Ohm's law allows us to find the equivalent resistance of the circuit. In fact:
[tex]V=IR_{eq}[/tex]
where
V is the voltage of the battery
I is the current in the circuit
[tex]R_{eq}[/tex] is the equivalent resistance
Re-arranging the equation and using the data of the problem, we get
[tex]R_{eq}= \frac{V}{I}= \frac{30 V}{2.0 A} =15 \Omega[/tex]
The three resistors are in series, so their equivalent resistance is just the sum of the three resistances:
[tex]R_{eq}= 2 \Omega + 3 \Omega + R_3[/tex]
Since we know the value of the equivalent resistance, [tex]R_{eq}=15 \Omega[/tex], we can find the value of R3:
[tex]R_3 = R_{eq} - 2 \Omega - 3 \Omega = (15-2-3) \Omega = 10 \Omega[/tex]
