ART. 226.-Watt; Joule. We have

1 W per T = (W per Q moved round) × Q passing round per T. A special unit of activity, called the watt, is obtained from this equivalence by making W per Q the volt, and Q per T the ampere.

Thus

1 watt volt by ampere,

=

= 107 ergs per second.

The joule is the corresponding unit of energy, 1 joule volt by coulomb.

These two denominations were proposed by Sir W. Siemens in his address to the British Association, 1882; they appear likely to be adopted, but they have not as yet the authoritative stamp of the other denominations defined in this section.

Also since

1 W per T = {(W per Q) / (Q per T)} × (Q per T) × (Q per T), = Rx (Q per T)2;

1 watt = ohm x (ampere).

The relation of the watt to the horse-power is

746 watts horse-power.

=

EXAMPLES.

Ex. 1. Find the multiplier for changing the electrostatic unit of potential from the centimetre, gramme, and second, to the metre, kilogramme, and second.

The old unit of potential is expressed by

Now

and

But

and

erg per Qe...

1 erg=gm. by cm. per sec. per sec. by cm.; 1 Q2.g...=dyne by cm.3,

=gm. by cm. per sec. per sec. by cm.2

hence, when we substitute instead of cm. and gm., we obtain

X

1 erg = '001 × 01 × 01 kgm. by metre per sec. per sec. by metre, = '0000001

Also

1 Q2.g..= .001 x 01 x (01) kgm. by metre per sec. per sec. by metre2, = '000000001

√kgm. by metre per sec. per sec. by metre.

The kilogrammetre here meant is the absolute not the gravitational unit.

Ex. 2. Find the current in a circuit of 50 ohms, generated by a dynamo machine having an internal resistance of 5 ohms, when the electromotive force of the dynamo is 450 volts.

1. Find the multipliers in the electrostatic system for changing the units of Quantity, Capacity, and Current from centimetre, gramme, and second, to metre, kilogramme, and second.

2. Find the corresponding multipliers when the units mentioned above belong to the electromagnetic system.

3. Find the multipliers for changing the electromagnetic units of Electromotive Force, Current, and Resistance, from the C.G.S. units to the F.P.S. units.

4. A table of electromotive forces is expressed in terms of the millimetre, milligramme, and second; find the factor for changing to the C.G.S. unit.

5. Compare the millimetre-milligramme-second unit of current with the

ampere.

6. Find the multiplier for changing cheval-vapeur to watt.

7. Compare the kilogrammetre with the joule.

8. If an electromotive force of 90 volts is maintained between the terminals of an incandescent lamp, and a current of 15 amperes flows through the lamp; what is the rate at which energy is supplied to the lamp?

9. A dozen incandescent lamps, each having a resistance of 2.75 ohms, are joined in a single circuit, and the resistance of the wires connecting the terminals with the terminals of the dynamo machine is 12 ohms. If the maximum electromotive force of the dynamo is 250 volts, what is the maximum current which can be sent through the lamps?

10. A single Grove's cell is employed to send a current through an external resistance of 100 ohms. What is the strength of the current taking the internal resistance of the cell at 25 ohm?

11. Calculate, in terms of the watt, the activity of the above circuit.

12. When the poles of a battery were connected with the terminals of a tangent-galvanometer, a current of 24 amperes was produced; and when the resistance of the circuit was increased by 1 ohm, all else remaining as before, the strength of the current was 11 amperes. Find the electromotive force of the battery.

13. A circuit is formed containing galvanometer, battery, and connecting wires, the total resistance of the circuit being 4.85 ohms; the galvanometer shows a deflection of 48. When a piece of platinum wire is introduced into the circuit, the deflection falls to 29°. Calculate the resistance of the platinum wire, given tan 48° 1.121, and tan 29° = 0·554.

=

14. The ratio of the electrostatic to the electromagnetic unit of quantity is 3 × 1010 in the C.G.S. system; what is it in the F.P.S. system?

15. A battery of 50 Grove cells, having a total internal resistance of 13.5 ohms, is joined by a short-circuit; find the current which will be given.

16. The Board of Trade, acting under the Electric Lighting Act, have adopted a unit of energy which is defined as "the energy contained in a current of 1,000 amperes flowing under an electromotive force of one volt during one hour." Compare this unit with the joule.

SECTION XLIX.-RESISTANCE.

ART. 227.-Resistance of a Substance. When a steady current of electricity flows along a wire of uniform material, having a uniform cross-section, the strength of the current is directly

proportional to the difference of the electro-motive forces at the ends and to the cross-section, and inversely proportional to the length of the circuit, depending otherwise only on the nature and the temperature of the material of the wire. Hence we have a

rate

kQ per T per S cross-section = (W per Q) difference per L length; it is called the electric conductivity of the substance. The idea is analogous to that of thermal conductivity; (W per Q) difference per L length takes the place of difference per L normal; it may therefore be called the gradient of electromotive force.

The reciprocal is

1/k (W per Q) diff. per L length = Q per T per S cross-section; it is called the electric resistance of the substance. It may be expressed in the equivalent form

1/k (W per Q) diff. per (Q per T) curt. = Llength per S cross-section. The C.G.S. unit for the resistance of a substance is

erg per Qc.gs. per cm. per (erg per sec. per cm.2).

It is equivalent to 109 of the above C.G.S. units.

ART. 228. Relative Resistance; Specific Resistance. Let the electric resistance of a substance A be

(W per Q) per L of A = (Q per T) per S of A ;

this can be put into the form

71 (S per L) of A = (Q per T) per (W per Q).

Similarly for a substance B,

TM1⁄2 (S per L) of B = (Q per T) (per W per Q). Hence T2/T1 (S per L) of B = (S per L) of A; which denotes the resistance of B relatively to that of A.

If A is the standard substance (at a standard temperature) with which other substances are compared, then the relative resistance becomes the specific resistance. In the table below the standard

substance is mercury at 0°; it is the most suitable substance for the purpose as the ohm is now defined in terms of a column of mercury (Art. 225).

Observe-The word "specific" is used throughout in the sense which it has in the term "specific gravity."

SPECIFIC ELECTRIC CONDUCTIVITY.

=

k L length per S section of substance at 0° L length per S section of mercury at 0° C.

The resistance of mercury at 0° C. is 943 ohm = metre per sq. mm. cross section.

ART. 229.-Resistance in terms of Linear Density. resistance of a substance be

rS per L = (Q per T) per (W per Q).

Let its density be

P M = S by L,

Let the