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First of all let me give you the schematic diagram of the circuit,

enter image description here

Here, how to find the current passing through 150 ohm resistor?

These are the things I have tried,

Since I want to find the current through 150 ohm resistor I need the potential drop across that resistor , therefore in order to do find the potential drop across 150 ohm resistor what I did was I marked the nodes as shown below

enter image description here

After that I took the voltage through-out (green)node as 10V(10V battery is on green node) , likewise I took the voltage throughout red colour node as 5V.

Then I could find the voltage drop through 150 ohms, so let's assume it as K

Therefore 10V - k = 5V

So K=5V

Then the current through that resistor is ,

V=IR 5V/150 I = 0.03A

But the answer is wrong, The correct answer given is 0.1A.

What is the mistake I have done? Please help me

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    \$\begingroup\$ Which node did you assume as your ground (reference) node? \$\endgroup\$ – Roger Rowland May 21 '15 at 6:32
  • \$\begingroup\$ your first equation should be (10V +5V = k) , red node is 5V lower then the black node , which again is 10 V lower than the green node. \$\endgroup\$ – Password May 21 '15 at 6:41
  • \$\begingroup\$ There is 15V across 150 ohms. Do you see that? \$\endgroup\$ – Andy aka May 21 '15 at 7:14
  • \$\begingroup\$ @ Andy aka: Did you see the figure I draw with nodes? there in green node voltage is 10v while at red is 5v , therefore since the current goes through high potential to low potential potential across 150ohm would be 10v-5v = 5v, so what is the fault I have done? \$\endgroup\$ – On the way to success May 21 '15 at 7:27
  • \$\begingroup\$ VOLTAGE across CURRENT through \$\endgroup\$ – JonRB May 21 '15 at 12:22
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An alternative method to solve this would be using the superposition theorem. It isn't really required in this case, but in more complicated setups with multiple sources it is certainly useful. Here is how you do it.

  1. 10V source
  2. 5V source

1.

schematic

simulate this circuit – Schematic created using CircuitLab

Looking at the diagram above it is clear that the current through R3 is

$$ I_{R3}=\frac{V}{R_3} = \frac{10}{150}=\frac{1}{15}\text{Amps} $$


2.

schematic

simulate this circuit

Because of the short R2 and R1 are in parallel and become zero and therefore the current through R3 is:

$$I_{R3}=\frac{V}{R_3}=\frac{5}{150}=\frac{1}{30}\text{Amps}$$


Now because the direction of current assumed in part 1 and 2 is the same, both the currents are additive. Hence:

$$I = \frac{1}{15}+\frac{1}{30} = 0.1\text{Amps}$$

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  • \$\begingroup\$ How could you say "Because of the short R2 and R1 are in parallel and become zero " \$\endgroup\$ – On the way to success May 21 '15 at 10:34
  • \$\begingroup\$ Solve this for the equivalent resistance and see what you get R1||R2||0 \$\endgroup\$ – Sada93 May 21 '15 at 11:32
  • \$\begingroup\$ Okay your way seems easier, could you plz tell me what is the mistake I have done?Did you read what I have done? \$\endgroup\$ – On the way to success May 21 '15 at 11:35
  • \$\begingroup\$ You have not taken a ground reference. When solving circuits always define a ground point. So when you say some node is at 5V it is at 5V with respect to ground. You have not indicated where ground is so your initial voltage assumptions are wrong. In my circuit above i have clearly defined ground, you should do the same. \$\endgroup\$ – Sada93 May 21 '15 at 11:38
  • \$\begingroup\$ I want to clear another thing also, is the voltage within a lopp equal? \$\endgroup\$ – On the way to success May 21 '15 at 11:45
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Your voltage equation is incorrect. Both voltage sources have the same polarity with respect to the resistor, therefore they add.

