a, khi K mở \(=>[\left(R1ntR2\right)//\left(R3ntR4\right)]ntR5\)
ampe kế chỉ 0,5A\(=>I\left(A\right)=Im=I5=I1234=0,5A\)
\(=>U5=I5.R5=0,5.15=7,5V\)
\(=>U1234=Um-U5=12-7,5=4,5V\)
\(=>R1234=\dfrac{U1234}{I1234}=\dfrac{\left(R1+R2\right)\left(R3+R4\right)}{R1+R2+R3+R4}\)
\(< =>\dfrac{4,5}{0,5}=\dfrac{\left(12+R2\right)\left(4+8\right)}{24+R2}=9=>R2=24\left(om\right)\)
b, ta có R2=24
\(\Rightarrow\dfrac{R_1}{R_3}=\dfrac{R_2}{R_4}\) mạch cầu cb
cđ dđ qua K bằng 0
\(R_{tđ}=\dfrac{R_{12}.R_{34}}{R_{12}+R_{34}}+R_5+R_a=25\left(\Omega\right)\)
\(\Rightarrow I_a=\dfrac{12}{25}=0,48\left(A\right)\)
