Find the power dissipated in each resistor. Three resistors having resistances of R1 = 1.84 Ω , R2 = 2.28 Ω and R3 = 4.75 Ω respectively, are connected in series to a 28.5 V battery that has negligible internal resistance. Find (a) the equivalent resistance of the combination; (b) the current in each resistor; (c) the total current through the battery;(d) the voltage across each resistor; (e) the power dissipated in each resistor. (f) Which resistor dissipates the most power: the one with the greatest resistance or the least resistance? Explain why this should be.

a) R_eq = 8.87 Ω , b) i₁ = i₂ = i₃ = 3,213 A , c)  i = 3,213 A , d)   V₁ = 5,912 V,        V₂ = 7,326 V , V₃ = 15,262 V,  e)   P₁ = 19.0 W , P₂ = 23.5 W ,     P₃ = 49.0 W ,

Explanation:

a) and e) The power dissipated in a resistor is

     P = V I = I² R

In a series circuit the current is constant and the equivalent resistance is the sum of the resistances

    R_eq = R₁ + R₂ + R₃

     R_eq = 1.84 + 2.28 +4.75

     R_eq = 8.87 Ω

      V = i R_eq

       i = V / R_eq

       i = 28.5 /8.87

       i = 3,213 A

There we can calculate the power dissipated in each Resistor

        P₁ = i² R₁

        P₁ = 3,213² 1.84

        P₁ = 19.0 W

        P₂ = i² R₂

        P₂ = 3,213² 2.28

        P₂ = 23.5 W

        P₃ = i² R₃

        P₃ = 3,213² 4.75

        P₃ = 49.0 W

b) in a series circuit the current is constant

            i₁ = i₂ = i₃ = 3,213 A

c) The current is not lost so the current supplied by the battery must be equal to the current passing through the resistors

           i = 3,213 A

d) V = i R

       V₁ = 3,213 1.84

      V₁ = 5,912 V

      V₂ = 3,213 2.28

      V₂ = 7,326 V

      V₃ = 3,213 4.75

      V₃ = 15,262 V

f) the resistor that dissipates more power is the one with the highest resistance values             R₃

E that dissipates less power is R₁

This is because the current in a series circuit is constant through all resistors

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answer: b.

explanation: