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;

$I=\frac{V}{R}$

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:

$\begin{gathered} I=\frac{V}{R} \\ \\ I=\frac{12\text{ V}}{63\text{ \Omega}} \\ \\ I=0.19\text{ A} \end{gathered}$

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:

$P=\frac{V^2}{R}$

Where:

• P is the power in watts

,

• V is the voltage = 12 V

,

• R is the resistance = 63 Ω

Thus, we have:

$\begin{gathered} P=\frac{12^2}{63} \\ \\ P=\frac{144}{63} \\ \\ P=2.29\text{ W} \end{gathered}$

Therefore, the power in the circuit is 2.29 W.

ANSWER:

• (a). 0.19 A

• (b). 2.29 W