Question

If two resistors of 35 Ω and 15 Ω are joined together in series and then placed in parallel with a 40 Ω resistor, the effective resistance of the combination is :; A. 0.1 Ω; B. 10 Ω; C. 20 Ω; D. 40 Ω

Answer 1

Answer:

RESISTANCE IN SERIES=35Ω+15Ω=50Ω

RESISTANCE IN PARALLEL=1/50 + 1/40

                                              LCM=200

                                              4+5/200=9/200

1/Rp=9/200

Rp=200/9=22.2

THE CLOSEST OPTION IS 20Ω

                                             

Explanation:

Answer 2

Answer:

D. 20ohm

Explanation:

$when \: the \: resistors \: are \: connected \\ in \: series \: then \:$

$r1eff = 35 + 15 = 50ohm$

now when they are connected in parallel so

R2eff. =

$\frac{1}{50} + \frac{1}{40 } = \frac{40 + 50}{2000} = \frac{90}{2000}$

so now effective resistance will be

$r = \frac{2000}{90} = 22.22ohm$