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.

a)
equivalent resistance is given as
R_{eq} = R_{1} + R_{2} + R_{3} = 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)
V_{1} = Voltage across R_{1} = i R_{1} = (3.2) (1.52) = 4.86 Volts
V_{2} = Voltage across R_{2} = i R_{2} = (3.2) (2.32) = 7.42 Volts
V_{3} = Voltage across R_{3} = i R_{3} = (3.2) (4.94) = 15.8 Volts
e)
P_{1} = power dissipated in R_{1} = V_{1} i = 4.86 x 3.2 = 15.6 Watt
P_{2} = power dissipated in R_{2} = V_{2} i = 7.42 x 3.2 = 23.7 Watt
P_{3} = power dissipated in R_{3} = V_{3} i = 15.8 x 3.2 = 50.6 Watt
The one greatest resistance dissipates greatest energy.
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