Question

1. how many amps of current would flow through the circuit in figure 2?

2.5 amps
b. 3.24 amps
c. 4.14 amps
d. 5.21 amps

1. what is the total resistance of three resistors in parallel with values of 2 ohms, 8 ohms, and

14 ohms?
a. 1.44 ohms
b. 2.24 ohms
c. 2.32 ohms
d. 2.65 ohms

1. what is the resistance of a light bulb that uses 15 watts of power when placed across a 15-

volt battery?
a. 1.5 ohms
b. 15 ohms
c. 50 ohms
d. 150 ohms

1. the statement that total current entering a point must equal total current leaving that point

best describes
a. ohms law
b. coulombs total current law
c. the theory of relativity
d. kirchhoffs current law

Explanation:

3.

Step1: Calculate the equivalent resistance of \(R_2\) and \(R_3\) in parallel

The formula for two - resistors \(R_a\) and \(R_b\) in parallel is \(R_{eq}=\frac{R_a\times R_b}{R_a + R_b}\). For \(R_2 = 5\Omega\) and \(R_3=20\Omega\), we have \(R_{23}=\frac{5\times20}{5 + 20}=\frac{100}{25}=4\Omega\)

Step2: Calculate the total resistance of the circuit

The resistors \(R_1 = 15\Omega\), \(R_{23}=4\Omega\) and \(R_4 = 10\Omega\) are in series. The formula for resistors in series is \(R_{total}=R_1+R_{23}+R_4\). So \(R_{total}=15 + 4+10=29\Omega\)

Step3: Calculate the total current using Ohm's law \(I=\frac{V}{R}\)

Given \(V = 120V\) and \(R = 29\Omega\), \(I=\frac{120}{29}\approx4.14A\)

Step1: Use the formula for resistors in parallel \(\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

For \(R_1 = 2\Omega\), \(R_2 = 8\Omega\) and \(R_3=14\Omega\), \(\frac{1}{R_{total}}=\frac{1}{2}+\frac{1}{8}+\frac{1}{14}\)

Step2: Find a common denominator

The common denominator of \(2\), \(8\) and \(14\) is \(56\). Then \(\frac{1}{R_{total}}=\frac{28 + 7+ 4}{56}=\frac{39}{56}\)

Step3: Solve for \(R_{total}\)

\(R_{total}=\frac{56}{39}\approx1.44\Omega\)

Step1: Use the power formula \(P=\frac{V^{2}}{R}\)

We can re - arrange it to solve for \(R\), \(R=\frac{V^{2}}{P}\)

Step2: Substitute \(V = 15V\) and \(P = 15W\)

\(R=\frac{15^{2}}{15}=15\Omega\)

Answer:

c. 4.14 amps

4.