Question

Two resistors with values of 18 Ω and 45 Ω, respectively, are connected in series and hooked to a 12 V battery.(a) How much current is in the circuit? A(b) How much power is expended in the circuit? W

Answer 1

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:

`V=IR`

Where:

• V is the voltage

,

• I is the current

,

• R is the resistance.

Rewrite the formula for I;

`I=(V)/(R)`

Since the resistors are connected in series, the equivalent resistance, R will be:

Now, to find the current in the circuit, we have:

`\begin{gathered} I=(V)/(R) \\ \\ I=\frac{12\text{ V}}{63\text{ \Omega}} \\ \\ I=0.19\text{ A} \end{gathered}`

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:

`P=(V^2)/(R)`

Where:

• P is the power in watts

,

• V is the voltage = 12 V

,

• R is the resistance = 63 Ω

Thus, we have:

`\begin{gathered} P=(12^2)/(63) \\ \\ P=(144)/(63) \\ \\ P=2.29\text{ W} \end{gathered}`

Therefore, the power in the circuit is 2.29 W.

ANSWER:

• (a). 0.19 A

• (b). 2.29 W