Question

The current in the circuit shown is 2.0 A.

Answer 1

Ohm's law allows us to find the equivalent resistance of the circuit. In fact:
$V=IR_{eq}$
where
V is the voltage of the battery
I is the current in the circuit
$R_{eq}$ is the equivalent resistance

Re-arranging the equation and using the data of the problem, we get
$R_{eq}= \frac{V}{I}= \frac{30 V}{2.0 A} =15 \Omega$

The three resistors are in series, so their equivalent resistance is just the sum of the three resistances:
$R_{eq}= 2 \Omega + 3 \Omega + R_3$
Since we know the value of the equivalent resistance, $R_{eq}=15 \Omega$, we can find the value of R3:
$R_3 = R_{eq} - 2 \Omega - 3 \Omega = (15-2-3) \Omega = 10 \Omega$