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Parallel RL Circuits

### The Circuit

With an ac signal applied to it, the parallel RL circuit shown to the right offers a significant impedance to the flow of current. This impedance will change with frequency, since that helps determine XL, but for any given frequency, it will not change over time.

As you would expect, Ohm's Law still applies, just as it has in other circuits. Voltage, being the same for all components, is our reference. Current, however, is the sum of the currents through R and L, keeping in mind that the coil opposes any change in current through itself, so its current lags behind its voltage by 90°. Therefore, our basic equation for current must be:

 I = V = V - j V Z R XL

If we move the "j" to the denominator of its fraction, we must change its sign. This is also in keeping with the fact that jωL = jXL. As with the parallel RC circuit, we can divide the entire equation by V and solve for the complex impedance of this circuit. Our resulting initial equation is:

 1 = 1 + 1 Z R jXL

### The Complex Number Calculations

To calculate the total circuit impedance, we go back to the general equation that we introduced when discussing RC parallel circuits:

 Z = R × j(XL - XC) R + j(XL - XC)

This time, however, we only have R and L, so the XC factors drop out of the equation. This leaves us with:

 Z = R × (jXL) R + jXL

We complete the calculation using the same relationship we applied with the parallel RC circuit, to remove the "j" from the denominator:

 Z = R × jXL × R - jXL R + jXL R - jXL = (jRXL)(R - jXL) (R + jXL)(R - jXL) = jR²XL - j²RXL² R² + XL² = RXL² + jR²XL R² + XL²

As before, this gives us an entirely real number in the demoninator, which in turn makes the necessary computations possible and practical. Our parallel RL impedance is still a complex number, which can be written as:

 Z = RXL² + j R²XL R² + XL² R² + XL²

This expression can now be used to calculate the parallel impedance of any resistor and any inductor, provided the signal frequency is known. If you wish to attempt the same sort of proof that we used for the parallel RC circuit, you will find that your results still match exactly.

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