1. 50 mA through 120-ohm resistor, 0 mA through 330-ohm resistor
2. 13 mA through both resistors

1. Your mistake here is that you're assuming that the full 6V is applied across the 120 ohm resistor.
2. Your answer treats the diode as if it weren't there.
Start by treating the circuit as if the diode weren't there, and then add the diode and see what changes you have to make.

Ok, so here's my new answer:
1. A zener diode acts as a regular diode would in forward bias. A geranium diode usually has a forward voltage of about 0.3 V. The 330-ohm resistor in parallel would also have a voltage drop of 0.3 V across it due to beeing in parallel with the zener diode, where the current here would be about 0.00091 A or 0.91 mA from . The current across the 120-ohm reisistor would be the difference of the voltage of the voltage source and the zener diode voltage, which is 5.7 V.
= 0.0475 A or 47.5 mA
Answer: The current flowing through the 120-ohm resistor is 47.5 mA and the current flowing through the 330-ohm resistor is 0.91 mA mA.
2.
4.4 V > 2.4 V -> Since the voltage across the diode is greater than its breakdown voltage, it will experience avalanche breakdown and current will flow through the diode. This current is used to regulate the voltage across the diode so that the voltage across is close to its breakdown voltage.
The current across the 120-ohm resistor can be represented by the expression , since the node where current first flows through the zener diode reads 2.4 volts when there is a ground node set at the node on the other side. This current has a value of about 0.03 A, or 30 mA. Since the 330-ohm resistor is in parallel with the zener diode, the two both have the same voltage. From here, we just do , giving us a current value of about 0.0073 A or 7.3 mA.
Answer: The current flowing through the 120-ohm resistor is 30 mA and the current flowing through the 330-ohm resistor is 7.3 mA.

Last edited by azboy1910 on October 11th, 2020, 9:18 am, edited 1 time in total.

1. 50 mA through 120-ohm resistor, 0 mA through 330-ohm resistor
2. 13 mA through both resistors

1. Your mistake here is that you're assuming that the full 6V is applied across the 120 ohm resistor.
2. Your answer treats the diode as if it weren't there.
Start by treating the circuit as if the diode weren't there, and then add the diode and see what changes you have to make.

Ok, so here's my new answer:
1. A zener diode acts as a regular diode would in forward bias. A geranium diode usually has a forward voltage of about 0.3 V. The 330-ohm resistor in parallel would also have a voltage drop of 0.3 V across it due to beeing in parallel with the zener diode, where the current here would be about 0.00091 A or 0.91 mA from . The current across the 120-ohm reisistor would be the difference of the voltage of the voltage source and the zener diode voltage, which is 5.7 V.
= 0.0475 A or 47.5 mA
Answer: The current flowing through the 120-ohm resistor is 47.5 mA and the current flowing through the 330-ohm resistor is 0.91 mA mA.
2.
4.4 V > 2.4 V -> Since the voltage across the diode is greater than its breakdown voltage, it will experience avalanche breakdown and current will flow through the diode. This current is used to regulate the voltage across the diode so that the voltage across is close to its breakdown voltage.
The current across the 120-ohm resistor can be represented by the expression , since the node where current first flows through the zener diode reads 2.4 volts when there is a ground node set at the node on the other side. This current has a value of about 0.03 A, or 30 mA. Since the 330-ohm resistor is in parallel with the zener diode, the two both have the same voltage. From here, we just do , giving us a current value of about 0.0073 A or 7.3 mA.
Answer: The current flowing through the 120-ohm resistor is 30 mA and the current flowing through the 330-ohm resistor is 7.3 mA.

You got the 120 ohm resistor ones correct, but you're something important with the 330. What does a diode act like when a sufficient voltage is applied for charge to flow? In other words, what is the effective "resistance" of an ideal diode?

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I believe the resistance of an ideal diode is 0 ohms, but what does this have to do with the current flowing through the 330-ohm resistor? I am confused here.

From the discussion on discord, azboy1910 is correct, though if you want to verify this in lab you'll need to either make sure your multimeter can measure low currents (less than 100uA for a good reading), or measure voltage across the resistors and compute the currents using ohm's law.

Good job! Your turn.

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Use correct significant figures and metric units for the following questions.

