The resistance ( k2) of a 1 cm' pure silicon crystal.[n-1x10 cm, le-1250 cm² V-'s, H-350 cm2 V-ls] 390 39 347 34.7
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![The resistance (k2) of a 1 cm pure silicon crystal.[n,-1x10" cm, He=1250 cm² V-'s',
-350 cm2 V-ls]
390
39
347
34.7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff3b15a9-8fe3-44e3-b508-ed5134b40672%2Fd6a3c5b2-29d6-48f2-9b3b-b74a0223e7a2%2F65d6tji_processed.jpeg&w=3840&q=75)
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- Solve it asap1. If electricity costs 15 Bz per kilowatt-hour, how much does it cost to operate a 220 V water heater that draws 5 A of current for 3 hours per day for a month of thirty days? The Total Electrical energy consumed for a month = kwh and the Total cost of electricity consumption = riyal 2. If current of I = 3.00 A flows through a wire, How many number of electrons flow through a point in a wire in 3.00 s ? Number of electrons flow through a point in wire in 3.00 s = electronsA rod of semiconducting material of length L 2 m and cross-sectional area A - 4.5 mm lies along the x-axis between z 0 and z - L The material obeys Ohm's law, and its resistivity varies along the rod according to p-p(1 where p-4x 10 n-m. The end of the rod at z=0 is at a potential V, - 15 V greater than the end at z- L C) What is the electric potential, in units of Volt, in the rod at zL/2? Answer
- 1. Calculate the potential due to a dipole as shown in the figure at positions A and B in millivolts. 2. What would be the potential at Bif the distance from the dipole center to B is doubled,. 3. What would be the potential at A if the intensity of the current source is doubled. +2mA B =12em F 10mm 20 degree 90 degree r95 cm A. Resistivity = 75 Ohm-cm -2mA DipoleA potentiometer wire of length 4 m has a resistance of 4 Q. What resistance must be connected in series with the wire and an accumulator of e.m.f 2 volts, so as to get a potential drop of 10-3 volts/cm of the wire.Aluminum at 20°C has a resistivity of 2.65×10°8 №∙m²¹ if the temperature is increased to 60°C estimate the new resistivity.
- II "wire table" states that 12-gauge copper wire at 20°C has a diameter of 2.053 mm, a cross-sectional area of 3.309 mm2, and a resistance per length of 5.211 ohms per kilometer. Calculate the resistivity of the copper in this wire at 20°C three significant figures.Recall that mm2 = (10-3 m)2 = 10-6 m2 and that km = 103 m. ge %3D %3D ogle rome 1.70 X 10-8 ohm-m 1.72 x 10-8 om-m Mw Horizo... 2.00 x 10-8 ohm-m O 2.02 x 10-8 ohm-m Microsoft Click Save and Submit to save and submit. Click Save All Answers to save all answers. Teams Save All Answers Save and P Type here to search5) The resistivity of a certain material is 0.02 22.m whilst the Hall coefficient is 5x10-4m³/C. Assuming an electric field of 1 V/m, deduce the (i) conductivity (ii) The electron concentration (iii) the intercollision time (iv) the mobility and (v) the drift velocity.12- At 20 C, a copper wire has a diameter of 4 mm, a length of 20m, a resistivity of 1.7x10 ° Om and has a uniform current density /-0.1×10A/m2. What is the potential difference between the two ends of the wire? a) 0.034 V b)-0.0.024 V 00 0.084 V 0.048 V 0.063 V Leave blank Next> Close
- A rod of semiconducting material of length L 2 m and cross-sectional area A- 4.5 mm? lies along the x-axis between z0 and a- L The material obeys Ohm's law, and its resistivity varies along the rod according to p= p(1 where po4x 10 n m. The end of the rod at z=0 is at a potential V- 15 V greater than the end at z-L L2 A) What is the total resistance, in units of fn, of the rod?In a conductor, if the number of conduction electrons per unit volume is 8.5×10²8 and mean free time is 25 fs (femto -3 m second), it's approximate resistivity is 9.1×10-³1 kg) (Take, m₂ = 9.1 x 107 me (a) 10 Q-m (c) 10¯n-m (b) 10¯5 №-m (d) 108 №-m160% P 2:35 Asiacell I. H.W -2.pdf H.W -2 6. For the circuit board in Fig. 7.66: a. Find the total resistance Ry of the configuration. b. Find the current drawn from the supply if the applied voltage is 48 V. c. Find the reading of the applied voltmeter. 6.8 kf V 1.2 k? 3.3 kN 2 kn Rr - 2.4 k2 48 V FIG. 7.66 Problem 6.