Why does the potential drop in this case?

In summary: PERES = VOLTAGE / OHMSI = 12V / 10Ω = 1.2AI = 12V / (1000Ω + 40Ω) = 0.0119AV = IR = 0.0119A * 1000Ω = 11.9VIn summary, the student selected four resistors of value 10 Ω and set them up in a series circuit, with a 12 V supply and a 1000 Ω alarm. The voltmeter indicated a potential drop of 3.0 V across each resistor, but when the alarm circuit was connected, the potential dropped to near zero. This was likely
  • #1
rachael
55
0
8 A voltage of 3.0 V is required to operate an alarm
circuit. Only a 12 V supply is available, together with a
number of resistors. The resistance of the alarm circuit is
about 1000 Ω. A student selects four resistors of value
10 kΩ and sets them up in a series circuit. The voltmeter
indicates a potential drop across each resistor of 3.0 V.
However, as soon as the alarm circuit is connected, the
potential drops to near zero.
a Comment on why this happens.
b Check the student’s calculations.
c Should the student have selected resistors of higher
or lower value?
 
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  • #2
You should really show some working and thoughts
 
  • #3
rachael said:
8 A voltage of 3.0 V is required to operate an alarm
circuit. Only a 12 V supply is available, together with a
number of resistors. The resistance of the alarm circuit is
about 1000 Ω. A student selects four resistors of value
10 kΩ and sets them up in a series circuit. The voltmeter
indicates a potential drop across each resistor of 3.0 V.
However, as soon as the alarm circuit is connected, the
potential drops to near zero.
a Comment on why this happens.
b Check the student’s calculations.
c Should the student have selected resistors of higher
or lower value?
I think you mean that the four resistors have a value of 10 Ω, not kΩ.
a. With just the 4 10 Ω resistors, what is the current? When you add the 1000 Ω load, what is the current? Since voltage drop, V = IR what is the voltage across the 1000 Ω alarm?

AM
 

1. Why does the potential drop occur in this case?

The potential drop occurs in this case due to the presence of an electric field. When a charge moves through an electric field, it experiences a change in potential energy, resulting in a potential drop.

2. What factors affect the potential drop in this case?

The potential drop is affected by the strength of the electric field, the distance the charge travels through the field, and the magnitude of the charge itself. Additionally, the presence of other charges or conductors in the field can also impact the potential drop.

3. How is the potential drop calculated?

The potential drop can be calculated using the equation V = Ed, where V is the potential drop, E is the electric field strength, and d is the distance the charge travels through the field. This equation assumes a constant electric field and neglects any potential changes due to other factors such as charge buildup.

4. Can the potential drop be negative?

Yes, the potential drop can be negative if the charge is moving in the opposite direction of the electric field. In this case, the charge would be gaining potential energy instead of losing it, resulting in a negative potential drop.

5. How does the potential drop affect the behavior of the charge?

The potential drop affects the behavior of the charge by influencing its movement. A larger potential drop will result in a greater change in potential energy for the charge, causing it to accelerate or decelerate accordingly. Additionally, the direction of the potential drop can determine the direction of the charge's movement within the electric field.

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