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serg [7]
2 years ago
6

A car’s headlights dissipate 40 W on low beam and 50 W on high beam. Is there more or less resistance in the high-beam filament?

Physics
1 answer:
love history [14]2 years ago
3 0

Answer:

The high-beam filament has more resistance than low beam filament.

Explanation:

To determine if the 50 W dissipated by a car’s headlights on high beam, has more or less resistance, formula for power dissipated in the filament is used.

Power = IV

Voltage =IR

∴ Power = I*IR = I²R

where;

I is the current

V is the voltage

R is the resistance

From the equation above, Power dissipated is directly proportional to Resistance.

Hence, the high-beam filament has more resistance than low beam filament.

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efficiency is given by

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dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

4 0
3 years ago
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