EXAMPLE--If, in fig 83, the resistance of R = 10 ohms, and R′ = 20 ohms, the current through R will be to the current through R′ as 1/10 to 1/20; or, as 2:1, or, in other words, 2/3 of the total current will pass through R and 1/3 through R′. The joint resistance of the two branches between A and B will be less than the resistance of either branch singly, because the current has increased facilities for travel. In fact, the joint conductivity will be the sum of the two separate conductivities. Taking again the resistance of R = 10 ohms and R′ = 20 ohms, the joint conductivity is 1/10 + 1/20 = 3/20 and the joint resistance is equal to the reciprocal[7] of 3/20 or 6-2/3 [Illustration: FIG. 83.--Divided circuit with two conductors _in parallel_.] In most cases the resistance of the different branches will be alike. This simplifies the calculations considerably. Take, for instance, two branches of 100 ohms resistance each and find the joint resistance. SOLUTION: 1/100 + 1/100 = 2/100; the reciprocal is 100/2 = 50 ohms, or, in other words, the joint resistance is one-half of the resistance of a single branch, and each branch, of course, will carry one-half of the total current in amperes. With three branches of equal resistance, the joint resistance will be 1/3; with four branches 1/4; with 100 branches 1/100 of the resistance of a single branch. [Illustration: FIG. 84.--Hydraulic analogy for divided circuits. In the system of pipes shown, water flows from A B to C D through the six vertical pipes 1 to 6, the greatest amount going through the one which offers the least resistance. If pipes 1 to 6 all have the same dimensions, equal quantities of water will flow through them. It follows that the resistance which the water encounters diminishes with the increase in the number of pipes between A B and C D. The electrical circuit presents the same conditions: the greater the number of parallel connections (corresponding to the pipes 1 to 6) the less is the resistance encountered by the current.] If, for instance, the resistance of an incandescent lamp hot be 180 ohms, the joint resistance of 100 such lamps connected in multiple is 180/100 = 1.8 ohms. If the electromotive force of the system is to be, say 110 volts, then, according to Ohm’s law, the current for 100 lamps is: 110/1.8 = 61.11 amperes. giving for each lamp a current of 110/180 = .61 ampere. In the case of two branches only, the following rule may be applied also: _Multiply the two resistances and divide the product by their sum._ Written as a formula: Joint resistance = (R × R′)/(R + R′) Again, assuming that R = 10 ohms and R′ = 20 ohms: Joint resistance (10 × 20)/(10 + 20) = 200/30 = 6-2/3 ohms. This rule _cannot_ be employed for more than two branches at a time.