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The voltage sources are both in the same direction Therefore your calculation for K is wrong. It should be:

K = 10V + 5V = 15V

from U = R*I we derive I = U/R and end up with:

I = 15 V / 150 Ohm = 0.1 A

The special case of this circuit is that all resistors are connected directly to a voltage source and the voltage drop over each resistor is given straightforward from the sum of connected voltage sources (assuming unrestricted currents from sources).

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  • \$\begingroup\$ It means , you are saying that the potential drop across 150 ohm resistor is 15 v? \$\endgroup\$ – On the way to success May 21 '15 at 6:58
  • \$\begingroup\$ Yes, exactly. The voltage drops are 10V over both 100 Ohm resistors and 15V over the 150 Ohm resistor. this is because they are all directly connected to a voltage source. because a voltage source always provides a fixed voltage, independent of the current, the drops over the resistors are directly the voltages on the nodes. \$\endgroup\$ – foehnx May 21 '15 at 7:50
  • \$\begingroup\$ Okay, but I am confused with a different thing. Let me clear that thing , in green node voltage is 10v while at red is 5v , therefore since the current goes through high potential to low potential ,potential across 150 ohm would be 10v-5v = 5v, \$\endgroup\$ – On the way to success May 21 '15 at 7:53
  • \$\begingroup\$ But why that way is wrong? \$\endgroup\$ – On the way to success May 21 '15 at 7:54
  • \$\begingroup\$ A potential always has a direction. If both direction are the same, they add up. If the direction are against each other, they subtract. In your sketch, both voltage sources force a current in the same direction (seen from the resistor), namely from top down through the 150 Ohm resistor (technical current direction). \$\endgroup\$ – foehnx May 21 '15 at 12:52
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You can analyze this circuit by inspection (and without having to resort to superposition) if you consider KVL around this loop:

enter image description here

The KVL equation would be

-5 V + (-10 V) + \$V_{150}\$ = 0

where \$V_{150}\$ is the voltage across the 150-ohm resistor (with positive reference terminal at the top).

Rearranging,

\$V_{150} = 10 \mathrm{V} + 5 \mathrm{V}\$.

But this is really just a fancy way of saying that by looking at the loop I outlined you can immediately see that 15 V is applied across the 150-ohm resistor.

And 15 V / 150 ohms is 0.1 A, the given answer.

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  • \$\begingroup\$ You drew that path by considering that mid 100 ohm resistor is shorted one? \$\endgroup\$ – On the way to success May 22 '15 at 0:41
  • \$\begingroup\$ @Onthewaytosuccess, KVL is true for any closed path through a circuit. I chose that one because it makes it easy to calculate the voltage across the 150 ohm resistor. \$\endgroup\$ – The Photon May 22 '15 at 0:50
  • \$\begingroup\$ Aha then you can neglect that 100 ohm resistance , isn't it? \$\endgroup\$ – On the way to success May 22 '15 at 1:49
  • \$\begingroup\$ You can write a KVL equation for any closed loop. Since the 100 ohm resistor is not part of the loop I chose, it has no effect on the equations for that loop. In some other circuit a single KVL equation might not be sufficient to solve for some desired variable, but in this case it was. \$\endgroup\$ – The Photon May 22 '15 at 1:52
  • \$\begingroup\$ Okay, I got it. \$\endgroup\$ – On the way to success May 22 '15 at 2:09
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Here's a more-or-less easy way to look at it:

enter image description here

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  • \$\begingroup\$ What you have mentioned in-front of 3? \$\endgroup\$ – On the way to success May 21 '15 at 11:47
  • \$\begingroup\$ @Onthewaytosuccess: 1) shows the circuit as you presented it, 2)removes R1 and R2 in order to show the two voltage sources in series with R3 and calculates the current through R3, 3)shows that R1 and R2 are in parallel and that the equivalent resistance of the parallel pair is 50 ohms, 4) shows a circuit identical with the one under 1), and 5) shows that there are two currents in the circuit, and that the addition of resistance across V1 won't have any effect on the current through the 150 ohm resistor. \$\endgroup\$ – EM Fields May 21 '15 at 13:33

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