Answer:
A) 2.0 A
Explanation:
First of all, we calculate the equivalent resistance of the two resistors in parallel.
The equivalent resistance of resistors in parallel is given by:
[tex]\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]
where [tex]R_1,R_2[/tex] are the individual resistances.
In this problem we have:
[tex]R_1=R_2=4.0\Omega[/tex]
Therefore the equivalent resistance of the two resistors is
[tex]\frac{1}{R_{eq}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \rightarrow R_{eq}=2.0\Omega[/tex] (1)
Then, the ammeter is connected in series with the rest of the circuit: this means that its resistance of
[tex]R_A=1.0 \Omega[/tex]
will add in series to resistance (1). Therefore, the total resistance of the circuit will be:
[tex]R=R_{eq}+R_A=2.0 + 1.0 = 3.0\Omega[/tex]
Now we can finally find the current in the circuit, given by Ohm's law:
[tex]V=RI[/tex]
where:
V = 6.0 V is the voltage provided by the batter
[tex]R=3.0 \Omega[/tex] is the total resistance of the circuit
I is the current in the circuit
Solving for I,
[tex]I=\frac{V}{R}=\frac{6.0}{3.0}=2.0A[/tex]
