In a two-source circuit, one source acting alone produces 10 ma through a given branch. the other source acting alone produces 8
ma in the opposite direction through the same branch. the actual current through the branch is
1 answer:
Refer to the figure below.
R = resistance.
Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R
Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R
Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.
Answer: 2 mA
R = resistance.
Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R
Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R
Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.
Answer: 2 mA
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Answer:
655128 ohm
Explanation:
C = Capacitance of the capacitor = 7.8 x 10⁻⁶ F
V₀ = Voltage of the battery = 9 Volts
V = Potential difference across the battery after time "t" = 4.20 Volts
t = time interval = 3.21 sec
T = Time constant
R = resistance
Potential difference across the battery after time "t" is given as
T = 5.11 sec
Time constant is given as
T = RC
5.11 = (7.8 x 10⁻⁶) R
R = 655128 ohm