2, \(R_đ=\dfrac{6^2}{3}=12\left(\Omega\right)\)

\(E=12=I.r+I\left(5+12\right)\Rightarrow I=\dfrac{2}{3}\left(A\right)\)

\(Q=I^2.R_đ.t=...\)

b,\(I_{đm}=\dfrac{3}{6}=0,5\left(A\right)\)

lại có \(E=12=0,5.1+0,5.\left(12+R\right)\Rightarrow R=11\left(\Omega\right)\)

1, a, \(F=k.\dfrac{\left|q_1q_2\right|}{0,06^2}=...\)

b, bạn tính 2 E bằng ct \(E=k.\dfrac{\left|q\right|}{0,03}\)

xét 2 E khi vẽ thấy có cùng phương ngược chiều \(E=\left|E_1-E_2\right|\)

\(=>\dfrac{R}{R'}=\dfrac{\dfrac{pl}{S}}{\dfrac{pl'}{S}}=\dfrac{\dfrac{pl}{S}}{\dfrac{p4l}{S}}=\dfrac{1}{4}=>R'=4R=>R'=48\left(om\right)\)

a, \(=>\dfrac{1}{R1234}=\dfrac{1}{\left(R1+R2\right)}+\dfrac{1}{R3}+\dfrac{1}{R4}=>\dfrac{1}{R1234}=\dfrac{1}{24}+\dfrac{1}{12}+\dfrac{1}{8}=>R1234=4\left(om\right)\)

\(=>Rab=R7+\dfrac{R6\left(R5+R1234\right)}{R6+R5+R1234}=8+\dfrac{12\left(20+4\right)}{12+20+4}=16\left(om\right)\)

b.,\(=>I7=123456=\dfrac{U}{Rab}=\dfrac{48}{16}=3A=>U7=I7.R7=24V=>U123456=48-24=24V\)

\(=>U123456=U6=U12345=24V=>I6=\dfrac{U6}{R6}=2A=>I12345=\dfrac{24}{R12345}=\dfrac{24}{24}=1A=I5=I1234\)

\(=>U5=I5R5=20V,=>U1234=24-20=4V=U12=U3=U4\)

\(=>I3=\dfrac{U3}{R3}=\dfrac{4}{12}=\dfrac{1}{3}A,=>I4=\dfrac{U4}{R4}=\dfrac{4}{8}=0,5A\)

\(=I12=I1=I2=\dfrac{4}{24}=\dfrac{1}{6}A=>U1=I1R1=\dfrac{2}{3}V=>U2=I2R2=\dfrac{10}{3}V\)

(bài dài nên bn bấm lại máy tính ktra cho chắc)