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how to calculate the resistance if the resistors are inclined between parallel wires
For such problems,
We will start solving from right side.
In circuit diagram,
R8 and R10 are in series with each other while R9 is in parallel to them
Thus,
Rs = 10 + 2 = 12 Ω
and
1/Rp = (1/ Rs) + (1/R9) = (1/12) + (1/6)
Thus,
Rp = 4 Ω
Let this combination of resistors R8, R9 and R10 be named as Ra
The resistor R7 is in resistor with Ra
Thus,
Ra + R7 = 4 + 8 = 12 Ω
R6 is in parallel to this whole combination and let us name this whole combination as Rb

Thus,
1/Rb = (1/R6) + (1/12) = (1/6) + (1/12)
Rb = 4 Ω
From circuit diagram
Rb ad R5 is in series,
Thus,
Rb + R5 = 4 + 4 = 8 Ω

R4 is in paralell to this combination and let the  resistance be Rc
Thus,
1/Rc = (1/8) + (1/ 8)
Rc = 4 Ω
Rc and R3 are in series
Thus,
Rc + R3 = 4 + 4 = 8 Ω
This 8 ohms is in parallel with R2 and can be named as R
Thus,
1/Rd = (1/8) + (1/R2) = (1/8) + (1/8)
Rd = 4 Ω
Rd is in series with R1
Thus,
Req = Rd + R1 = 4 + 6 = 10 Ω
Answered by Shiwani Sawant | 19 Oct, 2020, 02:45: PM

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