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Ksivusya [100]
2 years ago
9

A 16-Ω loudspeaker, an 8.0-Ω loudspeaker, and a 4.0-Ω loudspeaker are connected in parallel across the terminals of an amplifier

. Determine the equivalent resistance of the three speakers, assuming that they all behave as resistors.
Physics
1 answer:
kolbaska11 [484]2 years ago
6 0

Answer:

2.286 ohm

Explanation:

R1 = 16 ohm

R2 = 8 ohm

R3 = 4 ohm

They all are connected in parallel combination

Let the equivalent resistance is R.

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/16 + 1/8 + 1/4

1/R = (1 + 2 + 4) / 16

1/R = 7 / 16

R = 16/7 = 2.286 ohm

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Answer:

Therefore the required solution is

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Explanation:

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Solving equation (1) and (2) by cross multiplication method

\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

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U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

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