Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Vẽ sơ đồ mạch điện và tính cường độ dòng điện qua R2 khi Rx = 45 Ω.
b) Tìm Rx khi dòng qua R3 là 0,15 A.
b) Cường độ dòng điện qua từng điện trở thay đổi thế nào khi tăng Rx còn các
điện trở khác giữ nguyên giá trị.
![](https://rs.olm.vn/images/avt/0.png?1311)
Tiếp nè
Do \(R_{12}//R_3\)
\(\Rightarrow U_{12}=U_3=U_{123}=\frac{48a}{7}\left(V\right)\)
\(\rightarrow I_3=\frac{U_3}{R_3}=\frac{\frac{48a}{7}}{4a}=\frac{12}{7}\left(A\right)\)
\(\rightarrow I_{12}=\frac{U_{12}}{R_{12}}=\frac{\frac{48a}{7}}{3a}=\frac{16}{7}\left(A\right)\)
Do \(R_1ntR_2\)
\(\Rightarrow I_1=I_2=I_{12}=\frac{16}{7}\left(A\right)\)
Vậy:
\(I_4=4\left(A\right)\)
\(I_3=\frac{12}{7}\left(A\right)\)
\(I_1=I_2=\frac{16}{7}\left(A\right)\)
Ta có : [(R\(_1\) nt R\(_2\) ) \(//\) R\(_3\) ] nt R\(_4\)
\(\Rightarrow\) R\(_{12}\) = R\(_1\) + R\(_2\) = a+2a = 3a (\(\Omega\) )
\(\Rightarrow R_{234}=\frac{R_{12}.R_3}{R_{12}+R_3}=\frac{3a.4a}{3a+4a}=\frac{12a^2}{7a}=\frac{12a}{7}\left(\Omega\right)\)
\(\Rightarrow R_{td}=R_{123}+R_4=\frac{12a}{7}+a=\frac{19a}{7}\left(\Omega\right)\)
Do \(R_{123}ntR_4\)
\(\Rightarrow I_{123}=I_4=I=4\left(A\right)\)
\(\Rightarrow U_{123}=I_{123}.R_{123}=\frac{12a}{7}.4=\frac{48a}{7}\left(V\right)\)
Do
![](https://rs.olm.vn/images/avt/0.png?1311)
\(R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{12\cdot6}{12+6}=4\Omega\)
\(R_{34}=R_{tđ}-R_{12}=10-4=6\Omega\)
\(\dfrac{1}{R_{34}}=\dfrac{1}{R_3}+\dfrac{1}{R_4}=\dfrac{1}{24}+\dfrac{1}{R_4}=\dfrac{1}{6}\)
\(\Rightarrow R_4=8\Omega\)
![](https://rs.olm.vn/images/avt/0.png?1311)
theo hình vẽ giả sử dòng điện từ A có chiều từ C đến D qua R5
\(=>I4=I2+I5=I2+2\left(A\right)\)
\(=>I3=I1-I5=I1-2\left(A\right)\)
\(=>Uab=U1+U3=U2+U4\)
\(=>2.I1+2.I3=10.I2+2.I4\)
\(< =>\)\(2I1+2\left(I1-2\right)=10I2+2\left(I2+2\right)\)
\(< =>2I1+2I1-4=10I2+2I2+4\)
\(< =>4I1=12I2+8=>I1=\dfrac{12I2+8}{4}=3I2+2\left(A\right)\)
\(=>60=2I1+2I3+3,2\left(I1+I2\right)\)
\(=>60=2I1+2\left(I1-2\right)+3,2\left(I1+I2\right)\)
\(=>60=2\left(3I2+2\right)+3,2\left(3I2+2+I2\right)=>I2=4\left(om\right)\)
\(=>I1=3.2+2=8\left(om\right)\)
