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madreJ [45]
3 years ago
12

A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

ting at 50.0 Hz. The maximum current in the circuit is 100 mA. Calculate (a) the inductive reactance, (b) the capacitive reactance, (c) the impedance, (d) the resistance in the circuit, and (e) the phase angle between the current and the generator voltage.
Physics
1 answer:
yanalaym [24]3 years ago
5 0

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

\Delta V_{max} = 240 V

frequency, f = 50Hz

I_{max} = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, \omega =  2\pi f = 2\pi \times50 = 100\pi

a). Inductive reactance,  X_{L} is given by:

\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega

X_{L} = 47.12\Omega 

b). The capacitive reactance,  X_{C} is given by:

\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega

X_{C} = 0.636\Omega

c). Impedance, Z = \frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega

Z = 2400\Omega

d). Resistance, R is given by:

Z = \sqrt {R^{2} + (X_{L} - X_{C})}

2400^{2} = R^{2} + (47.12 - 0.636)^{2}

R = \sqrt {5757839.238}

R = 2399.5\Omega

e). Phase angle between current and the generator voltage is given by:

tan\phi = \frac{X_{L} - X_{C}}{R}

\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})

\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}

\phi = 1.11^{\circ}

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