Respuesta :
Answer:
Part a)
R = 9.93 ohm
Part b)
P = 2375 W
Explanation:
Part a)
Resistance of the toaster is given as
[tex]R = \frac{V^2}{P}[/tex]
here we know that
P = power of device
V = voltage across the device
now we have
[tex]R = \frac{120^2}{1450} = 9.93 ohm[/tex]
Part b)
when microwave fails and replaced
then here the current flowing in toaster is given as
[tex]i = \frac{P}{V}[/tex]
[tex]i = \frac{1450}{120} = 12.1 A[/tex]
similarly current in the microwave is given as
[tex]i_2 = \frac{923}{120} = 7.7 A[/tex]
now total current in it is
[tex]i = 12.1 + 7.7 = 19.8 A[/tex]
So power rating is given as
[tex]P = Vi = 120(19.8) = 2375 Watt[/tex]
The maximum power rating of a device that can be used in place of the
microwave depends on the power ratings of the circuit and the toaster.
Correct Responses;
- (a) The toaster's resistance, R ≈ 9.93 Ω
- (b) The maximum power rating that can be used without tripping the circuit is 1,550 W.
Methods by which the above values are obtained;
Given Parameters;
Power rating of the toaster = 1,450 W
Power rating of the microwave = 923 W
Current in the circuit = 25.0 A
Voltage of the circuit, V = 120 V
Solution;
(a) The resistance of the toaster is given as follows;
Power, P = I²·R = V·I
[tex]I = \mathbf{\dfrac{V}{R}}[/tex]
Therefore;
[tex]P = \dfrac{V^2}{R}[/tex]
Which gives;
[tex]R = \mathbf{\dfrac{V^2}{P}}[/tex]
For a circuit arranged in parallel, V is constant, therefore;
[tex]The \ resistance \ of \ the \ toaster, \ R = \dfrac{(120120 \, V) ^2}{1,450 \, W} \approx \mathbf{9.93 \ \Omega}[/tex]
(b) The power rating of the circuit, [tex]P_{circuit}[/tex] = 25.0 A × 120 V = 3,000 W
Power rating of the microwave = 1,450 W
The maximum power rating (in W) of a device that can be used without tripping the 25.0 A circuit breaker is therefore;
3,000 W - 1,450 W = 1,550 W
- The maximum power rating (of the device) that can be used without tripping the circuit is 1,550 W.
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