Đáp án:

\(\begin{array}{l}
a.\\
R = 80\Omega \\
b.\\
{I_3} = 2,5A\\
{I_1}{R_1} = {I_2}{R_2}\\
{I_1} = \frac{5}{6}A\\
{I_2} = \frac{5}{3}A\\
{U_1} = {U_2} = 100V\\
{U_3} = 100V\\
c.\\
{P_1} = \frac{{250}}{3}W\\
P = 500W\\
d.\\
{Q_2} = \frac{{500}}{3}.t\\
Q = 500t\\
e.\\
d = 3,{9894.10^{ - 4}}m
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
R = {R_3} + {R_{12}} = {R_3} + \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = 40 + \frac{{120.60}}{{120 + 60}} = 80\Omega \\
b.\\
{I_3} = I = \frac{U}{R} = \frac{{200}}{{80}} = 2,5A\\
{I_1}{R_1} = {I_2}{R_2}\\
120{I_1} = 60{I_2}\\
{I_1} + {I_2} = I = 2,5\\
{I_1} = \frac{5}{6}A\\
{I_2} = \frac{5}{3}A\\
{U_1} = {U_2} = {I_1}{R_1} = \frac{5}{6}.120 = 100V\\
{U_3} = U - {U_1} = 200 - 100 = 100V\\
c.\\
{P_1} = {U_1}{I_1} = 100.\frac{5}{6} = \frac{{250}}{3}W\\
P = UI = 200.2,5 = 500W\\
d.\\
{Q_2} = {R_2}I_2^2t = 60.{\frac{5}{3}^2}t = \frac{{500}}{3}.t\\
Q = R{I^2}t = 80.2,{5^2}t = 500t\\
e.\\
{R_3} = \rho \frac{l}{s}\\
s = \frac{{\rho l}}{{{R_3}}} = \frac{{0,{{5.10}^{ - 6}}.10}}{{40}} = 1,{25.10^{ - 7}}{m^2}\\
s = \pi \frac{{{d^2}}}{4}\\
d = \sqrt {\frac{{4s}}{\pi }}  = \sqrt {\frac{{4.1,{{25.10}^{ - 7}}}}{\pi }}  = 3,{9894.10^{ - 4}}m
\end{array}\)

Giải thích các bước giải:

a.R=R3+R12=R3+R1R2R1+R2=40+120.60120+60=80Ωb.I3=I=UR=20080=2,5AI1R1=I2R2120I1=60I2I1+I2=I=2,5I1=56AI2=53AU1=U2=I1R1=56.120=100VU3=U−U1=200−100=100Vc.P1=U1I1=100.56=2503WP=UI=200.2,5=500Wd.Q2=R2I22t=60.532t=5003.tQ=RI2t=80.2,52t=500te.R3=ρlss=ρlR3=0,5.10−6.1040=1,25.10−7m2s=πd24d=√4sπ=√4.1,25.10−7π=3,9894.10−4m