Question

. Three resistors having resistances of 1.60 Ω, 2.40 Ω, and 4.80 Ω
are connected in parallel to a 28.0-V battery that has negligible internal resistance. Find (a) the equivalent resistance
of the combination.

Answer 1

Answer:

0.8Ω

Explanation:

Let the equivalent resistance be R

1/R = 1/1.6 + 1/2.4 + 1/4.8

1/R = 5/4

R= 4/5 = 0.8Ω

The equivalent resistance is 0.8Ω.

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Answer 2

Answer:

1.25 ohm

Explanation:

when the resistors are connected in parallel the formula becomes

1/R = 1/R1+ 1/R2+ 1/R3

and it goes on when you add the respective values

here we've to find the equivalent resistance and R1= 1.6ohm R2= 2.4 ohm and R3= 4.8ohm

So 1/R= 1/1.6+ 1/2.4 1/4.8

1/R= 10/16+10/24+10/48

take LCM and evaluate. Your answer is 1.25ohm

hope this helps✌️