Part 47 (1/2)
8. The resistance of a hot incandescent lamp is 100 ohms. The current used is 1.1 amperes. Find the E.M.F. applied.
9. What is the resistance of the wires in an electric heater if the current used is 10 amperes, the voltage being 110?
10. The resistance of 1000 ft. of No. 36 copper wire is 424 ohms. How many feet should be used in winding a 200 ohms relay?
11. The resistance of No. 00 trolley wire is 0.80 ohm per 1000 ft. What is the resistance of a line 1 mile long?
12. A wire has a resistance of 20 ohms. It is joined in parallel with another wire of 6 ohms, find their combined resistance.
13. The separate resistances of two incandescent lamps are 200 ohms and 70 ohms. What is their combined resistance when joined in parallel? When joined in series?
(4) METHODS OF GROUPING CELLS AND MEASURING RESISTANCE
=275. Internal Resistance of a Voltaic Cell.=--The current produced by a voltaic cell is affected by the resistance that the current meets in pa.s.sing from one plate to another through the liquid of the cell. This is called the _internal resistance_ of the cell. A Daniell cell has several (1-5) ohms internal resistance. The resistance of dry cells varies from less than 0.1 of an ohm when new to several ohms when old.
If cells are joined together their combined internal resistance depends upon the method of grouping the cells.
[Ill.u.s.tration: FIG. 256.--The four cans exert four times the water pressure that one can will exert.]
=276. Cells Grouped in Series and in Parallel.=--When in _series_ the copper or carbon plate of one cell is joined to the zinc of another and so on. (See Fig. 251.) The effect of connecting, say four cells, in series may be ill.u.s.trated by taking four cans of water, placed one above another. (See Fig. 256.) The combined water pressure of the series is the sum of the several pressures of the cans of water, while the opposition offered to the movement of a quant.i.ty of water through the group of cans is the sum of the several resistances of the cans. In applying this ill.u.s.tration to the voltaic cell, we make use of Ohm's law. Let _E_ represent the e.m.f. of a single cell, _r_ the internal resistance of the cell, and _R_ the external resistance or the resistance of the rest of the circuit. Consider a group of cells in series. If _n_ represents the _number_ of cells in _series_, then Ohm's law becomes
_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 ill.u.s.trate 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_).
[Ill.u.s.tration: FIG. 257.--Four cells connected in parallel.]
[Ill.u.s.tration: FIG. 258.--The water pressure of the group in parallel is the same as that of one.]
=277. Ill.u.s.trative 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? Subst.i.tute 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, subst.i.tuting 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.
[Ill.u.s.tration: FIG. 259.--Diagram of a Wheatstone bridge.]
Important Topics