Find the power dissipated in 1.40 resistor.Three resistors having resistances of

Question

Find the power dissipated in 1.40 resistor.Three resistors having resistances of 1.40 , 3.00 , and 4.90 are connected in series to a 26.0 V battery that has negligible internal resistance.

Find the power dissipated in 1.40 resistor.

Find the power dissipated in 3.00 resistor.

Find the power dissipated in 4.90 resistor.

Which resistor dissipates the most power: the one with the greatest resistance or the least resistance?

Which resistor dissipates the most power: the one with the greatest resistance or the least resistance?

- the power P is largest for the one with the least resistance. - the power P is largest for the one with the greatest resistance.

Explanation / Answer

Here,

for the equivalent resistance

for resistances in series

Req = 1.40 + 3 + 4.90

Req = 9.30 Ohm

current in the circuit is I

Using Ohm's law

I = V/Req

I = 26/9.30

I = 3.87 A

power dissipated in 1.40 Ohm = I^2 * R

power dissipated in 1.40 Ohm = 1.40 * 3.87^2

power dissipated in 1.40 Ohm = 21 W

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power dissipated in 3 Ohm = 3.87^2 * 3

power dissipated in 3 Ohm = 44.9 W

-----------------------------

power dissipated in 4.90 Ohm = 3.87^2 * 4.90

power dissipated in 4.90 Ohm = 73.4 W

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for the resistor disspating most power

it is 4.90 ohm

the correct option is

the power P is largest for the one with the greatest resistance.

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