Question

module 4 test
ex. 7 (3 pts)
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?
what is the total power used b
1 ω
3 ω
4.5 ω
67 ω
10,000,000,0000,0000000 ω

Explanation:

Step1: Determine voltage across R2

In a parallel circuit, the voltage across each resistor is equal to the source voltage. So \( V_{R2} = 9\space V \) (same as the battery voltage).

Step2: Find total current and currents through R1, R3

First, current through R1: Using \( I = \frac{V}{R} \), \( I_{R1} = \frac{9\space V}{6\space \Omega} = 1.5\space A \) (matches the given value).
Current through R3: Using \( P = VI \), \( I_{R3} = \frac{P_{R3}}{V} = \frac{4.5\space W}{9\space V} = 0.5\space A \).
Total current \( I_{total} = 5\space A \) (given). So current through R2: \( I_{R2} = I_{total} - I_{R1} - I_{R3} = 5 - 1.5 - 0.5 = 3\space A \).

Step3: Calculate resistance of R2

Using Ohm's Law \( R = \frac{V}{I} \), \( R_{2} = \frac{9\space V}{3\space A} = 3\space \Omega \).

Answer:

3 Ω