Equivalent Resistance of a Ciruit

In summary, the equivalent resistance of the circuit can be found by combining R3 and R4 in parallel, then adding that to R2 in series. R6 and R5 can be added together in series. Then, the parallel combination of R1, R7, and the series combination of R5 and R6 can be added to the first equivalent resistance found. The whole circuit can also be simplified by lining up the resistors in a straight line.
  • #1
GeorgeCostanz
31
0

Homework Statement



Find the equivalent resistance of the circuit as shown in the diagram below; where, R1 = 2 Ω, R2 = 1 Ω, R3 = 2 Ω, R4 = 1 Ω, R5 = 4 Ω, R6 = 4 Ω, and R7 = 4 Ω.

http://i.imgur.com/OsAs2.gif

Homework Equations




The Attempt at a Solution



R3 and R4 are in parallel, found R of those 2 then added that to R2 (in series)

R6 and R5 in series, added those together

this is where i get lost

I added R1 + R7 + (R6+R5) in parallel.

then added that to the 1st R-eq i found - in parallel

wrong.

i'm clearly confused about the relationships

thanks
 
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  • #2
GeorgeCostanz said:

Homework Statement



Find the equivalent resistance of the circuit as shown in the diagram below; where, R1 = 2 Ω, R2 = 1 Ω, R3 = 2 Ω, R4 = 1 Ω, R5 = 4 Ω, R6 = 4 Ω, and R7 = 4 Ω.

http://i.imgur.com/OsAs2.gif

Homework Equations




The Attempt at a Solution



R3 and R4 are in parallel, found R of those 2 then added that to R2 (in series)

R6 and R5 in series, added those together

this is where i get lost

I added R1 + R7 + (R6+R5) in parallel.

then added that to the 1st R-eq i found - in parallel

wrong.

i'm clearly confused about the relationships

thanks

The R 5&6 combination [in series with each other] is in parallel to the R2,3,4 combination you established. Then that whole combination is in series with R 1&7.
 
  • #3
GeorgeCostanz said:

Homework Statement



Find the equivalent resistance of the circuit as shown in the diagram below; where, R1 = 2 Ω, R2 = 1 Ω, R3 = 2 Ω, R4 = 1 Ω, R5 = 4 Ω, R6 = 4 Ω, and R7 = 4 Ω.

http://i.imgur.com/OsAs2.gif

Homework Equations

The Attempt at a Solution



R3 and R4 are in parallel, found R of those 2 then added that to R2 (in series)

R6 and R5 in series, added those together

this is where i get lost

I added R1 + R7 + (R6+R5) in parallel.

then added that to the 1st R-eq i found - in parallel

wrong.

i'm clearly confused about the relationships

thanks

It sometimes helps to draw the resistor set up in a straigh line, rather than 3 sides of a square.

This would start with R1 , then divide to two branches, with R5&R6 on the bottom, and R2, along with a parallel R3&R4 on the top, the the branches re-joining to get to R7.
 
Last edited:
  • #4
hmm, i never thought to look at it that way.
thanks guys
 
  • #5
for any help
To find the equivalent resistance of this circuit, we can use the following formula:

Req = (R1 + R2) || (R3 + R4) + R5 + R6 + R7

Where || represents the parallel combination of resistors.

Plugging in the values given, we get:

Req = (2 + 1) || (2 + 1) + 4 + 4 + 4

Simplifying, we get:

Req = 3/2 + 3/2 + 4 + 4 + 4

Req = 8 Ω

Therefore, the equivalent resistance of the circuit is 8 Ω.
 

What is meant by the equivalent resistance of a circuit?

The equivalent resistance of a circuit is the total resistance that a single resistor would have if it were replacing all the resistors in the circuit. It takes into account both the individual resistances and the way they are connected in the circuit.

Why is it important to calculate the equivalent resistance of a circuit?

Calculating the equivalent resistance of a circuit is important because it helps us understand how the circuit behaves as a whole. It also allows us to determine the amount of current flowing through the circuit and the voltage drops across each resistor.

How do you calculate the equivalent resistance of a series circuit?

In a series circuit, the equivalent resistance is equal to the sum of all the individual resistances. This is because the current flows through each resistor in sequence, and the total resistance is the sum of the resistance values.

How do you calculate the equivalent resistance of a parallel circuit?

In a parallel circuit, the equivalent resistance is calculated differently. The inverse of the equivalent resistance is equal to the sum of the inverses of each individual resistance. This is because the current splits up and flows through each resistor separately, resulting in a lower overall resistance.

Can the equivalent resistance of a circuit be lower than the lowest individual resistance?

Yes, it is possible for the equivalent resistance of a circuit to be lower than the lowest individual resistance. This can happen in a parallel circuit, where the total resistance decreases as more resistors are added in parallel. However, the equivalent resistance can never be lower than 0, as this would mean there is no resistance in the circuit.

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