Bridge Circuit (resistors, whoo)

In summary: Applying it to the bottom junction (B):(VB-VA)/R = (VB-VA)/10 + (VB-VA)/2Net current out of something must be positive, or else charge is beingreleased from that something.In summary, David found that if the potentials on the right and left are the same, then the net current into and out of something must be zero. If the potentials on the right and left are different, then the net current out of something must be positive.
  • #1
GDGirl
50
0

Homework Statement


The resistor R in the above figure has a resistance of 21.7 Ω.
a) Find the current in the resistor R. (Enter a positive number for a current up and to the right, and negative for down and to the left.
b) Find the current in the 10-Ω resistor. (Enter a positive number for a downward current, negative for an upward current.)
https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/Knox/phys130a/spring/homework/10/03/P26_29.jpg

Homework Equations


Current Leaving Battery= Voltage/total resistance
V=IR (Ohm's Law)


The Attempt at a Solution


Well, I tried to find the total resistance and ended up with 2.692 (I'm not sure this is right because of how the circuit is set up), then I divided the voltage by this, and got 2.228 as the current from the battery. Then I used this current to find the voltage at the point after the resistor R with Ohm's Law, which gives me 60.378V. I use the voltage drop between the points (which is -54.3788) and the resistance of the resistor to find the current (from Ohm's Law again), this gives me -2.007. This is wrong, as I expected. I'm not sure how to go about doing this. :/
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
There is no quick fix for this.
It is a full blown exercise in applying Kirchoff's laws.
 
  • #3
I didn't think about Kirchhoff's laws. :/ Dang. I suppose it's because my prof only ever used it in problems with multiple power sources. Alright, thanks
 
  • #4
I like to work with potentials at the junctions.
(Doing this trivially satisfies the law about potential drops around a closed
loop summing to zero).

Let the potential on the right be 0V and so on the left it is 6V.
Let the potential of the top junction be VA and the bottom be VB.

All the currents can be expressed easily in terms of VA and/or VB by
Ohm's Law. (e.g. the current through the 10 ohms is (VA-VB)/10)

Now apply "what flows in must flow out" to the top and bottom junctions,
and we get two simultaneous equations in VA and VB. Just solve them and
you have all you need.

David
 
Last edited:
  • #5
I think I understand how to do everything you're describing well enough- except the last bit. What do you mean by "apply 'what flows in must flow out' to the top and bottom junctions"? Like, I understand the what flows in part, obviously if you have a current leaving a battery, the same current should be going back into it, but I'm not sure what you mean about applying it to the junctions.
 
  • #6
Applying it to the top junction (A):

(6-VA)/R = (VA - VB)/10 + VA/2

Net current into something must be zero, or else charge is accumulating
in that something.
 
  • #7
Alright, I think I've got it now, thanks.
 

1. What is a bridge circuit?

A bridge circuit is an electrical circuit consisting of four resistors in a diamond shape, with a power source and a galvanometer (a device used to measure small electric currents) connected to the opposite corners. It is commonly used to measure unknown resistances or to compare two resistances.

2. How does a bridge circuit work?

A bridge circuit works by using a null measurement technique. The four resistors are arranged in a way that when the bridge is balanced, there is no current flowing through the galvanometer. This balance is achieved by adjusting the values of the resistors until the bridge is balanced, and the unknown resistance can be calculated using a simple formula.

3. What are the advantages of using a bridge circuit?

One major advantage of using a bridge circuit is its high level of accuracy. It can measure resistances with a precision of up to 0.01%. It is also relatively simple and inexpensive to construct, making it a popular choice for many applications. Additionally, it is a versatile circuit that can be used for measurement, comparison, and testing purposes.

4. What are the limitations of a bridge circuit?

One limitation of a bridge circuit is that it requires a power source, which can introduce errors in the measurement. It also relies on the assumption that the resistors used are linear, which may not always be the case. Additionally, the bridge circuit may not work well with very small or very large resistances, as it may not be sensitive enough to detect the changes in balance.

5. What are some common applications of a bridge circuit?

A bridge circuit has many practical applications, such as measuring the resistance of a strain gauge, determining the purity of a metal, and testing the accuracy of resistors. It is also commonly used in electronic devices and systems, such as audio amplifiers, to adjust and balance the output signals. Additionally, it can be used in laboratory experiments to measure unknown resistances and calibrate other instruments.

Similar threads

Why do the headlights on the car become dim when the car is starting?
    • Introductory Physics Homework Help
Replies
4
Views
263
Understanding example in Wikipedia entry for open circuit voltage
    • Introductory Physics Homework Help
Replies
3
Views
561
Analyzing Bridge Circuit w/ Loop Currents
    • Introductory Physics Homework Help
Replies
6
Views
593
Ohm's Law - Finding Source Voltage
    • Introductory Physics Homework Help
Replies
3
Views
531
A circuit involving two non-linear resistances (incandescent lamps)
    • Introductory Physics Homework Help
Replies
14
Views
605
Current through a circuit when the key is open
    • Introductory Physics Homework Help
2
Replies
42
Views
1K
Two concentric conducting spherical shells and resistor in between
    • Introductory Physics Homework Help
2
Replies
44
Views
896
Parallel circuit with internal resistance
    • Introductory Physics Homework Help
Replies
6
Views
1K
Finding the value of this load resistor
    • Introductory Physics Homework Help
Replies
2
Views
822
Finding the Current Through a Resistor (Working With Parallel and Series)
    • Introductory Physics Homework Help
Replies
3
Views
926

Hot Threads

Recent Insights

Back
Top