Question

Given three resistors of resistance R1=10Ω, R2=3Ω, R3=10Ω . The combination of R1 , R2,

R3 which are in parallel, is connected in series with R2. The resultant resistance Rf is

(a) 5Ω b) 3Ω c) 2Ω d) 8Ω
​

Answer 1

Answer:

Rf=8ohm

Explanation:

First, in parallel=

1/Rp=1/10+1/10

1/Rp=2/10

1/Rp=1/5

5ohms=Rp

Now, they are connected in series with R2

R=R2+Rp

R=3+5

R=8ohms