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Ostrovityanka [42]
3 years ago
5

A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr

ent leads the emf of the generator by 0.850 rad. A) What is the impedance of the circuit? B) What is the resistance of the circuit? C) Is the circuit predominantly inductive, or capacitive?
Physics
1 answer:
weqwewe [10]3 years ago
3 0

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

\phi = 0.85 rad = 48.72°

(a). Impedance of the circuit is given by

Z = \frac{V}{I}

Z = \frac{150}{2.75}

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }

Put the values of Z & \phi in above formula we get

R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

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The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha
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Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t
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Answer:

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\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega

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and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

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So now we know that the potential drop across the parellel resistors must be:

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and with this info, we can calculate the current through R1 using Ohm's Law:

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