Solving Wheatstone Bridge w/ R1=2000, R2=3000, R3=4000 & 12V Battery

In summary: I'll try to do that. In summary, if the voltage of the battery is 12 volts and the resistors are 2000 ohms, 3000 ohms, and 4000 ohms, the following is true: -The voltage between nodes 1 and 2 is 12 volts-The voltage between nodes 2 and 3 is 12 volts-The voltage between nodes 3 and 4 is 0 volts-The voltage between nodes 4 and 5 is 12 volts
  • #1
control
14
0

Homework Statement


Set voltage between 1. a-b; 2- b-c; 3 c-d 4. a-c; if voltage of battery is 12V and R_1=2000 ohms R_2=3000 ohms; R_3= 4000 ohms. (see attached file)

Homework Equations


voltage=final voltage- innitial voltage
resistors in series R_1,2=R_1+R_2
parallel resistors: R_1,2=(R_1*R_2)/(R_1+R_2)

The Attempt at a Solution


I have no idea how to solve such problem. I can use Kirchhoff laws in circuits, I can use all equations I wrote in "relevant equations", I watched video how to set equivalent resistance of those resistors through "delta to ypsilon transformation" () but I have absolutely no idea how to apply my knowledge in such problem.
 

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  • #2
control said:
I can use Kirchhoff laws in circuits, I can use all equations I wrote in "relevant equations"
Good. So that means you can write the KVL loop equations for the circuit (or the KCL node equations). What do you get?
 
  • #3
Well, resistance of equivalent resistor would be R_eq=2461.44 ohms.
KVL loop looks: +voltage of battery-(current*equivalent resistor)=0 => 12volts-current*2461.44ohms=0 => current=0.004875ampers
So what should I do next?
 
  • #4
control said:
Well, resistance of equivalent resistor would be R_eq=2461.44 ohms.
KVL loop looks: +voltage of battery-(current*equivalent resistor)=0 => 12volts-current*2461.44ohms=0 => current=0.004875ampers
So what should I do next?
You have not solved the circuit. You need to either solve for loop currents (using KVL loop equations) or for node voltages (KCL node equations). My own preference would be to write node equations for the two essential nodes (b and c) since the problem is asking for potential difference between nodes.
 
  • #5
You mean:
-I_3*R_1-I_2*R_3+I_1*R_2=0
-I_2*R_3-I_4*R_1+I_5*R_2=0
I_1+I_2=I_4
I_3=I_2+I_5
?
upload_2017-5-26_1-38-10.png
 
  • #6
That's a start. You need one more loop, one involving the voltage source.
 
  • #7
There should be two loops involving source:
ε-R_1*I_3-R_2*I_5=0
ε-R_2*I_1-R_1*I_4=0
 
  • #8
control said:
There should be two loops involving source:
ε-R_1*I_3-R_2*I_5=0
ε-R_2*I_1-R_1*I_4=0
Okay, but you'll only need one of them; you need just enough loops so that every component is included in at least one of the loops. After that any additional loops will not introduce any new (independent) information to the system of equations.
 
  • #9
Ok, so now I choose 5 equations with 5 unknown currents, find them and using Ohm´s law I will find voltage on resistors?
 
  • #10
control said:
Ok, so now I choose 5 equations with 5 unknown currents, find them and using Ohm´s law I will find voltage on resistors?
Yes. Three loop equations and two node equations (KCL at two of the nodes).

If I may make a suggestion, you can reduce the number of current unknowns if you are careful about how you introduce new currents when you label your circuit. Proceed systematically and only introduce the minimum number of new currents required at junctions. For example, if one current (say ##i_1##) enters a junction and two leave, then only introduce one new current (say ##i_2##) leaving the junction by one of the paths so that ##i_1 - i_2## leaves by the other. If you do it this way you're effectively doing the KCL sums ahead of time and you'll end up with the minimum number of current variables required and you'll only require loop currents to solve.

upload_2017-5-26_6-49-59.png
 
  • #11
OK, thanks a lot.
 

Related to Solving Wheatstone Bridge w/ R1=2000, R2=3000, R3=4000 & 12V Battery

1. How do I calculate the total resistance of the Wheatstone Bridge?

The total resistance of a Wheatstone Bridge can be calculated using the formula: R = (R1 * R3) / (R2 + R3). In this case, the total resistance would be (2000 * 4000) / (3000 + 4000) = 8000 / 7000 = 1.14 ohms.

2. What is the voltage across each resistor in the Wheatstone Bridge?

The voltage across each resistor can be calculated using Ohm's Law (V = IR) and Kirchhoff's Voltage Law (the sum of the voltage drops in a closed loop is equal to the applied voltage). In this case, the voltage across R1 would be (1.14 ohms * 1 amp) = 1.14 volts, across R2 would be (1.14 ohms * 0.86 amps) = 0.9864 volts, and across R3 would be (1.14 ohms * 0.71 amps) = 0.8094 volts.

3. How do I determine the current through each resistor in the Wheatstone Bridge?

The current through each resistor can be calculated using Ohm's Law (I = V/R) and Kirchhoff's Current Law (the sum of the currents entering a node is equal to the sum of the currents leaving the node). In this case, the current through R1 would be (12 volts / 2000 ohms) = 0.006 amps, through R2 would be (12 volts / 3000 ohms) = 0.004 amps, and through R3 would be (12 volts / 4000 ohms) = 0.003 amps.

4. What is the power dissipated by each resistor in the Wheatstone Bridge?

The power dissipated by each resistor can be calculated using the formula: P = VI. In this case, the power dissipated by R1 would be (1.14 volts * 0.006 amps) = 0.00684 watts, by R2 would be (0.9864 volts * 0.004 amps) = 0.0039456 watts, and by R3 would be (0.8094 volts * 0.003 amps) = 0.0024282 watts.

5. How do I know if the Wheatstone Bridge is balanced?

A Wheatstone Bridge is considered balanced when the voltage across R1 is equal to the voltage across R2 and the voltage across R3 is equal to the voltage across the battery. In this case, the voltage across R1 is 1.14 volts and the voltage across R2 is 0.9864 volts, which are not equal. Therefore, the Wheatstone Bridge is not balanced.

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