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3 years ago

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Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 2

40-V AC. What is the ratio of the resistance of 240V device to the resistance of the 120V device i.e. \frac{R_{240}}{R_{120}} R 120 R 240 ?

1 answer:

kupik [55]3 years ago

8 0

Answer:4

Explanation:

For same Power one device is operated at $V_1=120 V$ and another at $V_2=240\ V$

Suppose $R_1$ and $R_2$ is the resistance of first and second device respectively

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

For first device

$P=\frac{(120)^2}{R_1}$

$R_1=\frac{120^2}{P}$

For second Device

$R_2=\frac{240^}{P}$

Ratio of resistances are

$\frac{R_2}{R_1}=\frac{240^2}{120^2}$

$\frac{R_2}{R_1}=(2)^2$

$\frac{R_2}{R_1}=4$

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insens350 [35]

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Since $\tau$ is at the order of less than milliseconds.

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(c) Yes

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Substituting into (2) we find

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$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

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After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago