Doubt about potential difference through a resistor

circuit analysis

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

Best Answer

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.

10V source
5V source

schematic

^{simulate this circuit – Schematic created using CircuitLab}

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

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

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:

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

Related Solutions

Electronic – Developing an intuition about this circuit

Other people have given good quantitative methods. If you want a more qualitative method, one place you can start is to try to figure out which way the currents are flowing in each branch. That will usually help give you some intuition as to what's happening with the voltages, too.

There are three branches -- The V1/R1 branch, the R2 branch, and the I1 branch. The I1 branch has a constant current source, so you know which way the current is flowing. Next, look at the R2 branch. All of your sources are pointing towards the top of R2, which tells us that the current will be flowing from top to bottom.

That leaves the V1/R1 branch. The question here is, under what conditions will current flow into the branch (from the top node towards V1) and out of the branch (from V1 towards the top node)? Well, it depends on the voltage at the top node, which I'll call Vtop. When Vtop < V1, V1 sources current. When Vtop > V1, V1 sinks current. When V1 = Vtop, no current flows through V1.

Let's imagine that V1 = Vtop, so there's no V1 current. All of the I1 current goes through R2. What would Vtop be, then? \$3A * 5\Omega = 15V\$. Do we have anything close to 15V in V1's branch? Nope! Therefore, V1 must be sinking some of I1's current -- probably a fair bit since it's a small voltage source.

In this circuit, the resistors were all roughly the same size. That's not always the case. If one resistor is much larger than the others, anything in series with it will have a small effect on the rest of the circuit. Likewise, a small resistance will tend to short out whatever's in parallel with it. (You'll see this principle at work in passive filters once you get to AC circuit analysis.)

Lastly, there are two special cases that are very rare but may help with your intuition. The first is an ideal voltage source in parallel with an ideal current source. In this case, the current source has no effect at all. Likewise, when an ideal voltage source is in series with an ideal current source, the voltage source has no effect at all. You should make sure you understand why. Feel free to post a comment if you want to confirm anything.

Electronic – Homework: DC circuit help

By applying the superposition theorem to your circuit I have noticed that the current you are looking for can only be dependent from voltage source E, because when you are looking at another source, E is shorted, so your current gets 0A. So now it depends on E alone (all other sources shorted or open).

Ig is open
Ig4 is open
E4 is shorted

So R1 gets in serial to R4 and R2 and R3 get just serial, too. Finally the serial resistance groups are in parallel.

R14=2 Ω + R4
R23=8 Ω
=> R14|23 = (16 Ω + 8R4)/(10 Ω + R4)
I=-(E / R14|23)
 =-(50 Ω + 5R4)/(16 Ω + 8R4) A
=> R4=(-50 ΩA - 16I Ω )/(8I + 5 A)

When I=-3 A:
R4=2/19 Ω ≈ 0.1 Ω
When I=-1 A:
R4=34/3 Ω ≈ 11.3 Ω

As R4 is effectively rising with I, it must be that R4∈[0.1 Ω; 11.3 Ω].

Hope this helped a bit, though late....

Best Answer

Related Solutions

Electronic – Developing an intuition about this circuit

Electronic – Homework: DC circuit help

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