A circuit consists of a cell of electromotive force (emf) 6.0V and negligible internal resistance connected to two resistors of 4.0Ω (parallel).
The resistance of the ammeter is 1.0 Ω.
What is the reading of the ammeter?
A. 2.0A
B. 3.0A
C. 4.5A
D. 6.0A

Please show the work so I can understand...

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:

where are the individual resistances.

In this problem we have:

Therefore the equivalent resistance of the two resistors is

(1)

Then, the ammeter is connected in series with the rest of the circuit: this means that its resistance of

will add in series to resistance (1). Therefore, the total resistance of the circuit will be:

Now we can finally find the current in the circuit, given by Ohm's law:

where:

V = 6.0 V is the voltage provided by the batter

is the total resistance of the circuit

I is the current in the circuit

Solving for I,

answer and explanation;