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ziro4ka [17]
3 years ago
5

An oil bath maintained at 50.5°C loses heat to its surroundings at the rate of 4.68 kJ/min. Its temperature is maintained by an

electrically heated coil with a resistance of 60 operated from a 110 V line. A thermoregulator switches the current on and off. What fraction of the time will the current be turned on?
Physics
1 answer:
NISA [10]3 years ago
8 0

Answer:

The fraction of the time is 38.67%.

Explanation:

Given that,

Energy = 4.68 KJ

Resistance = 60

Voltage =110 V

If the rate of heat energy supplied  by the coil to the oil bath = Q

Q=4.68\ kJ/min

We need to calculate the power released by the resistor at voltage

P=\dfrac{V^2}{R}

Put the value into the formula

P=\dfrac{110^2}{60}

P=201.7\ W

P=12.102\ KJ/min

We need to calculate the fraction of the time

T=\dfrac{Q}{P}

Put the value into the formula

T=\dfrac{4.68}{12.102}

T=0.3867

The percentage of time is

T = 38.67\%

Hence, The fraction of the time is 38.67%.

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Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

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Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

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Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

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