Question

Calculate the equivalent resistance R_ab of the circuit shown. The resistorvalues are: R₁ = 7 ohms, R₂ = 4ohms,R₃ = 13 ohms, R₄ = 20 ohms, R₅ =50 ohms, R₆ = 10 ohms, R₇ = 25 ohms,R₈ = 22 ohms,R₉ = 5 ohms, R₁₀ =15 ohms and R₁₁ = 5 ohms.

Answer 1

Secret: 1. Current will divide at the junction where R8 and R9is meeting.

2. Current will take a resistance-free path wherever possible.

Let's start now:

R10 & R11 are in series, so the net R there is(R10+R11=) 20 ohms. Also, R8 & R9 are parallel, so the current will get divided there. Find R8\\R9 = 4.07 Ohm, right! R7is in series with both (R8//R9), So the equivalent Rent(here) = R7 + (R8\\R9) = 29.047, okay! Now:

[R7 + (R8\\R9)] \\ [R10+ R11] = 11.849 Ohms (so far to the left bigger box on right). Are you getting it... I am just writing, don't have drawings, so please try to get it like this ... also you can now draw simplified diagrams!

Come to the next guy now: We see the current move-in a pretty simple way .. it will never mess up its own route, or the circuit is going to blow off .. so here in this circuit, the current is entering from R1 and going through R4, but it is returning through R6, R5, R3, and R2.

Do you know why? Just look to the circuit. If not, then the current will cross itself in some loops, and that can never happen... it's straightforward.. going straight, simple from one end and coming straight, simple from another end.

So the whole R (bulk) of the right circuital loop (= 11849 ohms)is in series to R6.

Therefore, R6+R(so far) = 21.849 ohms.

Now at the juncture of R4 & R5, the current is being divided, so R5 is in parallel with R (so far = 21.849 ohms). Therefore R5\\ R(so far) = 15.205 ohms.

Now the R4 is in series with R(so far) = R4 + R(so far)= 35.205 ohms. Now R3 is parallel with R(so far) =>

R(so far) will be now => R3 \\ R(sp far, previous) = 9.494ohms.

Now R(so far) is in series with both R1 and R2 =>R1+ R2 + R(so far) = 7+ 4 + 9.494 =20.494 ohms.