Answer:
The fraction of time for turn on is 0.3852
Solution:
As per the question:
Temperature at which oil bath is maintained, [tex]T_{o} = 50.5^{\circ}[/tex]
Heat loss at rate, q = 4.68 kJ/min
Resistance, R = [tex]60\Omega[/tex]
Operating Voltage, [tex]V_{o} = 110 V[/tex]
Now,
Power that the resistor releases, [tex]P_{R} = \frac{V_{o}^{2}}{R}[/tex]
[tex]P_{R} = \frac{110^{2}}{60} = 201.67 W = 12.148 J/min[/tex]
The fraction of time for the current to be turned on:
[tex]P_{R} = \frac{q}{t}[/tex]
[tex]12.148 = \frac{4.68}{t}[/tex]
t = 0.3852
