1) The equivalent resistance of two resistors in parallel is given by:
![(1)/(R_(eq))= (1)/(R_1)+ (1)/(R_2)](https://img.qammunity.org/2019/formulas/physics/high-school/3hqhoj03hyrbp28t9bt0ly2mr9tm35mxny.png)
so in our problem we have
![(1)/(R_(eq)) = (1)/(15 \Omega)+ (1)/(40 \Omega)=0.092 \Omega^(-1)](https://img.qammunity.org/2019/formulas/physics/high-school/h16nffd2bxk2tpx7mq10p338705xdfpz4y.png)
and the equivalent resistance is
![R_(eq) = (1)/(0.092 \Omega^(-1))=10.9 \Omega](https://img.qammunity.org/2019/formulas/physics/high-school/eotbpdsqra0tuhm5561ofpk9nqnfoxd3v3.png)
2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
![I= (V)/(R_(eq))= (12 V)/(10.9 \Omega)=1.1 A](https://img.qammunity.org/2019/formulas/physics/high-school/jozyumi8l72mpdbzffpefdkmqrnrsxei32.png)