Tóm tắt :
\(U=70V\)
\(R_1=15\Omega\)
\(R_2=30\Omega\)
\(R_3=60\Omega\)
________________________
\(R_{tđ}=?\)
\(I_1=?;I_2=?;I_3=?\)
\(U_{23}=?\)
GIẢI :
*TH1 : R1nt R2 nt R3
Điện trở tương đương toàn mạch là :
\(R_{tđ}=R_1+R_2+R_3=15+30+60=105\left(\Omega\right)\)
Cường độ dòng điện qua mạch chính là :
\(I=\dfrac{U}{R_{tđ}}=\dfrac{70}{105}=\dfrac{2}{3}\approx0,67\left(A\right)\)
Vì R1 ntR2 ntR3 => \(I=I_1=I_2=I_3=\dfrac{2}{3}\approx0,67A\)
R23 là:
\(R_{23}=R_2+R_3=30+60=90\left(\Omega\right)\)
I2 = I3 = \(\dfrac{2}{3}\) (A)
=> \(U_{23}=I_{23}.R_{23}=\dfrac{2}{3}.90=60\left(V\right)\)
* TH2 : R1//R2//R3
\(R_{tđ}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}=\dfrac{1}{\dfrac{1}{15}+\dfrac{1}{30}+\dfrac{1}{60}}=\dfrac{1}{\dfrac{7}{60}}=\dfrac{60}{7}\left(\Omega\right)\)
Vì R1//R2//R3 => U =U1 =U2 =U3 = 70V
+) \(\left\{{}\begin{matrix}I_1=\dfrac{U_1}{R_1}=\dfrac{70}{15}=\dfrac{14}{3}\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{70}{30}=\dfrac{7}{3}\left(A\right)\\I_3=\dfrac{U_3}{R_3}=\dfrac{70}{60}=\dfrac{6}{7}\left(A\right)\end{matrix}\right.\)
+) U23 =U2 =U3 = 70V
