I = nE/(nr + R).
Cells are grouped in series when large E.M.F. is required to force a current through a large external resistance such as through a long telegraph line. Cells are connected in parallel when it is desired to send a large current through a small external resistance. To connect cells in parallel all the copper plates are joined and also all the zinc plates. (See Fig. 257.) To illustrate the effect of this mode of grouping cells, suppose several cans of water are placed side by side (Fig. 258). It is easily seen that the pressure of the group is the same as that of a single cell, while the resistance to the flow is less than that of a single cell. Applying this reasoning to the electric circuit we have by Ohm's law the formula for the current flow of a group of
n cells arranged in parallel I = E/((r/n) + R).
Fig. 257.—Four cells connected in parallel.
Fig. 258.—The water pressure of the group in parallel is the same as that of one.
277. Illustrative Problems.—Suppose that four cells are grouped in parallel, each with an E.M.F. of 1.5 volts and an internal resistance of 2 ohms. What current will flow in the circuit if the external resistance is 2.5 ohms? Substitute in the formula for cells in parallel the values given above, and we have I = 1.5/(0.5 + 2.5) = 1.5/3 = 0.5 ampere. Suppose again that these four cells were grouped in series with the same external resistance, substituting the values in the formula for cells in series we have I = 4(1.5)/(4 × 2 + 2.5) = 6/10.5 = 0.57 ampere.
278. Volt-ammeter Method for Finding Resistance.—Measurements of the resistance of conductors are often made. One of these methods depends upon an application of Ohm's law. It is called the volt-ammeter method since it employs both a voltmeter and an ammeter. If the conductor whose resistance is to be measured is made a part of an electric circuit, being connected in series with the ammeter and in shunt with the voltmeter, the resistance may easily be determined, since R = E/I. (See Fig. 250.) If, for example, the difference in E.M.F., or as it is often called, the fall of potential between the ends of the wire as read on the voltmeter is 2 volts, and the current is 0.5 ampere, then the resistance of the wire is 4 ohms. This method may be readily applied to find the resistance of any wire that is a part of an electric circuit.
279. The Wheatstone Bridge.—To find the resistance of a separate wire or of an electrical device another method devised by an Englishman named Wheatstone is commonly employed. This method requires that three known resistances, a, b, c, in addition to the unknown resistance x be taken. These four resistances are arranged in the form of a parallelogram. (See Fig. 259.) A voltaic cell is joined to the parallelogram at the extremities of one diagonal while a moving-coil galvanometer is connected across the extremities of the other diagonal. The known resistances are changed until when on pressing the keys at E and K no current flows through the galvanometer. when this condition is reached, the four resistances form a true proportion, thus a: b = c: x.
Since the values of a, b, and c are known, x is readily computed. Thus if a = 10, b = 100, and c = 1.8 ohms, then x, the unknown resistance, equals 18 ohms, since 10: 100 = 1.8: 18. This method devised by Wheatstone may be employed to find the resistance of a great variety of objects. It is the one most commonly employed by scientists and practical electricians.