Chapt. 27Currents and Resistance211A to B. Using the two results find the formula for ratioR. What does one actually needto evaluate the formula?h) Say one had the special case of exact cross sectional symmetry along an axis:i.e., theconductor cross section was constant along the axis and the E-field was uniform everywherein the conductor and its direction was parallel to the axis of symmetry. In this case, thereis no guide charge. The cross sectional area isAand the conductor length isℓ.027 qfull 00410 2 3 0 moderate math: power integrated for heater6. A 1250 W radiant heater operates with a potential difference across it of 115 V.a) What is the current in the heater?b) What is the resistanceR0of the heater coil?c) How much energy (in MKS units) does the heater absorb and emit (it’s the same amount)in a timet0= 1 hour?d) Now say that due to a defect the resistor starts to increase in resistance from its propervalueR0calculated in part (b). The resistance increases according to the expressionR(t) =R0(
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