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