Two resistors of resistance 1.0 Ω and 2.0 Ω connected in parallel are linked in series to a 3.0 Ω resistor. All three of this is in parallel with a fourth resistor. If the total effective resistance is 1.0 Ω, what is the resistance of the fourth resistor?
Answers
Given - total resistance - 1 ohm
R1, R2 and R3 - 1 ohm, 2 ohm and 3 ohm respectively.
Find - resistance of fourth ohm
Solution - First two resistances are commented in parallel. Hence, combined resistance R12 is - R1*R2/(R1+R2)
R12 - 1*2/(1+2)
R12 - 2/3
Third resistor is connected in series with combination of R1 and R2. Hence,
R123 - 2/3 + 3
R123 - 11/3
Now, R123 is stated to be connected in parallel with fourth resistor.
Hence, Rt = R123*R4/(R123+R4)
Since, total resistance is 1 ohm, and R123 is 11/3, keeping the values in equation-
1 = (11/3*R4)/(11/3+R4)
11/3+R4 = (11/3*R4)
(11+3R4)/3 = 11/3*R4
(11+3R4) = 11R4
8R4 = 11
R4 = 11/8 ohm
Hence, fourth resistor has resistance 11/8 ohm
resistance of the fourth resistor 11/8 Ω
Explanation:
Two resistors of resistance 1.0 Ω and 2.0 Ω connected in parallel
=> Req = 1/(1/1 + 1/2) = 2/(2 + 1) = 2/3
Connected in Series with 3 Ω resistor
=> Req = 2/3 + 3 = 11/3
Now let say connected in parallel with R & effective resistance is 1.0 Ω,
=> 1 = 1/ (1/(11/3) + 1/R)
=> 1 = 1/(3/11 + 1/R)
=> 1 = 11R/(3R + 11)
=> 3R + 11 = 11R
=> 8R = 11
=> R = 11/8
resistance of the fourth resistor = 11/8 Ω
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