The current supplied to an air conditioner unit is 5 amps. The air conditioner is wired using a 10-gauge (diameter 2.588 mm) wire. The charge density of the wire is n= 8.48·1028  electrons/m3. Find the magnitude of a) current density and b) the drift velocity.

Wires A and B, having equal lengths of 40.0 m and equal diameters of 2.94 mm, are connected in series. A potential difference of 62.9 V is applied between the ends of the composite wire. The resistances are RA = 0.0996 Q and RB = 0.570 2. For wire A, what are (a) magnitude J of the current density and (b) potential difference V? (c) Of what type material is wire A made (see the table below)? For wire B, what are (d) J and (e) V? (f) Of what type material is B made? Resistivities of Some Materials at Room Temperature (20°C) Resistivity, p (N m) Temperature Coefficient of Resistivity, a (K¯') Material Typical Metals 4.1 x 10-3 4.3 x 10-3 Silver 1.62 x 10-8 Copper 1.69 × 10-8 4.0 x 10-3 4.4 x 10-3 Gold 2.35 x 10-8 Aluminum 2.75 x 10-8 Manganin" Tungsten Iron 4.82 x 10-8 0.002 x 10-3 5.25 x 10-8 4.5 x 10-3 6.5 x 10-3 3.9 × 10-3 9.68 x 10-8 Platinum 10.6 x 10-8 Typical Semiconductors 2.5 x 103 Silicon, pure Silicon, n-type Silicon, p-type -70 x 10-3 8.7 x 10-4 2.8 x 10-3 Typical Insulators…

The electromotive force from a battery drives a current around the circuit shown in the figure. Measurements indicate that 6250 μC of charge pass point A in 10.0 s. A Device (a) What is the current in the circuit? i (b) How many electrons pass point A per second? i (c) What direction are the electrons flowing around the circuit? Save for Later x 10-4 A x 1015 electrons/s Attempts: 0 of 3 used Submit Answer

Suppose 1.67 Cof charge passes through the filament of a light bulb in 2.00 seconds. How many electrons pass through the filament in 5.00 s? Express all numerical values in three significant figures. The current in the light bulb is: milliamperes The total amount of charge passing through the filament in t=5.00 s is: coulombs Since a single electron has a charge of -1.60 × 10¬19 C, the number of electrons passing through the filament in t=5.00 s is: N= x 1019 electrons