Thomson meter the starting current is 0.14 ampere, the maximum current 10 amperes. The curves are obtained from data given by Mr. G. W. D. Ricks, in a paper read before the British Association, August 1897; see the Electrician' of August 27, 1897.

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If the torque curve is not straight, a further error will be introduced, tending to make the meter run slower or faster according as the curve is concave or convex to the current axis. In all these meters the curves are concave to the current axis, and for small speeds the error due to the starting torque is the greater of the two. For high speeds, when this error is very small, the error due to the bend of the curve becomes apparent, and the accuracy curve turns down again.

Hitherto we have supposed M to be constant. If M varies, the speed will depend on the product of M and c, so that if c is fixed and M increased, the speed will increase with M, or if both c and M increase, the speed will increase with their product.

If the magnetising coil is wound as a shunt across the main line terminals of the circuit in which the current is flowing, the current in the shunt will be proportional to the tension, and if the magnetic circuit contains no iron the induction curve will be a straight line, i.e. the value of M will be proportional to the tension, hence the speed will be proportional to the watts in the circuit. This is the principle of the Thomson watt-meter. For a given current the speed increases directly in proportion to the tension.

We shall now consider the case when two dynamos are mechanically coupled so that they rotate at the same speed, and have their main terminals connected in

parallel to the same line. The magnets should be shunt wound and separately excited, and provided with rheostats so that the induction factor of each dynamo can be adjusted to any desired value and kept constant. The connections will be made so that the dynamos tend to turn in the same direction when both are acting as motors.

We have further to suppose that by means of a pulley fitted on to one of the dynamo shafts and connected by a belt to a third dynamo, or by simply coupling a third dynamo direct on to the shaft line, we can produce a load of any required amount or of any required sign; in other words, we suppose that we can vary the mechanical torque on the shaft, making it either positive so that the dynamos have to act as motors, or negative so that they act as generators. Any arrangement will act provided the motion can be assisted or retarded at will.

The two dynamos may be distinguished by the letters A and B. Suppose that dynamo A has an induction factor of 6 and a resistance of 1.2 ohms, and that dynamo B has an induction factor of 4 and a resistance of 1.09 ohms. Let the tension of the line be 120 volts.

Set off ab in Fig. 18 equal to 100 amperes, the maximum current in A. Set off ad equal to 110 amperes, the maximum current in B. Let af represent 1,200 r.p.m., the speed of A when its induced tension is equal to the tension of the line. Let ag represent 1,800 r.p.m., the speed of B when its induced tension is equal to the tension of the line. Join fb and gd. curves of the two dynamos on a armature.

These will be the speed base of current in the

Since the dynamos are mechanically coupled, we can at once find from the diagram the current that each is

taking for any given speed. When the speed is nothing, the currents are 100 and 120 amperes. When the speed is 1,200 r.p.m. the current in A is nothing, and that in B is 36.7 amperes. When the speed is 1,800 r.p.m. the current in B is nothing, and A is now generating a current of 50 amperes. If the speed is above 1,800 both dynamos are acting as generators, if it is below 1,200 both are acting as motors, between these speeds A is acting as a generator and B as a motor.

If horizontal lines are drawn through g and f to cut the two speed curves in k and h, and if a line is drawn through the points h and h, this line is the combined speed curve, and will give the speed for any current from the line. The horizontal ordinates of this curve represent the current from or into the line. Where the combined speed curve cuts the axis of speed, the current from the line is nothing, and the current from A is the same as that into B.

Now let by represent the torque in A for a current of ab amperes, and let bp represent the torque in B for the same current; these distances will be related to one another in the ratio of the two induction factors. Join aq and ap, and produce in both directions. These lines will then be the torque curves for the two dynamos, on a base of amperes.

From k draw a vertical line to cut the torque curve of A in the point ï. Then since at the speed ag there is no current in B, the ordinate of the torque curve of A at r represents the combined torque of the two dynamos, and hence is a point on the curve of combined torque, the horizontal ordinates of which would give the current from or into the line. A second point on this curve may be found by drawing a vertical line from h to cut the