[Vật lý 11] Dòng điện không đổi

V

vuthienthien

(R2//R3)ntR1//R4(R_2 // R_3) nt R_1 // R_4

a) R23=R2.R3R2+R3=2R_{23} =\frac{R_2. R_3}{R_2+ R_3} = 2 Ω

R123=R1+R23=8+2=10R_{123} = R_1 + R_{23} = 8+2 =10 Ω

Rn=R123.R4R123+R4=207R_{n} =\frac{R_{123}.R_4}{R_{123}+R_4}=\frac{20}{7} Ω

\RightarrowI=ERn+rb=359A I=\frac{E}{R_n + r_b}=\frac{35}{9} A

b) số chỉ vôn kế = IRnIR_{n} =359.207=1009V\frac{35}{9}.\frac{20}{7} =\frac{100}{9}V

c) Q=CU=C.I.Rn=3.106.1009=3,3.105CQ= CU = C.I.R_{n}=3.10^{-6}.\frac{100}{9}=3,3.10^{-5} C
 
L

lorddragon

(R2//R3)ntR1//R4(R_2 // R_3) nt R_1 // R_4

a) R23=R2.R3R2+R3=2R_{23} =\frac{R_2. R_3}{R_2+ R_3} = 2 Ω

R123=R1+R23=8+2=10R_{123} = R_1 + R_{23} = 8+2 =10 Ω

Rn=R123.R4R123+R4=207R_{n} =\frac{R_{123}.R_4}{R_{123}+R_4}=\frac{20}{7} Ω

\RightarrowI=ERn+rb=359A I=\frac{E}{R_n + r_b}=\frac{35}{9} A

b) số chỉ vôn kế = IRnIR_{n} =359.207=1009V\frac{35}{9}.\frac{20}{7} =\frac{100}{9}V

c) Q=CU=C.I.Rn=3.106.1009=3,3.105CQ= CU = C.I.R_{n}=3.10^{-6}.\frac{100}{9}=3,3.10^{-5} C

Đọc mạch sai rồi bạn ơi
Phải là :
[TEX]R_1 nt (R_2//R_3) nt R_4[/TEX]

a) [TEX]R_b=R_1+R_4+\frac{R_2R_3}{R_2+R_3}[/TEX]

[TEX]I_b=\frac{E}{R_b+r}[/TEX]

[TEX]I_1=I_4=I_b[/TEX]

[TEX]U_2=U_3[/TEX]

[TEX]\Leftrightarrow R_2I_2=R_3I_3[/TEX]

[TEX]\Leftrightarrow R_2I_2=R_3(I_b-I_2)[/TEX]

[TEX]\Rightarrow I_2= [/TEX];[TEX]I_3=I_b-I_2= [/TEX]

b) [TEX]U_V=E-I_br[/TEX]

c) [TEX]U_C=U_1+U_{23}=I_1R_1+I_2R_2[/TEX]

[TEX]\Rightarrow Q=U_CC[/TEX]


Tự ráp số
 
V

vuthienthien

((R2//R3)ntR1)//R4((R_2 // R_3) nt R_1) // R_4

a) R23=R2.R3R2+R3=2R_{23} =\frac{R_2. R_3}{R_2+ R_3} = 2 Ω

R123=R1+R23=8+2=10R_{123} = R_1 + R_{23} = 8+2 =10 Ω

Rn=R123.R4R123+R4=207R_{n} =\frac{R_{123}.R_4}{R_{123}+R_4}=\frac{20}{7} Ω

\RightarrowI=ERn+rb=359A I=\frac{E}{R_n + r_b}=\frac{35}{9} A

b) số chỉ vôn kế = IRnIR_{n} =359.207=1009V\frac{35}{9}.\frac{20}{7} =\frac{100}{9}V

c) Q=CU=C.I.Rn=3.106.1009=3,3.105CQ= CU = C.I.R_{n}=3.10^{-6}.\frac{100}{9}=3,3.10^{-5} C
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