Solution.
l a = 5 m ; l_a=5m; l a = 5 m ;
d a = 2 m m = 0.002 m ; d_a=2mm=0.002m; d a = 2 mm = 0.002 m ;
ρ a = 2.6 ⋅ 1 0 − 8 O m ⋅ m ; \rho_a=2.6\sdot 10^{-8}Om\sdot m; ρ a = 2.6 ⋅ 1 0 − 8 O m ⋅ m ;
l c = 3 m ; l_c=3m; l c = 3 m ;
ρ c = 1.7 ⋅ 1 0 − 8 O m ⋅ m ; \rho_c=1.7\sdot10^{-8} Om\sdot m; ρ c = 1.7 ⋅ 1 0 − 8 O m ⋅ m ;
I = 4 A ; I=4A; I = 4 A ;
I a = 2.5 A ; I_a=2.5A; I a = 2.5 A ;
U a = U c ⟹ I a R a = I c R c ; U_a=U_c\implies I_aR_a=I_cR_c; U a = U c ⟹ I a R a = I c R c ;
R a = ρ a l a S a ; R_a=\rho_a\dfrac{l_a}{S_a}; R a = ρ a S a l a ; S a = π d 2 4 ; S_a=\pi\dfrac{d^2}{4}; S a = π 4 d 2 ;
S a = 3.14 ⋅ 0.00 2 2 4 = 3.1 ⋅ 1 0 − 6 m 2 ; S_a=\dfrac{3.14\sdot0.002^2}{4}=3.1\sdot 10^{-6}m^2; S a = 4 3.14 ⋅ 0.00 2 2 = 3.1 ⋅ 1 0 − 6 m 2 ;
R a = 2.6 ⋅ 1 0 − 8 5 3.1 ⋅ 1 0 − 6 = 4.2 ⋅ 1 0 − 2 O m ; R_a=2.6\sdot10^{-8}\dfrac{5}{3.1\sdot10^{-6}}=4.2\sdot 10^{-2}Om; R a = 2.6 ⋅ 1 0 − 8 3.1 ⋅ 1 0 − 6 5 = 4.2 ⋅ 1 0 − 2 O m ;
R c = I a R a I c ; R_c=\dfrac{I_aR_a}{I_c}; R c = I c I a R a ;
I c = I − I a = 4 − 2.5 = 1.5 A ; I_c=I-I_a=4-2.5=1.5A; I c = I − I a = 4 − 2.5 = 1.5 A ;
R c = 2.5 ⋅ 4.2 ⋅ 1 0 − 2 1.5 = 7 ⋅ 1 0 − 2 O m ; R_c=\dfrac{2.5\sdot 4.2\sdot 10^{-2}}{1.5}=7\sdot 10^{-2}Om; R c = 1.5 2.5 ⋅ 4.2 ⋅ 1 0 − 2 = 7 ⋅ 1 0 − 2 O m ;
R c = ρ c l c S c ⟹ S c = ρ c l c R c ; R_c=\rho_c\dfrac{l_c}{S_c}\implies S_c=\dfrac{\rho_cl_c}{R_c}; R c = ρ c S c l c ⟹ S c = R c ρ c l c ;
S c = 1.7 ⋅ 1 0 − 8 ⋅ 3 7 ⋅ 1 0 − 2 = 0.7 ⋅ 1 0 − 6 m 2 ; S_c=\dfrac{1.7\sdot 10^{-8}\sdot3}{7\sdot 10^{-2}}=0.7\sdot 10^{-6} m^2; S c = 7 ⋅ 1 0 − 2 1.7 ⋅ 1 0 − 8 ⋅ 3 = 0.7 ⋅ 1 0 − 6 m 2 ;
S c = π d c 2 4 ⟹ d c = 2 S c π ; S_c=\pi\dfrac{d_c^2}{4}\implies d_c=2\sqrt{\dfrac{S_c}{\pi}}; S c = π 4 d c 2 ⟹ d c = 2 π S c ;
d c = 2 0.7 ⋅ 1 0 − 6 3.14 = 1 ⋅ 1 0 − 3 m = 1 m m ; d_c=2\sqrt{\dfrac{0.7\sdot10^{-6}}{3.14}}=1\sdot 10^{-3}m=1mm; d c = 2 3.14 0.7 ⋅ 1 0 − 6 = 1 ⋅ 1 0 − 3 m = 1 mm ;
Answer: d c = 1 m m . d_c=1mm. d c = 1 mm .
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