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A coil having a resistance of 50.0 Omega...

A coil having a resistance of `50.0 Omega` and an inductance of `0.500 henry is connected to an `AC` source of `110 volts`, `50.0` cycle//s. Find the rms value of the current in the circuit.

Text Solution

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The angular frequency `ometga = 2piv = 10pi s^(-1)`
the impedance of the coil `= sqrt(R^2 + L^2 omega^2)`
`= sqrt((50 Omega)^2 + (0.50 H xx 100pi s^(-1))^2`
`= sqrt(2500 Omega^2 + 2500 pi^2 Omega^2) = 164.8 Omega`.
The rms current is `(epsilon_(rms)/(Z) = (110 V)/(164.8 Omega) = 0.667 A`.
The peak current `= sqrt(2) (rms current) `= 0.943 A`.
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