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bearhunter [10]
3 years ago
12

A particular circuit toolbox contains only 80 Ω resistors and switches, which can be open or closed. Construct a circuit, fillin

g in all four boxes, such that the overall equivalent resistance between A and B is 40 Ω?

Physics
1 answer:
Strike441 [17]3 years ago
3 0

Answer:

Here we need to make parallel connection of two 80 ohm resistors to achieve 40 ohm net resistance.

Explanation:

As we know that the resistances in series add up directly and here we are given with only the resistors of 80 Ω.

So when we connect two resistors of 80 ohm in parallel we get the resultant of 40 ohm.

Mathematically:

\frac{1}{R_p} =\frac{1}{R} +\frac{1}{R}

\frac{1}{40} =\frac{1}{R} +\frac{1}{R}

\frac{1}{40} =\frac{2}{R}

R=80\Omega gives us the only combination of two resistors in parallel.

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Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.
Sati [7]

Answer:

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836, z^{\frac{1}{3} }= -1.967 - i\cdot 2.265, z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]

Where angles are measured in radians and k represents an integer between 0 and n - 1.

The magnitude of the complex number is 27 and the equivalent angular value is 1.817\pi. The set of cubic roots are, respectively:

k = 0

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836

k = 1

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]

z^{\frac{1}{3} }= -1.967 - i\cdot 2.265

k = 2

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]

z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

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