1) The equivalent resistance of two resistors in parallel is given by:
[tex] \frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2} [/tex]
so in our problem we have
[tex] \frac{1}{R_{eq}} = \frac{1}{15 \Omega}+ \frac{1}{40 \Omega}=0.092 \Omega^{-1} [/tex]
and the equivalent resistance is
[tex]R_{eq} = \frac{1}{0.092 \Omega^{-1}}=10.9 \Omega [/tex]
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:
[tex]I= \frac{V}{R_{eq}}= \frac{12 V}{10.9 \Omega}=1.1 A [/tex]
