Question

Two resistances are expressed as R1=(4±0.5)Ω and R2=(12±0.5)Ω. what is the net resistance when they are connected 1) series 2) in parallel, with percentage error ?

Answer 1

Given :

Two resistance :

R1=(4±0.5) Ω

R2=(12±0.5) Ω

To Find :

The net resistance when they are connected in series .

Solution :

We know , resistance in series is given by :

R=R1+R2

So , putting values in above equation we get :

R=(4±0.5) + (12±0.5) Ω

R=(16±1) Ω

Therefore , net resistance when they are connected in series is (16±1) Ω .

Learn More :

Resistance in series and parallel

https://brainly.in/question/4443234