Question

Three resistors having resistances of R1 = 1.52 Ω , R2 = 2.32 Ω and R3 = 4.94 Ω respectively, are connected in series to a 28.4 V battery that has negligible internal resistance.

Find the equivalent resistance of the combination.

Find the current in each resistor.

Find the total current through the battery.

Find the voltage across each resistor.

Find the power dissipated in each resistor.

Which resistor dissipates the most power, the one with the greatest resistance or the one with the least resistance? Explain why this should be.

Answer 1

a)

equivalent resistance is given as

R_eq = R₁ + R₂ + R₃ = 1.52 + 2.32 + 4.94 = 8.78 ohm

b)

In series , current flows same through each resistor . Current flowing is given as

i = V/R_eq = 28.4/8.78 = 3.2 A

c)

i_total = total current = i = 3.2 A

d)

V₁ = Voltage across R₁ = i R₁ = (3.2) (1.52) = 4.86 Volts

V₂ = Voltage across R₂ = i R₂ = (3.2) (2.32) = 7.42 Volts

V₃ = Voltage across R₃ = i R₃ = (3.2) (4.94) = 15.8 Volts

e)

P₁ = power dissipated in R₁ = V₁ i = 4.86 x 3.2 = 15.6 Watt

P₂ = power dissipated in R₂ = V₂ i = 7.42 x 3.2 = 23.7 Watt

P₃ = power dissipated in R₃ = V₃ i = 15.8 x 3.2 = 50.6 Watt

The one greatest resistance dissipates greatest energy.