Question

use the laminated chart to help you solve for the values below.
i = 5 a
9 v
r1
6 ω
r2
r3
p = 4.5 w
what is the current through r1? 1.5 a
what is the resistance of r2? 3 ω
what is the total power used by this circuit?
12.82 w
13.5 w
45 w
67 w
10,000,000.0000,0000000 w

Explanation:

Step1: Recall Power Formula

The formula for power in a circuit is \( P = VI \), where \( V \) is voltage and \( I \) is current. For the total circuit, we know the voltage \( V = 9 \, \text{V} \) and the total current \( I = 5 \, \text{A} \).

Step2: Calculate Total Power

Using the formula \( P_{\text{total}} = V \times I \), substitute \( V = 9 \, \text{V} \) and \( I = 5 \, \text{A} \). So, \( P_{\text{total}} = 9 \times 5 = 45 \, \text{W} \)? Wait, no, wait. Wait, maybe we can also calculate by summing powers of each resistor. Let's check:

For \( R_1 \): \( V = 9 \, \text{V} \), \( R_1 = 6 \, \Omega \). Current through \( R_1 \) is \( I_1 = \frac{V}{R_1} = \frac{9}{6} = 1.5 \, \text{A} \) (matches the given). Power for \( R_1 \): \( P_1 = VI_1 = 9 \times 1.5 = 13.5 \, \text{W} \)? Wait, no, \( P = \frac{V^2}{R} \), so \( P_1 = \frac{9^2}{6} = \frac{81}{6} = 13.5 \, \text{W} \)? Wait, but \( R_3 \) has \( P_3 = 4.5 \, \text{W} \). Wait, maybe I made a mistake. Wait, the total current is \( 5 \, \text{A} \), voltage is \( 9 \, \text{V} \), so total power should be \( P = VI = 9 \times 5 = 45 \, \text{W} \)? But let's check the power of each resistor:

\( R_1 \): \( I_1 = 1.5 \, \text{A} \), \( P_1 = 9 \times 1.5 = 13.5 \, \text{W} \)

\( R_3 \): \( P_3 = 4.5 \, \text{W} \)

Now, let's find current through \( R_2 \). Total current \( I = 5 \, \text{A} \), \( I_1 = 1.5 \, \text{A} \), let \( I_3 \) be current through \( R_3 \). \( P_3 = VI_3 \), so \( I_3 = \frac{P_3}{V} = \frac{4.5}{9} = 0.5 \, \text{A} \). Then current through \( R_2 \): \( I_2 = I - I_1 - I_3 = 5 - 1.5 - 0.5 = 3 \, \text{A} \). Then power for \( R_2 \): \( P_2 = VI_2 = 9 \times 3 = 27 \, \text{W} \)? Wait, but \( R_2 \) resistance is given as \( 3 \, \Omega \), so \( I_2 = \frac{9}{3} = 3 \, \text{A} \), which matches. Then \( P_2 = 9 \times 3 = 27 \, \text{W} \). Now sum the powers: \( P_1 + P_2 + P_3 = 13.5 + 27 + 4.5 = 45 \, \text{W} \). Ah, so total power is \( 45 \, \text{W} \). Wait, but the options have 45 W. Wait, but earlier when I did \( VI = 9 \times 5 = 45 \, \text{W} \), that's correct. So the total power is 45 W.

Wait, but let's recheck:

Total current \( I = 5 \, \text{A} \), voltage \( V = 9 \, \text{V} \). Power \( P = VI = 9 \times 5 = 45 \, \text{W} \). That's the simplest way. So the correct answer is 45 W.

Answer:

45 W (from the options, it's the third option: 45 W)