![unknowntay04](/avatars/48314.jpg)
unknowntay04
20.08.2019 •
Physics
You connect a 250-2 resistor, a 1.20-mh inductor, and a 1.80-uf capacitor in series across a 60.0-hz, 120-v (peak) source. the approximate impedance of your circuit a) 1.49 k22 b) 1.47 ko2 c) 0.4522 d) 250 22 e) none of these is correct.
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Ответ:
The impedance is 1.49 kΩ.
(A) is correct option.
Explanation:
Given that,
Resistor = 250
Inductor![L= 1.20\ mH](/tpl/images/0182/8769/4f560.png)
Capacitor![C= 1.80\ \muF](/tpl/images/0182/8769/b1274.png)
Frequency f = 60.0 Hz
Voltage = 120 V
We need to calculate the![\omega](/tpl/images/0182/8769/a8223.png)
Using formula of![\omega](/tpl/images/0182/8769/a8223.png)
Put the value into the formula
We need to calculate the![X_{L}](/tpl/images/0182/8769/dc8af.png)
Using formula of![X_{L}](/tpl/images/0182/8769/dc8af.png)
Put the value into the formula
We need to calculate the![X_{C}](/tpl/images/0182/8769/1cc9f.png)
Using formula of![X_{C}](/tpl/images/0182/8769/1cc9f.png)
Put the value into the formula
We need to calculate the impedance
Using formula of impedance
Hence, The impedance is 1.49 kΩ.
Ответ:
Wavelength
Explanation: