how to calculate the resistance if the resistors are inclined between parallel wires

Asked by acshanadhana15 | 18th Oct, 2020, 07:21: PM

Expert Answer:

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 | 19th Oct, 2020, 02:45: PM