Resistor with two cuboidal parts. A resistor is formed from two rectangular cubo

Question

Resistor with two cuboidal parts. A resistor is formed from two rectangular cuboids of the same size: with sides a = 8 mm. b = 2 mm. and c - 4 mm The cuboids are made out from different resistive materials, with conductivities sigma_1 = 10^5 S/m and = 4 times 10T5 S/m. The voltage between the resistor terminals is V- 50 mV. Find the electric field intensity, current density, current intensity, and power of Joule's losses in each of the two cuboids if they are connected as in (a) Fig. 3.30(a) and (b) Fig. 3.30(b). Two rectangular cuboids made out from different resistive materials connected in (a) series and (b) parallel; for Problem

Explanation / Answer

current densityJ= I/A = (sigma)E here sigme is conductivity

here I is the current and A is the area

Electric field E=V/d

here V represents the voltage supplied by the battery and d is the distance between the plates.

Electric field E for (a) series

is E=( 50V/16mm) as given V=50V and d=a+a=8+8=16mm

E=3.125* 10-3 V/m

Electric field E for (b) Parallel

E= 50/8, because here d=a=8mm

E=6.25* 10-3 V/m

in series part (a)

effective conductivity= 2{(sigma)1 (sigma)2}/ {(sigma)1+sigma2}={2*105 *4*105 /5*105 }=1.6*105 S/m

So current densityJ= sigma*E= 1.6*105 *3.125*10-3 =5*102 A/m2

in parallel part (b)

effective cnductivity = ({ sigma)1+(sigma)2/2} =5*105 /2=2.5*105 S/m

So current density J=2.5*105 *6.25* 10-3 =15.625*102 A/m2

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