\(=>U5=10.2-8.2=4V=>R5=\dfrac{4}{2}=2\left(om\right)\)
*TK: http://nghiachi.com/16179/611.pdf
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :R12=R1+R2=10+10=20\(\Omega\)
Có :(R1nt R2)//R3 :
\(\Rightarrow\)R123=\(\frac{R_{12}.R_3}{R_{12}+R_3}=\frac{20.5}{20+5}=4\Omega\)
Có : R4nt(R1ntR2)//R3):
\(\Rightarrow\)Rtđ=R4+R123=6+4=10\(\Omega\)
\(\Rightarrow\)Ic=\(\frac{U}{R_{tđ}}=\frac{12}{10}=1,2A\)
\(\Rightarrow\)Ic=I4=I123=1,2A
\(\Rightarrow\)U4=I4.R4=1,2.6=7,2V
Có :R4nt((R1ntR2)//R3)
\(\Rightarrow\)U=U4+U123
\(\Rightarrow\)U123=U-U4=12-7,2=4,8V
mà (R1ntR2)//R3
\(\Rightarrow\)U12=U3=U123=4,8V
\(\Rightarrow\)I12=\(\frac{U_{12}}{R_{12}}=\frac{4,8}{20}=0,24A\)\(\Rightarrow\)I1=I2=I12=0,24A\(\Rightarrow\)\(\left\{{}\begin{matrix}U_1=R_1.I_1=10.0,24=2,4V\\U_2=R_2.I_2=10.0,24=2,4V\end{matrix}\right.\)
\(\Rightarrow\) I3=\(\frac{U_3}{R_3}=\frac{4,8}{5}=0,96\)A
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
R45 = R4 + R5 = 10 + R5 (Ω)
R453 = \(\frac{R_{45}.R_3}{R_{45}+R_3}\) = \(\frac{\left(10+R_5\right).10}{10+R_5+10}\) = \(\frac{100+10R_5}{20+R_5}\) (Ω)
R4532 = R2 + R453 = 10 + \(\frac{100+10R_5}{20+R_5}\) = \(\frac{10\left(20+R_5\right)+100+10R_5}{20+R_5}\) = \(\frac{300+20R_5}{20+R_5}\) (Ω)
R45321 = R1 + R4532 = 10 + \(\frac{300+20R_5}{20+R_5}\) = \(\frac{10\left(20+R_5\right)+300+20R_5}{20+R_5}\) = \(\frac{500+30R_5}{20+R_5}\) (Ω)
⇒ I = I1 = I2 = I453 = \(\frac{U}{R}\) = 63 : \(\frac{500+30R_5}{20+R_5}\) = \(\frac{63\left(20+R_5\right)}{500+30R_5}\) = \(\frac{1260+63R_5}{500+30R_5}\) (A)
⇒ U453 = U45 = U3 = I453.R453 = \(\frac{1260+63R_5}{500+30R_5}\).\(\frac{100+10R_5}{20+R_5}\) = \(\frac{63.\left(10+R_5\right)}{50+3R_5}\) (V)
Ta có: U2 = I2.R2 = \(\frac{1260+63R_5}{500+30R_5}\).10 = \(\frac{1260+63R_5}{50+3R_5}\) (V)
Theo đề bài ta có:
U23 = U2 + U3 (R2 nt R3)
⇒ 40,5 = \(\frac{1260+63R_5}{50+3R_5}\) + \(\frac{63.\left(10+R_5\right)}{50+3R_5}\)
⇒ 40,5 = \(\frac{63\left(20+R_5\right)+63\left(10+R_5\right)}{50+3R_5}\)
⇒ 40,5 = \(\frac{63\left(30+2R_5\right)}{50+3R_5}\)
⇔ 40,5.(50 + 3R5) = 63.(30 + 2R5)
⇔ 2025 + 121,5R5 = 1260 + 126R5
⇔ 765 = 4,5R5
⇒ R5 = 170 (Ω).
Vậy R5 = 170Ω.