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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?

Mar 14

Electricity

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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.