Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R2R2R_2). Answer
in terms of given quantities, together with the meter readings I1I1I_1 and I2I2I_2 and the current I3I3I_3.
1 answer:
Answer:
-I₁ + I₂ + I₃ = 0
I₁ = I₂ + I₃
Step-by-step explanation:
The image of the circuit is obtained online and attached to the question.
The junction rule is essentially a law of conservation of current (charges). It applies to electrical circuits at steady state.
It explains that the for any given junction (node in an electrical circuit), the sum of current entering the junction is equal to the sum of current leaving the junction. That is, the net sum of current at any junction is zero.
Current entering a junction is assigned a positive sign and that leaving the junction is assigned a negative sign.
Σ I = 0
From the image of the circuit attached, I₁ is leaving the junction labelled number 1 and I₂ and I₃ are entering the junction.
Hence,
-I₁ + I₂ + I₃ = 0
I₁ = I₂ + I₃
Hope this Helps!!!
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see attached picture:
9i-j = sqrt (9^2 - 1^2) = sqrt(81-1) = sqrt80
now divide both terms in V by that:
u = 9/sqrt(80)i - 1/sqrt(80)j
see attached picture:
Answer:
2
Step-by-step explanation: