online activity 47 exercise 3 PDF

Preview

Summary

dsadsa...

Description

Name:

Alexis D. Lachica

DC Circuit Builder – Parallel Circuit

Goal: To analyze mathematical relationships between quantities for parallel circuits. Getting Ready: Using your computer, tablet or phone and navigate to: http://goo.gl/M4Ewmh Tap or click the link to open the DC Circuit Builder. Once opened, select the pencil icon and use the tools (at the bottom of the screen) to build a circuit. Simply select a bulb, resistor, wire or ammeter (the rectangular box) and tap or click in the workspace where you wish it to be located. You’ll get the hang of it quite quickly. Note that the electric potential values are listed on the diagram at the corner of every square on the grid. Current values are listed on the ammeters. To change a battery voltage or a resistor value, select the second icon at the bottom of the screen; a magnifying glass appears above the circuit element. Adjust the voltage or the resistance using the up/down arrows next to the digital meter. Build, Measure, Analyze: Build the circuit shown with three resistors, four ammeters and a battery. Determine the values of current (amps) and electric pressure (volts) at the indicated locations.

1.

For resistors 1, 2, and 3 and for the battery (B), calculate the electric potential difference and fill in the table below.

Element

Electric Potential

©The Physics Classroom, All Rights

Current (I)

Resistance (R)

This document should NOT appear on other websites.

Difference (∆V)

B

9V

2.7 A

3.33 Ω

1

9V

0.9 A

10 Ω

2

9V

0.9 A

10 Ω

3

9V

0.9 A

10 Ω

How does the electric potential difference across each resistor (∆V 1, ∆V2, ∆V3) compare to one another and to the electric potential difference across the battery (∆VB)? Because it is a parallel circuit, the voltage drops across each branch are equal to the voltage gain in the battery, and the electric potential difference across each resistor remains constant for each component. 3. How does the current in the battery (IB) compare to the summative current in the three resistors (I1 + I2 + I3)? The total current flowing in each direction of the circuit equals the current in the battery. 4. Write the above relationship as an equation: IB = I1+I2+I3 2.

5.

Calculate the ratio of electric potential difference to current for the battery. 9V/2.7 A = 3.33 Ω

∆VB/IB =

How does this ratio compare to the resistance values of the resistor? Attempt to write an equation relating the ∆VB/IB ratio to R1, R2, and R3 values. (This is a challenge!) 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛: ∆V𝐵 𝑅1𝑅2𝑅3 = 𝐼𝐵 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅3𝑅1 6.

Alter the values of the battery voltage and the resistance of the resistors so that each resistor has a different resistance. Then make measurements and complete the table.

Element

Electric Potential Difference (∆V)

Current (I)

Resistance (R)

B

6V

1.8 A

3.33 Ω

1

6V

0.6 A

10 Ω

2

6V

0.6 A

10 Ω

3

6V

0.6 A

10 Ω

©The Physics Classroom, All Rights

This document should NOT appear on other websites.

7.

Using values from the circuit analyzed in Question #6, identify as many mathematical equations as you can that relate ∆V, I and R for individual circuit elements or for the circuit as a whole. For each equation that you write, demonstrate its validity by substituting in values from the table above. Equation

a. IB = I1+I2+I3

Demonstration of Equation’s Validity 𝐼𝐵 = 𝐼1 + 𝐼2 + 𝐼3 𝐼𝐵 = 0.6 𝐴 + 0.6 𝐴 + 0.6 𝐴 𝐼𝐵 = 1.8 𝐴 The 1.8 A tends to be the same as the value shown in table 2, question number 6. 1

1

1

1

= + + 𝑅𝐵 𝑅1 𝑅2 𝑅3 11 = b. 1/RB = 1/R1 + 1/R2 + 1/R3

+ + 𝑅𝐵 10 1 = 0.3 Ω 𝑅𝐵 = 0.3 Ω 3. The 0.3 appears to be the same as the value shown in table 2, question number 6. ∆𝑉3 𝐼3 = 𝑅3

c. I3 = ∆V3/R3

6𝑉 𝐼3 = 10𝛺 𝐼3 = 0.6 𝐴

©The Physics Classroom, All Rights

This document should NOT appear on other websites.

The value shown in table 2, question number 6 appears to be the same as the 0.6 A.

∆𝑉2 𝑅2 = 𝐼2 d. R2 = ∆V2/I2

6𝑉 𝑅2 = 0.6 𝐴 𝑅2 = 10 Ω The 10 appears to be the same as the value shown in table 2, question number 6. ∆𝑉1 = 𝐼1𝑅1 ∆𝑉1 = (0.6 𝐴)(10 Ω) ∆𝑉1 = 6 𝑉

e. ∆V1 = I1R1 The 6 V tends to be the same as the value shown in table 2, question number 6.

©The Physics Classroom, All Rights

This document should NOT appear on other websites....