Sure! Here are the answers to your questions based on the simulations about static electricity, Ohm’s law, and circuit construction:
Part A: Balloons and Static Electricity Simulation
Question 1:
What happens when you rub the balloon on the jersey?
Answer: When you rub the balloon on the jersey, negatively charged electrons from the jersey transfer to the balloon. This leaves the balloon with a negative charge and the jersey with a positive charge.
Question 2:
Why does the balloon then stick to the jersey?
Answer: The balloon sticks to the jersey because it becomes negatively charged while the jersey is left positively charged. The opposite charges attract each other, creating a force that causes the balloon to stick to the jersey.
Question 3:
What happens when you then move the balloon to the wall?
Answer: When you move the negatively charged balloon to the wall, the electrons in the wall redistribute themselves. The area of the wall closest to the balloon becomes positively charged due to the repulsion of electrons. This attraction between the negatively charged balloon and the positively charged area of the wall causes the balloon to stick to the wall.
Part B: Ohm’s Law Simulation
Question 1:
Why does the voltage V start at 4.5v?
Answer: The voltage starts at 4.5V as it is set as the initial voltage for the circuit in the simulation. This can represent the power supplied by a battery within the circuit.
Question 2:
What happens to the current I when the voltage goes up? .....and then when voltage goes down?
Answer: When the voltage (V) goes up, the current (I) increases proportionally according to Ohm's law (I = V/R). Conversely, when the voltage goes down, the current also decreases.
Question 3:
What happens to the current I when the resistance R goes up?......and then when resistance goes down?
Answer: When the resistance (R) goes up, the current (I) decreases because I is inversely proportional to R (I = V/R). When the resistance goes down, the current increases.
Question 4:
What happens to the voltage when there is a change in resistance?
Answer: When resistance changes while the current is maintained, the voltage across the resistor changes proportionally according to Ohm's law. If resistance increases, voltage would increase for the same current. If resistance decreases, voltage would decrease.
Part C: Circuit Construction Kit: DC - Virtual Lab Simulation
Question 1:
Close the switch, which way do the electrons flow? (use + and - in your answer)
Answer: When the switch is closed, electrons flow from the negative terminal of the battery (-) towards the positive terminal (+), through the circuit.
Which way does conventional current flow? (use + and - in your answer)
Answer: Conventional current flows from the positive terminal of the battery (+) to the negative terminal (-).
What is the voltage measurement across the bulb?
Answer: The voltage measurement across the bulb would be equal to the battery voltage (for instance, 9V if that is what is set in the simulation).
What is the measured current flow on the ammeter?
Answer: The measured current flow on the ammeter would depend on the circuit's total resistance, but with a standard 9V battery, the expected current can be calculated using Ohm's law (I = V/R).
Question 2:
What happens to the light bulb and current flow when you increase the battery voltage?
Answer: When the battery voltage is increased, the current flow also increases (assuming the resistance stays the same). The light bulb will glow brighter as a result of the increased current.
Question 3:
Return the battery voltage to 9V and add another light bulb in series with the first one. What has happened now to the current flow?
Answer: When another bulb is added in series, the total resistance in the circuit increases. The current flow decreases as more resistance is added to the circuit.
What is the voltage across each bulb?
Answer: The total voltage (9V) will be divided across both bulbs in series. If the bulbs are identical, each bulb would receive approximately 4.5V.
Question 4:
How can we get the 2 bulbs to glow as brightly as before? (demonstrate it in your simulation)
Answer: To make both bulbs glow as brightly as before, you could either increase the voltage of the battery (which raises the overall voltage supply to the circuit) or change the configuration to parallel rather than series so that each bulb receives the full battery voltage.
Imagine we had a string of bulbs all in series, what would happen if one bulb blew?
Answer: If one bulb in a series circuit blows, the circuit becomes open (broken), and no current can flow. As a result, all bulbs in the string would go out because they are all dependent on the same circuit path for current flow. This is similar to older string lights where one broken bulb would lead to the failure of the entire string.