A conductor with an active resistance of 15 ohms and a coil with an inductance of 50 MHz are connected in series

| January 9, 2021 | education

A conductor with an active resistance of 15 ohms and a coil with an inductance of 50 MHz are connected in series to an alternating current network with a frequency of 50 Hz. Determine the effective voltage if it is known that the amplitude of the current in the circuit is 7 A.

To calculate the value of the effective voltage in the network, we use the formula: U = I * Z = Imax / √2 * √ (R ^ 2 + (ω * L) ^ 2) = Imax / √2 * √ (R2 + (2 * Π * ν * L) ^ 2).
Variables: Imax – current amplitude (Imax = 7 A); R is the resistance of the switched on conductor (R = 15 Ohm); ν – current frequency (ν = 50 Hz); L is the inductance of the connected coil (L = 50 mH = 0.05 H).
Let’s perform the calculation: U = Imax / √2 * √ (R ^ 2 + (2 * Π * ν * L) ^ 2) = 7 / √2 * √ (152 + (2 * 3.14 * 50 * 0.05 ) ^ 2) = 107.5 V.
Answer: The effective voltage is 107.5 V.



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