# Confused

Discussion in 'Homework Help' started by Aeolus, Apr 23, 2008.

1. ### Aeolus Thread Starter New Member

Apr 23, 2008
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I was wondering if anyone could take me step by step through this series-parallel circuit. I'm able to calculate the total resistance and current, and the voltage and current of R1 (the current wasn't much of a calculation, just threw that in there so you know what I'm trying to find for the rest of the resistors). Oh, and I'm not quite sure if the battery is facing the right way.

2. ### mik3 Senior Member

Feb 4, 2008
4,846
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Where is the circuit?

There isnt an attachment.

Nov 17, 2003
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Feb 4, 2008
4,846
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5. ### Caveman Senior Member

Apr 15, 2008
471
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No problem. First of all we need to figure out the total resistance that the baterry is seeing. Redraw the circuit after every step so you know what you've got.
1. R6 and R5 are in parallel, so combine them using the parallel equation.
2. Now you can see that the new resistor, R3, and R4 is in series. So you just add the resistances.
3. Now the new resistor you just got is in parallel with R2. Calculate that.
4. And the newest resistor resulting from step 3 is in series with R1. So combine that.

The result is the total effective resistance across the battery. Using ohms law you can calculate the current out of the battery.

Now notice that in the original circuit all of the current coming out of the battery must also go through resistor R1. So they have equal currents. Now using ohms law calculate the voltage across R1.

Got it?

6. ### Caveman Senior Member

Apr 15, 2008
471
1
It is the word "this" in the OP (original posting).

7. ### mik3 Senior Member

Feb 4, 2008
4,846
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Ohhh yes you are right. I am a bit sleepy and i didnt realized it!!!

8. ### flubbo Member

Apr 21, 2008
25
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I'm going to expand on what Caveman said, just to test my own memory.
(It's been at least 15 years since I tinkered with electronics theory)

Before you begin, you need to know about Kirchoff's laws.

1 ) The current is the same at any point in a series circuit.

2 ) The sum of the voltage drops in a series circuit is equal to the applied voltage.

3 ) The voltage is the same across the elements in a parallel circuit.

4 ) The sum of the currents in each branch of a parallel circuit is equal to the total current.

So let's get started.

First, combine R5 with R6, using the rule for total resistance in parallel.

Rt = 1 / (1/Ra + 1/Rb +...)

for two resistors in parallel, the formula reduces to

Rt = (Ra x Rb) / (Ra + Rb)

(30 x 6) / (30 + 6) = 180 / 36 = 5 ohms

Now, the circuit reduces to R3 + R4 + the 5 ohms from the previous calculation. So, we combine R3 with R4 and add 5 ohms, because resistors are simply added for total resistance in series.

7 + 12 + 5 = 24 ohms.

Now, the circuit reduces to R1 + the parallel combination of R2 and 24 ohms. So, we combine R2 with 24 ohms

(8 x 24) / (8 + 24) = 192 / 32 = 6 ohms

So, the entire circuit reduces down to R1 + 6 ohms.

6 + 6 = 12 ohms.

12 ohms is the total resistance.

Now, using Ohm's law, we can compute the total current.

I = E / R = 48 / 12 = 4 amperes of current.

Now, we can compute the voltage drop across each resistor. Since R1 and the total resistance of the parallel network are the same, we will have 24 volts dropped across each resistor. We can check our work using Kirchoff's laws.

E = I x R = 6 x 4 = 24 volts.

24 + 24 = 48. (Q.E.D.)

Since we now know that the voltage drop across the series-parallel network made up of R2 through R6 is 24 volts, we can apply Kirchoff's law and Ohm's law to compute the voltages and currents in the other parts of the circuit.

First, we'll compute the current through R2.

I = 24 / 8 = 3 amps.

Since our total current is 4 amps, 1 amp of current is flowing thru the remainder of the circuit. (R3,R4, and the parallel combination of R5 and R6)

R3 is dropping 7 volts.

R4 is dropping 12 volts.

The parallel combination of R5 and R6 is dropping 5 volts, which we can prove by both ohm's law and Kirchoff's law.

R5 has 5/6 amperes of current.

R5 has 1/6 amperes of current.

To answer you question about the polarity of the battery, resistors are not polarized, so the polarity does not affect the magnitudes, just the signs of the voltage drops. (-24 + -24 = -48) The currents do flow in the opposite direction, however.

If multiple sources were present, this would not be the case, because the voltages / currents would interact differently. For multiple sources, You would have to reduce each source to it's thevenin equivalent (Voltage sources are open circuits, current sources are short circuits, IMSC), and calculate the voltages and currents for each "thevenized" representation and combine them using superposition, mesh or nodal analysis; depending on what method your instructor requires you to use.