I have to calculate the voltage across R5 using the voltage divider formula but the problem is the answer is V5=2.23V (Here V5 is the voltage across R5) but according to my solution,

Solution:

R234567=1.80Ω
On Applying Voltage Divider Formula on R234567,
(Other resistor in series with R234567 is R1)
V234567=6.428V

So this voltage will be same across R2345 and R7. Since these two resistors are in series with one another we can apply voltage divider formula on R2345 to get the voltage across R5.
(R2345=1.528Ω when it's solved)
On Applying Voltage Divider Formula on R2345 we get,
V2345=2.16V

Also V5=V2345
Therefore, V5=2.16V

But according the the given answer V5=2.23V.
I want to know that the answer given with the question is correct or not? • "V234567=6.428V" does not make much sense, you have a voltage between two spots, I suggest you label the nets and express your voltages in terms of those, maybe that will already clear up the confusion. – PlasmaHH Sep 20 '18 at 9:39
• Please rearrange your schematic. There is a free schematic editor in the toolbox. – Long Pham Sep 20 '18 at 9:46
• Has any professional EE ever had to do anything like this, or is it just to torture students? – Dirk Bruere Sep 20 '18 at 9:58
• Is this your homework? – Anuradha Sep 20 '18 at 10:31
• Yep. That one hit closer to home than I expected. – winny Sep 20 '18 at 13:00

The best way to do this is to re-draw the circuit completely so it makes sense. The circuit is drawn that way to deliberately confuse.

You can see that R2 is in parallel with R3, this is in series with R4, and that is all in parallel with R5: simulate this circuit – Schematic created using CircuitLab

Right, now all you need to do is solve the R2//R3 in parallel and add R4 into it: simulate this circuit

Now you can see that this is all in parallel with R5. Remember that the voltage drop across R5 will also be the same as the drop across the (R2//R3)+R4 resistor. So we can turn this into a single element and call this Rp: simulate this circuit

Now you just have to solve the value of Rp, which is quite simple to do. Now you have this, you can see that Rp+R7 are in parallel with R6. Solve this and you can get the voltage at the R1/R7/Rp junction.

Use this voltage and use the voltage divider law to find the voltage between Rp and R7. You now have the voltage levels both sides of Rp (which is both sides of R5 on the original schematic).

Simply take one away from the other and you have your voltage across R5, which I calculated to be 2.24V, although that was likely rounding error! So yes, the answer given with the question is correct.

Just to show this is indeed the correct method and answer, I used my calculated value of Rp and simulated on Proteus. I covered my calculated value of Rp so I would not give it away, so OP would have to do it themselves. As you can see, the answer 2.23V is correct. NOTE: As this looks like a homework question, I have not given away everything. I have shown OP the method used, so they can do the working out themselves.

• Thank you! Yeah I can understand why didn't you gave away the working but the way you drew the schematic helped a lot.Also the explanation was spot on. – Muhammad Ali Khan Sep 20 '18 at 11:25
• @MuhammadAliKhan Yeah in these cases, re-drawing the schematic really simplifies things. I find with these types of questions it's better to show the method and allow people to actually solve it themselves as then they get to learn from it. Glad this helped. – MCG Sep 20 '18 at 11:31