Question

Judge the equivalent resistance when the following are connected in parallel – (1Ω, 108 Ω, 106 Ω)

and (48Ω, 0.5Ω).​

Answer 1

Answer:

equivalent resistance will be 39.6Ω

Explanation:

first resistance 1Ω, 108Ω and 106Ω

where 48Ω and 0.5Ω are in series

and are parallel to eachother

first we find the resistance in series

Rs = R1+R2+...+Rn

= 1+108+106

=215Ω. .....(1)

let

=48+0.5

=48.5Ω. .....(2)

let

find the resistance in parallel (Rp)

1/Rp=1/R1+1/R2+......+1/Rn

=1/215+1/48.5

=263.5/10427.5

=39.57Ω

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