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By S. A. Knight (Auth.)

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C. source of 15 V. If a current of 25 m A flows in the circuit, calculate the supply frequency. C. c. circuit is given by the product of voltage and current. c. circuit the instantaneous power is similarly expressed by the product of the instantaneous voltage and the instantaneous current, so that P = vi W Since instantaneous values are continually changing, this expression for power is of no practical value. What we want to know is an average value for the power dissipated in the circuit over a period of time.

Now that we have all three basic elements in the circuit, the phase angle may be positive or negative according to whether the supply voltage leads or lags on the current, and this in turn is dependent upon which reactive element is dominant at the particular frequency con­ cerned. 11 tan φ = XL — XÇ R Alternating current: series circuits $9 The direct subtraction of one reactance from the other may not at first look like our customary procedure for dealing with phasors, but as the diagram shows, XL and XQ are in direct opposition to each other and their phasor resultant is clearly the algebraic difference between them.

Thus we see that the output voltage is proportional to the time derivative of the input, and with a 'gain' magnitude of CR. This means that the more rapid the rate of change of the input, the higher the output voltage. 24(b). 25(a). The input voltage V1 will now attempt to charge the capacitor plate connected to the input through the resistor. Because of the phase inversion taking place in the amplifier, however, the output voltage V0 will try to charge the other capacitor plate in a direction which tends to neutralise the charge on the input plate.

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