Physics, 12.07.2019 23:30 tangia

An 1450 w toaster and an 923 w microwave oven are connected in parallel to the same 25.0 a, 120 v circuit. hint (a) find the toaster's resistance r (in ohms). (b) if the microwave fails and is replaced, what maximum power rating (in w) can be used without tripping the 25.0 a circuit breaker?

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Answer from: lily3934

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

$I = \mathbf{\dfrac{V}{R}}$

Therefore;

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

Which gives;

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

For a circuit arranged in parallel, V is constant, therefore;

$The \ resistance \ of \ the \ toaster, \ R = \dfrac{(120120 \, V) ^2}{1,450 \, W} \approx \mathbf{9.93 \ \Omega}$

(b) The power rating of the circuit, $P_{circuit}$ = 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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Answer from: Averybeam300

Part a)

R = 9.93 ohm

Part b)

P = 2375 W

Explanation:

Part a)

Resistance of the toaster is given as

$R = \frac{V^2}{P}$

here we know that

P = power of device

V = voltage across the device

now we have

$R = \frac{120^2}{1450} = 9.93 ohm$

Part b)

when microwave fails and replaced

then here the current flowing in toaster is given as

$i = \frac{P}{V}$

$i = \frac{1450}{120} = 12.1 A$

similarly current in the microwave is given as

$i_2 = \frac{923}{120} = 7.7 A$

now total current in it is

i = 12.1 + 7.7 = 19.8 A

So power rating is given as

P = Vi = 120(19.8) = 2375 Watt

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