Question

3. The resistor should read 10.0 Ω. Tap on the battery so that the voltage adjustment window pops up. Adjust the voltage to 10.0 V. What is the current in the circuit? What is the voltage across the resistor? 

4. Now build the circuit with two resistors in series with 10.0 2 each (Series Circuit). What is the current through the circuit with a battery voltage of 10.0 V? How does this compare to the current with just one resistor of 10.0 2 in the circuit? What is the voltage drop across one resistor? What is the voltage drop across the other resistor? How do the total voltage drops compare to the battery voltage? 

5. Create a circuit with two resistors in parallel as shown below. What is the current through the resistor in the bottom branch of the circuit? What is the voltage across the resistor in the bottom branch of the circuit? Move the ammeter to the top branch and the voltmeter to the resistor in the top branch What is the current through that branch? What is the voltage across the resistor in the top branch? How does that voltage compare to the battery voltage? Move the ammeter to right before (or after) the battery. What is the current through that part of the circuit? How does it compare to the current in the bottom and top branches of the circuit 

Answer 1

3)Given,

R = 10 ohm ; V = 10 V ;

from Ohm's law, V = IR

I = V/R = 10/10 = 1 A

Hence, I = 1 A

Voltage V across the resistor is,

V = 10 V

As there is no other circuit element.

4)When two resistors of 10 ohm will be in series the equivalent resistance will be:

Req = 10 + 10 = 20 Ohm

I' = V/Req = 10/20 = 0.5 A

I' = 0.5 A

On comparing with I, we get:

I' = 1/2I

drop across one resistor is:

V1 = I1 R1 = 10 x 0.5 = 5 V

same current flows through the R connected in series

V2 = 10 x 0.5 = 5 V

V1 = V2 = 5 V

The total voltage drop is equal to the battery voltage.