1. What is the energy stored in a 50.0 mF capacitor connected in series with a 12.0-volt battery and a 25.0-ohm resistor after 1.10 time constants?
2. Calculate the net force exerted on a proton when there is one electron in the east direction 300.0 cm away from it and another electron in the west direction 250.0 cm away from it.
3. A transformer has a ratio of 50:200 of the number of windings from the primary coil to the secondary coil. There is a voltage of 3.00 kV and a current of 20.0 mA across the primary windings. What is the current across the secondary coil in microamperes?

Last edited by azboy1910 on October 14th, 2020, 6:45 pm, edited 1 time in total.

Use correct significant figures and metric units for the following questions.

1. What is the energy stored in a 50.0 mF capacitor connected in series with a 12.0-volt battery and a 25.0-ohm resistor after 1.10 time constants?
2. Calculate the net force exerted on a proton when there is one electron in the east direction 300.0 cm away from it and another electron in the west direction 250.0 cm away from it.
3. A transformer has a ratio of 50:200 of the number of windings from the primary coil to the secondary coil. There is a voltage of 3.00 kV and a current of 20.0 mA across the primary windings. What is the current across the secondary coil in microamperes?

Capacitors frustrate me

1. The voltage would be 12.0e^-1.10 = 3.99, the energy would be 0.5CV^2 = 0.399 J.
2. Using Coulomb's law twice, 1.536e-29 N west.
3. 20.0mA / 4 = 5.00 mA = 5000 microamperes.

Use correct significant figures and metric units for the following questions.

1. What is the energy stored in a 50.0 mF capacitor connected in series with a 12.0-volt battery and a 25.0-ohm resistor after 1.10 time constants?
2. Calculate the net force exerted on a proton when there is one electron in the east direction 300.0 cm away from it and another electron in the west direction 250.0 cm away from it.
3. A transformer has a ratio of 50:200 of the number of windings from the primary coil to the secondary coil. There is a voltage of 3.00 kV and a current of 20.0 mA across the primary windings. What is the current across the secondary coil in microamperes?

Capacitors frustrate me

1. The voltage would be 12.0e^-1.10 = 3.99, the energy would be 0.5CV^2 = 0.399 J.
2. Using Coulomb's law twice, 1.536e-29 N west.
3. 20.0mA / 4 = 5.00 mA = 5000 microamperes.

I agree with question 1, but I am not sure how you got your answers for 1 and 2. If you're willing, can you show how you got your answers? It may be possible that I am in the wrong, as I do sometimes get my own problems wrong, as you may have seen in other marathons.
Here's how I got mine.

Last edited by azboy1910 on October 15th, 2020, 9:22 am, edited 1 time in total.

Use correct significant figures and metric units for the following questions.

1. What is the energy stored in a 50.0 mF capacitor connected in series with a 12.0-volt battery and a 25.0-ohm resistor after 1.10 time constants?
2. Calculate the net force exerted on a proton when there is one electron in the east direction 300.0 cm away from it and another electron in the west direction 250.0 cm away from it.
3. A transformer has a ratio of 50:200 of the number of windings from the primary coil to the secondary coil. There is a voltage of 3.00 kV and a current of 20.0 mA across the primary windings. What is the current across the secondary coil in microamperes?

Capacitors frustrate me

1. The voltage would be 12.0e^-1.10 = 3.99, the energy would be 0.5CV^2 = 0.399 J.
2. Using Coulomb's law twice, 1.536e-29 N west.
3. 20.0mA / 4 = 5.00 mA = 5000 microamperes.

I agree with question 1, but I am not sure how you got your answers for 1 and 2. If you're willing, can you show how you got your answers? It may be possible that I am in the wrong, as I do sometimes get my own problems wrong, as you may have seen in other marathons.
Here's how I got mine.

For 1, I did 12e^-1.10 instead of 12(1 - e^-1.10), whoops should've known better.
For 2, it looks like I forgot to square the denominator, which is a mistake I really should not be making.

1. Explain how you could make a 0-10 V voltmeter using a 0-5 mA galvanometer and resistors.
2. Explain how you could make a 0-10 A ammeter using a 0-5 mA galvanometer and resistors.
3. What is the name for a resistor that provides a very low resistance in order to the limit the amount of electric current passing around the component it is put in parallel with?

Last edited by UTF-8 U+6211 U+662F on October 15th, 2020, 6:04 pm, edited 1 time in total.