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Chapter 14: Problem 2069

In an ac circuit, the current is given by $\mathrm{I}=5 \sin [100 \mathrm{t}-(\pi / 2)]\( and the ac potential is \)\mathrm{V}=200 \sin 100 \mathrm{t}$. Then the power consumption is, (a) 20 watts (b) 40 watts (c) 1000 watts (d) 0 watts

Short Answer

Expert verified
The given AC current and voltage equations are \(I(t) = 5\sin(100t - \frac{\pi}{2})\) and \(V(t) = 200\sin(100t)\). The instantaneous power \(P(t) = I(t) * V(t)\) can be found by multiplying these equations. To find the average power, integrate the instantaneous power over a full period and divide by the period. Since the current and voltage are 90 degrees (or π/2) out of phase, the average power is 0 watts. Therefore, the correct answer is (d) 0 watts.

Step by step solution

01

Identify the given AC current and voltage equations

The equations for the AC current and voltage are given as: \(I(t) = 5\sin(100t - \frac{\pi}{2})\) \(V(t) = 200\sin(100t)\)
02

Determine the instantaneous power

In an AC circuit, the instantaneous power is given by the product of the instantaneous current and instantaneous voltage: \(P(t) = I(t) * V(T)\) Substitute the given equations for I(t) and V(t): \(P(t) = 5\sin(100t - \frac{\pi}{2}) * 200\sin(100t)\)
03

Determine the average value of the power

To find the average power, we integrate the instantaneous power over a full period and divide by the period. The period of the function can be determined from the angular frequency of the sine functions (100t), which is 2π / period. Therefore, \( T = \frac{2\pi}{100} \) Now, let's find the average power: \(P_{ave} = \frac{1}{T} \int_{0}^{T} P(t) \, dt\)
04

Compute the integral

Now, we need to solve the integral: \(P_{ave} = \frac{1}{\frac{2\pi}{100}} \int_{0}^{\frac{2\pi}{100}} (5\sin(100t - \frac{\pi}{2}) * 200\sin(100t)) \, dt\) \(P_{ave} = \frac{100}{2\pi} \int_{0}^{\frac{2\pi}{100}} (1000\sin(100t - \frac{\pi}{2}) \sin(100t)) \, dt\) Since the integrand is a product of two sinusoids with the same frequency but different phase, the average value of the product will be zero. Alternatively, we can directly notice that the current and voltage are 90 degrees (or π/2) out of phase, resulting in zero average power. So, \(P_{ave} = 0\)
05

Find the correct answer

The average power is found to be 0 watts, which corresponds to option (d). Therefore, the correct answer is: (d) 0 watts

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Most popular questions from this chapter

The instantaneous value of current in an \(\mathrm{AC}\). circuit is \(\mathrm{I}=2 \sin [100 \pi \mathrm{t}+(\pi / 3)] \mathrm{A}\). The current will be maximum for the first time at, (a) \(\mathrm{t}=(1 / 100) \mathrm{sec}\) (b) \(\mathrm{t}=(1 / 200) \mathrm{sec}\) (c) \(t=(1 / 400) \mathrm{sec}\) (d) \(t=(1 / 600) \mathrm{sec}\)

An LCR series circuit with \(\mathrm{R}=100 \Omega\) is connected to a $200 \mathrm{~V}\(, \)50 \mathrm{~Hz}$ a.c source when only the capacitance is removed the current lies the voltage by \(60^{\circ}\) when only the inductance is removed, the current leads the voltage by \(60^{\circ}\). The current in the circuit is, (a) \(2 \mathrm{~A}\) (b) \(1 \mathrm{~A}\) (c) \((\sqrt{3} / 2) \mathrm{A}\) (d) \((2 / \sqrt{3}) \mathrm{A}\)

Same current is flowing in two alternating circuits. The first circuits contains only inductance and the other contains only a capacitor. If the frequency of the emf of ac is increased the effect on the value of the current will be. (a) Increase in the first circuit and decrease in other (b) Increase in both the circuit (c) Decrease in both the circuit (d) Decrease in the first and increase in other

A magnetic field \(2 \times 10^{-2} \mathrm{~T}\) acts at right angles to a coil of area \(200 \mathrm{~cm}^{2}\) with 25 turns. The average emf induced in the coil is \(0.1 \mathrm{v}\) when it removes from the field in time t. The value of \(\mathrm{t}\) is (a) \(0.1 \mathrm{sec}\) (b) \(1 \mathrm{sec}\) (c) \(0.01 \mathrm{sec}\) (d) \(20 \mathrm{sec}\)

An alternating voltage is represented as \(\mathrm{E}=20 \sin 300 \mathrm{t}\). The average value of voltage over one cycle will be. (a) zero (b) 10 volt (c) \(20 \sqrt{2}\) volt (d) \((20 / \sqrt{2})\) volt
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