# Series/Parallel DC circuit analysis training issue

I'm blocked in this exercise. I don't get how I can compute R1/R2/R3 voltage.

Here's the schema :

simulate this circuit – Schematic created using CircuitLab

So I was able to calculate some values :

$R_{T} = \dfrac{1}{\dfrac{1}{10k\Omega}+\dfrac{1}{7k\Omega+\dfrac{1}{\dfrac{1}{6k\Omega}+\dfrac{1}{1k\Omega+2k\Omega+3k\Omega}}}} = 5k\Omega$

$R_{eq_{1,2,3}} = R_{1}+R_{2}+R_{3} = 1k\Omega+2k\Omega+3k\Omega = 6k\Omega$

$R_{eq_{1,2,3,4}} = \dfrac{1}{\dfrac{1}{R_{4}}+\dfrac{1}{R_{eq_{1,2,3}}}} = \dfrac{1}{\dfrac{1}{6k\Omega}+\dfrac{1}{6k\Omega}} = 3k\Omega$

$I_{T} = \dfrac{V_{1}}{R_{T}} = \dfrac{10V}{5k\Omega} = 2mA$

$V_{R6} = V_{AppliedVoltage} = 10V$

$V_{R5} = \dfrac{V_{1} \times R_{5}}{R_{4}+R_{5}} = \dfrac{10V \times 7k\Omega}{3k\Omega+7k\Omega} = 7V$

$V_{R4} = (I_{T} - I_{R_{6}}) \times R_{eq_{1,2,3,4}} = (2mA - 1mA) \times 3k\Omega = 3V$

$R_{1}$ should be equal to 0.5V, $R_{2}$ should be equal to 1V and $R_{3}$ should be equal to 1.5V. But I can't find a way to compute them... I've cheated a bit and take a look at all the answers to try to "RE" based on the excepted result but can't figure out what I have to do... Any ideas ?

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Thévenin says: Remove the load that you are interested in and short all voltage sources:

simulate this circuit – Schematic created using CircuitLab

Notice that R6 is shorted by the voltage source and can be left out:

simulate this circuit

Now it is clear that when you look into the circuit to determine RTH = R4||R5.

Next we move the voltage source back in place and calculate VTH:

simulate this circuit

VTH can be calculated by the voltage divider rule: $V_{TH} = \dfrac{R_4}{R_4+R_5}\cdot V_1$

Now you have a simple 4 resistor voltage divider that allows you to calculate the voltage across every resistor:

simulate this circuit

$V_{R_1} = \dfrac{R_1}{R_1+R_2+R_3+R_{TH}}\cdot V_{TH}$

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I didn't know yet this theorem and it's really useful ! Will practice it on the next exercise, thank you ! –  Emmanuel Istace Dec 29 '13 at 13:37
@EmmanuelIstace Google a bit for Thévenin's theorem. –  jippie Dec 29 '13 at 14:12
Already done and found some exercises to learn how to factorize circuit in a simple one, I'm on actually. –  Emmanuel Istace Dec 29 '13 at 14:33

Applying Thevenin's theorem is often a good approach as another answer has shown.

Another approach is to redraw the circuit such that the solution can be written by inspection:

simulate this circuit – Schematic created using CircuitLab

Now, use voltage division to find the voltage across R4 and then voltage division again to find the voltage across R1, R2, and R3:

$$V_{R4}= 10V \dfrac{R_4||(R_1 + R_2 + R_3)}{R_5 + R_4||(R_1 + R_2 + R_3)}$$

$$V_{R1} = V_{R4}\dfrac{R_1}{R_1 + R_2 + R_3}$$

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If you know VR4, then you know VR1+2+3 (they are equal), so if you calculate R1+2+3 (=R1+R2+R3), you can calculate IR1+2+3 (=V/R), which is also equal to IR1 = IR2 = IR3, then you can calculate VR1 as R1 * IR1, and so on.

"IR1+2+3, which is also equal to IR1 = IR2 = IR3" means IR1+2+3 = IR1 = IR2 = IR3.

Sorry, but this is a very basic knowledge.

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So I need to device IR1,2,3 by 3 ? So to get the current of an individual resistor in a resistor series circuit I have to calculate the total current divided by the number of resistor ? So in series : Irx = (Vt/Rt) / RCount ? –  Emmanuel Istace Dec 29 '13 at 12:57
Basic knowledge but I'm a fresh hobbyist and we all have learned one day ;) –  Emmanuel Istace Dec 29 '13 at 13:12

You should be able to solve this circuit just by looking at it. The component values were chosen that way.

Ra = R1 + R2 + R3 = 6K

Note that Ra == R4, so current will divide equally between them.

Rb = Ra || R4 = 3K

Rc = Rb + R5 = 10K.

At that point, you have 10K across a 10V source. Current through Rc = 10V / 10K = 1 mA.

Rb and R5 are in series. so the same current flows through them. Rb (the parallel combination of R4 with the series combination of R1, R2, R3) sees 1 mA.

Because the current divides equally between the two arms, R1, R2, and R3 all see 0.5 mA.

V(R1) = R1 * 0.5 mA = 1K * 0.5 mA = 0.5V.

V(R2) = 2K * 0.5 mA = 1 V.

V(R3) = 3K * 0.5 mA = 1.5V.

It is usually easier to solve networks like this in pieces, rather than trying to set up one grand equation.

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