I agree with Brad that it's literally impossible.Hi i have been stuck on this circuit for a long time and it seems impossible how current labelled I4 on the circuit below is 4mA.
Rather than scanning various books for an exact solution, you should try to read about the general method to solve similar problems, as it's e.g. sketched in the DC network analysis chapter of this book https://www.allaboutcircuits.com/vol_1/chpt_10/index.htmli have searched 5 books and none of them do an example on more than 4 resistors connected in this way.
To analyze the above circuit, one would first find the equivalent of R2 and R3 in parallel, then add R1 in series to arrive at a total resistance. Then, taking the voltage of battery B1 with that total circuit resistance, the total current could be calculated through the use of Ohm's Law (I=E/R), then that current figure used to calculate voltage drops in the circuit. All in all, a fairly simple procedure.
Not quite right. R3 is series connected (resistance added) to the previously calculated "R456". So you get R3456 = 3k.Then i calculate the parallel resistance of R456 = (2k * 2k)/(2k + 2k) = 1k
And this is where i get confused i need to calculate the overall resistance of this circuit to calculate I1 but i don't know if i do as follows
R4 + R3 = 4k then R234 = (4k * 3k)/(4+3) = 1.7k
Hey osa1313,
If you want voltage drop across R1 you should do as follows;
I1*R1 = 4mA * 1k = 4V ;
or you could also do 10 - 6V (the voltage you calculated is the drop across R2 which is in parallel with the rest of the circuit)
Hope it helps
Regards
K
Hello guys,
I tried to solve this problem, but I don't know if it's correct. Could someone see it? Maybe will help osa1313.
View attachment 78385
Best regards.
Thank you.
But, now I am a little bite confused. You said in your post that the i1 is 4mA, but my calculations gave me 1,818mA. We can say, approximately 2mA. Right?
It looks like it is giving me half of the values.
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