Or, the root mean square current is numerically equal to the maximum value divided by

If the E.M.F. is represented by

e = E sin pt

2

the root mean square E.M.F. V, is similarly given by

We shall in future use the letters R.M.S. to indicate root mean square values.

It was proved, see § 24, that in a reactive circuit of selfinduction L and capacity C

is true both for maximum and R.M.S. values of current and E.M.F.

PROBLEMS ON CHAPTER IV.

1. What is the maximum current in a circuit having a capacity of 2 microfarads, and a resistance of 10 ohms, when an alternating E.M.F., whose maximum

value is 100 volts, and frequency 100 cycles, per second, is applied between its terminals?

Answer. 0.126 ampere.

2. What is the impedance of the circuit in Question 1 ?

Answer. 795.2 ohms.

3. What is the difference in phase between E.M.F. and current in Question 1 ? Draw curves respecting them.

Answer. 82° 50'.

4. What is the self-induction of a coil through which, when 100 R.M.S. volts is applied between its terminals, a current of 5 amperes passes, the frequency being 80 periods per second, and the resistance of the coil being 0.5 ohm ?

Answer. 00386 henry, nearly.

5. The maximum value of an alternating current is 100 amperes: what is its average value over half a period from zero to zero?

Answer.

200

П

amperes.

6. The R.M.S. value of an alternating E.M.F. is 100 volts: what is its maximum value?

Answer. 141.4 volts.

7. A coil having a self-induction of 0·05 henry allows a current of 3 R.M.S. amperes to pass when the frequency is 100: what is the P.D. between its terminals, the resistance of the coil being neglected?

Answer. 94.25 volts.

8. What is the P.D. in Question 7, if the resistance of the coil is 5 ohms? Answer. 95.4 volts.

9. What is the frequency when a current of 1 ampere is sent through a coil having a self-induction of 0.75 henry, a P.D. of 200 volts being applied between its terminals, the resistance of the coil being negligible?

Answer. 42.44.

10. What is the frequency in Question 9, if the resistance of the coil is 10 ohms? Answer. 42.4.

11. A condenser of capacity 10 microfarads is connected between the terminals of an alternator giving 1000 volts at a frequency of 50 periods per second what is the current?

Answer. π amperes.

12. If a resistance of 10 ohms is inserted in series with the condenser in Question 11, what current will pass?

Answer. 3.14 amperes.

13. If a self-induction of 0.05 henry is placed in series with a capacity of 1 microfarad, and a P.D. of 100 volts, at a frequency of 100 periods per second, be applied between the extreme terminals, what is the current?

Answer. 0.0641 ampere.

14. A circuit contains in series a resistance of 10 ohms, a self-induction of 0.5 henry, and a capacity of 0·5 microfarad : what is the current if the P.D. between the terminals of the arrangement is 100 volts, and the frequency 80 periods per second?

Answer. 0.0268 ampere.

15. What is the fall of potential (1) along the resistance, (2) the self-induction, and (3) the capacity in Question 14?

Answer. (1) 0·268 volt, (2) 674 volts, (3) 1066.3 volts.

16. What value of the capacity will make the circuit in Question 14 nonreactive?

Answer. 7.9154 microfarads.

17. What is the difference in phase between E.M.F. and current in Question 14?

Answer. 89° 51'.

18. Prove that the

R.M.S. value of a simple periodic function is the maximum value divided by 2, employing the simple trigonometrical method as on page 39.

CHAPTER V.

Expression for Power-Measurement of Power.

POWER GIVEN TO ALTERNATING-CURRENT CIRCUITS.

28. The power given to a circuit by a continuous current is the product of the current flowing in it and the P.D. between its terminals, and is given in watts when the current is given in amperes and the P.D. in volts. The power given to circuit by an alternating current cannot be determined in this fashion, for whereas in a continuous-current circuit the current and P.D. always act in the same direction round the circuit, in an alternatingcurrent circuit there are, in general, times occurring periodically when the current and P.D. act in opposition and the circuit is giving back energy to the source.

We have seen (see Chap. IV., § 24) that when an alternating P.D. e = E sin pt is applied between the terminals of a circuit containing resistance, self-induction, and capacity in series, the resulting current is given by

the notation being the same as in Chapter IV.

The power being given to the circuit at any instant of time t is the product of the P.D. and corresponding current at that instant, that is the product

EI sin pt sin (pt - 0)

The energy given to the circuit during a small interval of time dt is therefore

EI sin pt sin (pt — 0)dt

If we divide a complete period of the current into an infinitely large number of infinitely small times dt, and take the sum of the energies given to the circuit during those times, we shall obtain the total energy given to the circuit during a time equal to a periodic time of the current, and if, further, we divide the expression thus obtained by the periodic time, we shall have the average or mean power given to the circuit.

Denoting the mean power by P, we therefore have

That is the power given to an alternatingcurrent circuit is half the product of the maximum current and the maximum P.D. multiplied by the cosine of their phase difference. The expression for the mean power may be written

where E and I are the R.M.S. values of P.D. and current respectively.

Thus the power given to an alternating-current circuit is the product of the R.M.S. values of the current and P.D. multiplied by the cosine of their phase difference.

The R.M.S. values E and I are the quantities measured respectively by a voltmeter placed between the terminals of the circuit and an ammeter placed in the circuit. Since cos is always less than unity, we see that the product of amperes and volts is, unless = 0, greater than the power given to the circuit.

In contradistinction to the true power, El cos 0, the product EI is called the apparent power, and the ratio