Electronics - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 2)
2.
What is the power dissipated by R2, R4, and R6?
Discussion:
45 comments Page 3 of 5.
HAPPY said:
9 years ago
Leave all concepts, I want to know how the current is find in this question? We want to know the starting 3 steps. (i1, i2, i3)?
Srikanta roy chowdhury said:
9 years ago
Please explain this clearly.
S.L.Kulayappa said:
9 years ago
Please explain this clearly.
Amit kumar said:
9 years ago
Could you tell how can we find voltage drop when circuit in series.
Steve said:
10 years ago
Can you please tell how to get the current in every resistor?
Jorda said:
10 years ago
Pr2 = Vr2(I1-I2).
I1 = 0.01778 A.
I2 = 0.01058 A.
I3 = 0.00529 A.
= 57.9 V (0.01778 A-0.01058 A).
= 0.417 W.
Pr4 = Vr4 (I2-I3).
Pr6 = Vr6 (I3).
I1 = 0.01778 A.
I2 = 0.01058 A.
I3 = 0.00529 A.
= 57.9 V (0.01778 A-0.01058 A).
= 0.417 W.
Pr4 = Vr4 (I2-I3).
Pr6 = Vr6 (I3).
Lara said:
10 years ago
I didn't know how did you find the power for each one?
REETIKA said:
10 years ago
Please solve this circuit by using ohm's law.
Kim said:
10 years ago
Can somebody explain this step by step with units? So hard to understand this. Sorry guys. With total current, voltage drops, current and power for each resistors.
Kim said:
10 years ago
Guys can somebody explain this step by step like finding required, voltage drops, currents of each resistors and also the power of each things.
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