a) RTđ=\(\dfrac{60.90}{60+90}=36\Omega\)
b) R dây =p.\(\dfrac{l}{S}=1,7.10^{-8}.\dfrac{120}{0,2.10^{-6}}=10,2\Omega\)
Rtđ=36+10,2=46,2 ohm
=> I=I dây = I 2 đèn =\(\dfrac{46,2}{46,2}=1A\)
Vì R1//R2=>U1=U2=U12=I12.R12=1.36=36V
=>\(I1=\dfrac{36}{60}=0,6A;I2=\dfrac{36}{90}=0,4A\)
Vậy.............