Given:
• R1 = 18 Ω
,
• R2 = 45 Ω
,
• Voltage = 12 V
Given that the resistors are connected in series, let's solve for the following:
• (a). How much current is in the circuit?
To find the current in the circuit, apply the Ohm's Law:

Where:
• V is the voltage
,
• I is the current
,
• R is the resistance.
Rewrite the formula for I;

Since the resistors are connected in series, the equivalent resistance, R will be:
![\begin{gathered} R=R_1+R_2 \\ \\ R=18Ω+45Ω \\ \\ R=63\text{ \Omega} \end{gathered}]()
Now, to find the current in the circuit, we have:

Therefore, the current in the circuit is 0.19 A.
• (b). How much power is expended in the circuit?
To find the power, apply the formula:

Where:
• P is the power in watts
,
• V is the voltage = 12 V
,
• R is the resistance = 63 Ω
Thus, we have:

Therefore, the power in the circuit is 2.29 W.
ANSWER:
• (a). 0.19 A
• (b). 2.29 W