G
Greg Neill
Guest
Rich wrote:
a voltage potential difference across it of 10.1V . You cannot
arbitrarily choose to recognize and disregard this on a whim when
you analyze the circuit.
connected to the battery + terminal at one end, and R1 at the other.
Nowhere in sight is there a hint of a direct connection of the 1R
resistor to ground (which by common convention is taken to be the
negative terminal of the battery for a simple circuit like this).
+---------+--E2 = 0.1V
| |
[10R] R2 |
| |
+---E1
| |
[90R] R1 |
| [1R]
+--Ebat |
|+ |
[BAT] |
| |
| |
| |
+---------+
No, you can't, because BAT is a device that is *defined* to haveBAT is suppossed to be a device with zero resistance. So, you can
erroniously conclude there is 10V across 1R, thinking one part of 1R is at
ground.
a voltage potential difference across it of 10.1V . You cannot
arbitrarily choose to recognize and disregard this on a whim when
you analyze the circuit.
What's to be careful about? The circuit plainly shows the 1R resistorThe voltage stated is a voltage *reference to ground*. One end of 1R
is + 10V above ground or the negative terminal that's for sure. But in
fact
the other side of 1R is *not* connected to ground at all through a zero
resistance, so there is not + 10V across R1. This shows how careful one
must
be when associating a zero resaistance value to some ideal component. I
must
be careful in doing that.
connected to the battery + terminal at one end, and R1 at the other.
Nowhere in sight is there a hint of a direct connection of the 1R
resistor to ground (which by common convention is taken to be the
negative terminal of the battery for a simple circuit like this).
+-----------+--- 10V
| |
| [R1]
[1R| |
| |
| |
[BAT] [R2]
| |
| |
+-----------+
|
GND