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2 years ago

15

Ch 31 HW Exercise 31.10 7 of 15 Constants You want the current amplitude through a inductor with an inductance of 4.90 mH (part

of the circuitry for a radio receiver) to be 3.00 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor. Part A What frequency is required? ff = nothing Hz Request Answer Provide Feedback

1 answer:

sergey [27]2 years ago

5 0

Answer:

f = 130 Khz

Explanation:

In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.

For an inductor, this relationship can be expressed as follows:

VL = IL * XL (1) , which is a generalized form of Ohm's Law.

XL is called the inductive reactance, and is defined as follows:

XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and L is the value of the inductance, in H.

Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:

f = VL / 2*π*IL*L = 12 V / 2*π*(3.00*10⁻³) A* (4.9*10⁻³) H = 130 Khz

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Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

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Generally the heat lost by the copper cube is mathematically represented as

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$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

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