go to the website: http://phet.colorado.edu/en/simulation/circuit-construction-k

Question

go to the website: http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc 2.) Again, using the Circuit Construction Kit you used in the previous question, build the following circuit. The battery voltage should be exactly should be exactly 3 and 10 respectively 9 V and the resistances 3 10 a.) Once the circuit is set up, look at the flow of electrons in the program. Qualitatively, how does the current through the 3 resistor compare to the current through the 10 resistor? Why is this so? b.) Using the Circuit Construction Kit program, measure the following quantities Measured Current through the 3 Resistor (1) Measured Current through the 10 Resistor (1) Measured Voltage across the 3 Resistor (V Measured Voltage across the 10 Resistor (K) c.) Compute the equivalent resistance of this circuit from: V. total where V-j is the sum of the voltages across each resistor. Show vour work here please d.) Compute the equivalent resistance of this circuit from R=R1 + R2 and compare your answer here to your answer in Part (c) Show your work here please

Explanation / Answer

let,

battery voltage, V=9v

Resistance, R1=3ohms

Resistance, R2=10 ohms

current through R1 and R2 is I

a)

here,

R1 and R2 are in series combination,

then,

current through R1 and R2 is same

b)

Rnet=R1+R2

Rnet=3+10

Rnet=13 ohms

current,I=V/Rnet

I=9/13

I=0.692 A

the current through R1 and R2 is I=0.692 A

voltage across R1 is V1=I*R1

V1=0.692*3

V1=2.08 v

and

voltage across R2 is V2=I*R2

V2=0.692*10

V2=6.92 v

C)

Rnet=Vnet/I

Rnet=(v1+V2)/I

Rnet=(2.08+6.92)/0.692

Rnet=13 ohms

d)

Rnet=R1+R2

=3+10

=13 ohms

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