a) Ta có U1=I1.R1=6.\(\dfrac{1}{3}=2V\)
Vì R1//R2=>U1=U2=U12=2V
Ta có R3nt(R1//R2)=> Rtđ=R3+\(\dfrac{R1.R2}{R1+R2}=4+\dfrac{6.R2}{6+R2}=\dfrac{24+10R2}{6+R2}\Omega\)
=> I=\(\dfrac{U}{Rt\text{đ}}=6:\dfrac{24+10.R2}{6+R2}=\dfrac{6.\left(6+R2\right)}{24+10.R2}\)
Vì R3nt R12=>I3=I12=I=\(\dfrac{6.\left(6+R2\right)}{24+10R2}\)A
=> U12=I12.R12=\(\dfrac{6.\left(6+R2\right)}{24+10R2}\).\(\dfrac{6.R2}{6+R2}\)=\(\dfrac{6.6R2}{24+10.R2}=2V=>R2=3\Omega\)
Hazzzzzzzzz!
Câu c ) bạn tính tương tự câu A cho tới bước I3=I12=\(\dfrac{6.\left(6+R2\right)}{24+10R2}nh\text{é}\)
Vì R1//R2=>U1=U2=U12=I12.R12=\(\dfrac{6.6.R2}{24+10.R2}\)
=> I2=\(\dfrac{U2}{R2}=\dfrac{6.6R2}{24+10R2}:R2=\dfrac{2}{3}=>R2=3\Omega\)
Câu 1b nè : Ta có Ic=I12=1A
Vì R3nt(R1//R2)=> RTđ=R3+\(\dfrac{R1.R2}{R1+R2}=\dfrac{24+10R2}{6+R2}\Omega\)
=> I=\(\dfrac{U}{Rt\text{đ}}=6:\dfrac{24+10R2}{6+R2}=\dfrac{6.\left(6+R2\right)}{24+10R2}\)
\(V\text{ì}R3ntR12=>I3=I12=I=\dfrac{6.\left(6+R2\right)}{24+10R2}\)
Mà ta có Ic=I12=1A=>\(\dfrac{6.\left(6+R2\right)}{24+10R2}=1=>R2=3\Omega\)
Kết quả R2 abc giống nhau hết nhé ...chỉ khác cách làm thôi !