TC 9-62
Now that the total resistance is known, the circuit current can be calculated as follows:
ER
IT =
RT
20 V
IT =
2 Ω
I T = 10
A
The voltage drop across R1can be computed as follows:
ER1 = R1 x IT
ER1 = 1Ω X 10 A
ER1 = 10 V
The voltage at point A would be equal to the voltage of V1 minus the voltage drop of R1.
Compute this as follows:
Voltage at point A = V1 - ER1
Voltage at point A = +10 V- 10 V
Voltage at point A = 0 V
To check this result, compute the voltage drop across R2 and subtract this from the voltage
at point A. The result should be the voltage of V2. Compute this as follows:
ER2 = R2 x IT
ER2 = 1Ω x 10 A
ER2 = 10 V
V2 = (voltage at point A) - (ER2)
V2 = (0V) - (10 V)
V2 = -10 V
7-62. It is not necessary that the voltage supplied be equal to create a point of virtual
ground. In Figure 7-13, view (B), V1 supplies +1 volt to the circuit while V2 supplies -10
volts. The total difference in potential is 11 volts. The total resistance of this circuit (R1 +
R2) is 11 ohms. The total current (IT) is 1 ampere. The voltage drop across R1 (ER1 = R1 x
IT) is 1 volt. The voltage drop across R2 (ER2 = R2 x IT) is 10 volts. The voltage at point A
can be computed as follows:
Voltage at point A = V1 - ER1
Voltage at point A = (+1V) - (+1V)
Voltage at point A = 0 V
7-18
TC 9-62
23 June 2005
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