# Equivalent Resistance in partial parallel/series cicuit

• thatgirlyouknow
In summary, to find the equivalent resistance between points A and B, we first reduce the resistors in parallel (1/8 + 1/9 = 1/4.235), then in series (3 + 4 + 6 = 13), and finally in parallel again (1/10 + 72/17 = 737/170). This gives us a final equivalent resistance of 737/170 ohms.
thatgirlyouknow

## Homework Statement

Find the equivalent resistance between points A and B in the drawing.

## Homework Equations

Rseries = R1+R2+R3...
Rparallel: 1/Rp = 1/R1+1/R2+1/R3...

## The Attempt at a Solution

Reducing it down:
1/8+1/9 = 1/R1 = 4.235
That's in parallel with 20, so:
1/20+1/4.235 = 3.495 ohms

3, 4, and 6 are in series:
3+4+6 = 13 ohms
Add that to the previous and you have 16.495 ohms, which is incorrect.

Is one parallel and I'm not adding it as such? Or should I combine differently?

Thanks so much!

#### Attachments

• circuit2.gif
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thatgirlyouknow said:
Reducing it down:
1/8+1/9 = 1/R1 = 4.235
Good. R1 = 4.235 (not 1/R1).
That's in parallel with 20, so:
Nope. The 3 ohm resistor is in the way.

So then 20 and 3 are in series, right?
20+3=23 ohms

So now my drawing has:

23 ohms . 4 ohms

4.235 ohms

. 6 ohms

Would the 4 and 6 be considered series?

4+6 are in series
9+8 are in parallel
3 and 20 are not in series!

You need to do this step by step...

6+4 = 10
1/8 + 1/9 = 17/72
now you have

------3 -----------
--|--------|------|
-20------17/72--10
--|--------|------|
-------------------

Next you have 17/72 and 10 in parallel.

1/10+72/17 = 737/170

-- -- 3 -------
--|---------|
-20------737/170
--|---------|
--------------

Now the 3 and 737/170 are in series so just add them to get

-------------
--|---------|
-20-------1147/170
--|---------|
-------------

Now these are in parallel you can solve!

When you added for 17/72 and 10 in parallel, shouldn't it have been 1/4.235+1/10 = 1/R = 2.975?

so then with
------3--------
| |
20 2.975
| |
------------

So now which should I combine first?

first take this and then you shall have the feel of the question much better..""

General rules for doing the reduction process include:

1. Two (or more) resistors with their heads directly connected together and their tails directly connected together are in parallel, and they can be reduced to one resistor using the equivalent resistance equation for resistors in parallel.

2. Two resistors connected together so that the tail of one is connected to the head of the next, with no other path for the current to take along the line connecting them, are in series and can be reduced to one equivalent resistor.

Finally, remember that for resistors in series, the current is the same for each resistor, and for resistors in parallel, the voltage is the same for each one

## What is the definition of equivalent resistance in a partial parallel/series circuit?

Equivalent resistance in a partial parallel/series circuit refers to the total resistance of a circuit that is simplified to a single resistor. This single resistor would have the same current-voltage relationship as the original circuit.

## How is equivalent resistance calculated in a partial parallel/series circuit?

To calculate equivalent resistance in a partial parallel/series circuit, you need to first find the total resistance of the parallel resistors and the series resistors separately. Then, you can use the formula: 1/Req = 1/R1 + 1/R2 + ... + 1/Rn, where Req is the equivalent resistance and R1, R2, ... Rn are the individual resistances.

## What is the difference between partial parallel and partial series circuits?

In a partial parallel circuit, there are multiple paths for the current to flow, while in a partial series circuit, there is only one path for the current. In a partial parallel circuit, the voltage is the same across all resistors, while in a partial series circuit, the current is the same through all resistors.

## How does the addition of resistors affect the equivalent resistance in a partial parallel/series circuit?

In a partial parallel circuit, adding more resistors in parallel will decrease the equivalent resistance, as there are more paths for the current to flow. In a partial series circuit, adding more resistors in series will increase the equivalent resistance, as the current has to pass through each resistor in the path.

## Why is it important to calculate equivalent resistance in a partial parallel/series circuit?

Calculating equivalent resistance allows us to simplify complex circuits into a single resistor, making it easier to analyze and understand the circuit. It also helps in determining the total current and voltage in the circuit, which is essential for designing and troubleshooting electrical systems.

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