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A coil of negligible resistance and induction of100mH is connected in series with a capacitor 2NF and a resistance of 10 ohms across 50v variable Frequency supply. Determine resistance at Frequency?
Anonymous
Mar 14
A coil of negligible resistance and induction of100mH is connected in series with a capacitor 2NF and a resistance of 10 ohms across 50v variable Frequency supply. Determine resistance at Frequency?
Electricity
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ChatGPT
Mar 14
To calculate the resistance at a certain frequency, we need to first calculate the impedance of the circuit at that frequency. The impedance of the circuit is given by:
Z = R + j(ωL - 1/ωC)
Where:
Z = Impedance of the circuit
R = Resistance (10 ohms)
j = Imaginary unit
ω = Angular frequency (2πf)
L = Inductance (100mH = 0.1H)
C = Capacitance (2nF = 2 x 10^-9 F)
At a frequency of 50 Hz, the angular frequency ω = 2πf = 2π x 50 = 100π
Now, we can calculate the impedance Z at this frequency:
Z = 10 + j(100π x 0.1 - 1/(100π x 2 x 10^-9))
Z = 10 + j(10π - 1/(200π x 10^-9))
Z = 10 + j(10π - 5000/π)
Z = 10 + j(10π - 1591.5)
Z = 10 + j(31.42)
The resistance at this frequency is the real part of the impedance, so the resistance at 50 Hz is 10 ohms.
Z = R + j(ωL - 1/ωC)
Where:
Z = Impedance of the circuit
R = Resistance (10 ohms)
j = Imaginary unit
ω = Angular frequency (2πf)
L = Inductance (100mH = 0.1H)
C = Capacitance (2nF = 2 x 10^-9 F)
At a frequency of 50 Hz, the angular frequency ω = 2πf = 2π x 50 = 100π
Now, we can calculate the impedance Z at this frequency:
Z = 10 + j(100π x 0.1 - 1/(100π x 2 x 10^-9))
Z = 10 + j(10π - 1/(200π x 10^-9))
Z = 10 + j(10π - 5000/π)
Z = 10 + j(10π - 1591.5)
Z = 10 + j(31.42)
The resistance at this frequency is the real part of the impedance, so the resistance at 50 Hz is 10 ohms.
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