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This is an addendum to luchador's answerluchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta post.)

This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta post.)

This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta post.)

replaced http://meta.electronics.stackexchange.com/ with https://electronics.meta.stackexchange.com/
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This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta postthis Meta post.)

This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta post.)

This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta post.)

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Dave Tweed
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This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta post.)

This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

This is an addendum to luchador's answer.

The transient power dissipation in the two boxes is very different. The following simulation demonstrates this.

schematic

simulate this circuit – Schematic created using CircuitLab

Run the simulation for 40 seconds, and plot the expression "I(R1.nA)^2+I(R2.nA)^2", which represents the total instantaneous power in the two resistors.

As I said in my comment, box A will not only heat up more slowly while the pulse is on, it will exhibit a spike in temperature when the pulse ends, because the total instantaneous power dissipated in the resistors is doubled at that moment. Box B will not exhibit such a spike.

(NOTE: If you have trouble running the simulation, see this Meta post.)

Source Link
Dave Tweed
  • 178.3k
  • 17
  • 242
  • 418
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