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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

In general in an alternating current circuit. (a) The average value of current is zero. (b) The average value of square of current is zero. (c) Average power dissipation is zero. (d) The phase difference between voltage and current is zero.

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