1: A defibrillator passes 12 A of current through the torso of a person for 0.0055 s. Part (a) How much charge moves in C? 0 = Part (b) How many electrons pass through the wires connected to the patient? N= Otheexpertta.com Problem 2: Suppose you have a 2.6 V flashlight. What current, in amperes, flows through the bulb of the flashlight when its resistance is 3.95 0? 1 = Problem 3: The diameter of a specific copper wire is 8.389 mm, and has a resistivity of 1.72 x 10-8 02.m. Find the resistance of a 1.3 km length of such wire used for power transmission in 2. R = Problem 4: An extension cord that has a length of 9.95m9.95m is made from wire with a resistivity of 5.39x10-80.m5.39x10-80.m and a diameter of 0.522cm0.522cm. Part (a) What is the resistance, in ohms, of the extension cord? R = Part (b) If a current of 1.66A1.66A passes through the wire, then what is the voltage, in volts, between the ends of the extension cord? V = Problem 5: Suppose a flashlight has 5.2 x 102 C of charge pass through it during time 0.45 h. What is the rate of the flashlight's energy consumption, in watts, if it operates at a voltage of 3.00 V? P = Problem 6: Military aircraft use 400 Hz AC power, because it is possible to design lighter-weight equipment at this higher frequency. What is the time for one complete cycle of this power in ms? T = Problem 7: During surgery, a current as small as 20.0 uA applied directly to the heart may cause ventricular fibrillation. If the resistance of the exposed heart is 290 2, what is the smallest voltage that poses this danger in mV? V = Problem 8: In the figure, these three resistors are connected to a voltage source so that R2 = 5.75 2 and R3 = 10.5 2 are in parallel with one another and that combination is in series with R1 = 2.5 2. R Part (a) Calculate the power being dissipated by the third resistor R3, in watts. P3 = Part (b) Find the total power supplied by the source, in watts. (P_\\text { total} = Problem 9: If you try to measure the voltage of a battery with a voltmeter connected in series, as shown in the figure, you won't get a completely accurate measurement because of the internal resistance of the battery. To see how large this effect is, consider trying to measure the terminal voltage of a 3.200 V lithium cell having an internal resistance of 35 2 by placing a 0.65 kQ voltmeter across its terminals. Randomized Variables r = 35 0 R = 0.65 kQ Part (a) What current, amperes, does the cell supply? / = Part (b) Find the terminal voltage, in volts. V = Part (c) To see how close the measured terminal voltage is to the emf, calculate their ratio. V/ &=