29R

i_A=0.5i-(1.06332iR-0.5iR)/(1.29R)
i_A=0.5i-(0.56332/1.29)i
i_A=0.5i-0.43668217i
i_A=0.06331877iI=V/R
i=ε/R

so this should be the answer, but still..

Last edited:
You made it a bit complicated. ((R*R1)/(R+R1)+R/2)*i=ε.

It is simpler with the topright current.
iA=i/2-itopright.
itopright=Vt/r
Vt=ε-ir/2.
-->

iA=i-ε/r.

i=ε/Req, and Req=r(3r1+r)/(2(r+r1)). ..

ehild

Last edited:
1 person
Question: Have you studied Thevenin equivalents yet?

ehild said:
You made it a bit complicated. ((R*R1)/(R+R1)+R/2)*i=ε.

It is simpler with the topright current.
iA=i/2-itopright.
itopright=Vt/r
Vt=ε-ir/2.
-->

iA=i-ε/r.

i=ε/Req, and Req=r(3r1+r)/(2(r+r1)). ..

ehild

it is! Maybe I was too hanging out with R1. Now I got the right answer.
Even though I got the right answer, I really should trying to get answer with R1, though. So much appreciated for your devotion!

gneill said:
Question: Have you studied Thevenin equivalents yet?

I don't think I've heard of that. My class just started with Kirchhoff's two rules.

dhkdeoen said:
then it should be

i_A=0.5i-(((R*R1)/(R+R1)+R/2)*i-0.5iR)/(1.29R))

substitute R1=1.29R

i_A=0.5i-(1.06332iR-0.5iR)/(1.29R)
i_A=0.5i-(0.56332/1.29)i
i_A=0.5i-0.43668217i
i_A=0.06331877i

I=V/R
i=ε/R

so this should be the answer, but still..

You need the answer in term of ε/R, but i=ε/Req

By the way, the derivation is much simpler if you simplify whenever possible.

Vt= i(R1R)/(R1+R), Itopleft=Vt/R1=iR/(R1+R),
iA=i/2-iR/(R1+R)= i(R-R1)/(2(R+R1))...

ehild

dhkdeoen said:
I don't think I've heard of that. My class just started with Kirchhoff's two rules.
Okay. I was just wondering what level of theory you'd covered. While problems like this one can be solved with basic Kirchcoff's laws, they become easier to tackle when you've covered some additional techniques. I thought perhaps it might have been an exercise intended to evoke such an approach. Never mind, carry on

## 1. What is an ammeter?

An ammeter is a device used to measure the electric current in a circuit. It is typically connected in series with the circuit and has a low resistance, so it does not significantly affect the current being measured.

## 2. How do you get currents in an ammeter?

To get currents in an ammeter, you need to connect the ammeter in series with the circuit. This means that the current flowing through the ammeter will also flow through the rest of the circuit.

## 3. What is resistance?

Resistance is the measure of how much a material or component opposes the flow of electric current. It is measured in ohms (Ω) and can be affected by factors such as the material, length, and cross-sectional area of the component.

## 4. How do resistances affect the current in a circuit?

Resistances in a circuit can affect the current by either increasing or decreasing it. The higher the resistance, the less current will flow through the circuit. This is due to Ohm's law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R): I = V/R.

## 5. Why is it important to use an ammeter with low resistance?

It is important to use an ammeter with low resistance because it allows for an accurate measurement of the current in the circuit. If the ammeter had a high resistance, it would affect the current being measured and give an inaccurate reading. Additionally, a high resistance ammeter could potentially overload and be damaged by the current.

• Introductory Physics Homework Help
Replies
7
Views
543
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
851
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
9K
• Introductory Physics Homework Help
Replies
11
Views
2K