11. Turn the shunt to 0, and reverse the key, stopping at the discharge position.
12. Disconnect No. 2 conductor from ground. Disconnect No. 1 from the lead and connect up No. 2. Connect No. 1 to ground. It is not necessary to wait for No. 1 to be discharged completely before disconnecting it.
13. Proceed with No. 2 as with No. 1 and repeat with each conductor.
14. On the completion of the test all conductor ends should be carefully taped.
15. To determine the correct value of the insulation resistance it is essential that the negative pole of the battery be connected to the core of the cable, otherwise the products of electrolysis will tend to seal up any fault which may exist and will cause the conductor to appear better than it really is. With the negative pole of the battery to the core the tendency is to deposit copper on the core and thus to lay bare any fault. The insulation resistance of any conductor is therefore found by multiplying the corrected deflection at the end of one minute, with + of battery to earth, by the denominator of the shunt used, and then dividing the galvanometer constant by this product. The resistance of the ¹/₁₀-megohm box is neglected unless the insulation resistance determined is very low, say, under 1 megohm, when the 100,000 ohms should be subtracted from the above quotient.
16. To determine the insulation resistance per mile at 60° F., multiply the actual insulation resistance found by the length of the cable in miles, and this result by the multiplier furnished by the torpedo depot for the particular make of cable, corresponding to the temperature of the water in the tank observed during test.
Example.—Leakage of the leads found to be one-half division. Earth currents found to give 1½ divisions in a negative direction from 0 of the scale. Galvanometer throw at the end of one minute (+ to earth), 15 divisions. The corrected deflection is, 15 - ½ + 1½ = 16 divisions.
The galvanometer constant (450 divisions through ¹/₁₀ megohm, shunt at ¹/₁₀₀₀), 45,000 megohms. That is, the battery will give ¹/₁₀ of 450 divisions = 45 through 1 megohm, the shunt at ¹/₁₀₀₀; or, what is the same thing, one division through 45 megohms, the shunt at ¹/₁₀₀₀; therefore with the shunt at unity the battery will give one division through 45 × 1,000 = 45,000 megohms. The insulation resistances = 45,000 ÷ 16 = 2,813 megohms. If the cable is three-fourths mile long, the insulation resistance in megohms per mile is 2,813 × ¾ = 2,110 megohms.
Manufacturer, Safety Insulated Wire & Cable Co.
Temperature of water in tank, 80° F.