Electronics - Series-Parallel Circuits - Discussion

Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 2)

Series-Parallel Circuits

What is the power dissipated by R2, R4, and R6?

P2 = 417 mW, P4 = 193 mW, P6 = 166 mW

P2 = 407 mW, P4 = 183 mW, P6 = 156 mW

P2 = 397 mW, P4 = 173 mW, P6 = 146 mW

P2 = 387 mW, P4 = 163 mW, P6 = 136 mW

Answer: Option

Explanation:

No answer description is available. Let's discuss.

Discussion:

45 comments Page 3 of 5.

MeghA said: 1 decade ago

Since the voltage drop is same in parallel circuits why can't we take.

P = V^2/R formula.

Richard Msesi said: 9 years ago

I don't understand guys please can you put it in a simple way? i.e. use ohm's law.

Aseel said: 9 years ago

Yes, I am also not able to understand this, please repeat it and make it clearer.

Sue said: 9 years ago

By using the formula p = I^2/R.

WE CAN CALCULATE POWER IN R2, R4 AND R6.

Amit kumar said: 9 years ago

Could you tell how can we find voltage drop when circuit in series.

Avi said: 1 decade ago

How to determine current in this circuit. Will any body explain ?

EMEDIONG said: 9 years ago

Please, can someone please take it step by step for me to follow.

Steve said: 10 years ago

Can you please tell how to get the current in every resistor?

Ganesh said: 6 years ago

For individual resistance, I am getting R2 = 0.01773ohms.

Lara said: 10 years ago

I didn't know how did you find the power for each one?

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