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?

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 5 of 5.

Sankeerthana said:   7 years ago
Why the answer is different when we apply v^2/R?

Hanis said:   7 years ago
When you do the loop?

how do you get Equation 1: 11kI1-8kI2-111 = 0, where does the 11KI1 come from?

Ganesh said:   6 years ago
For individual resistance, I am getting R2 = 0.01773ohms.

Rahul said:   6 years ago
Simple solution:-

First find the total resistance of the circuit i.e., 6.25 kiloohm (hope you can find that).
Then find total current i.e., 111/6.25 = 17.73mA.

This current will only flow through 3kiloohm resistor as it is in series with the voltage source.
So the voltage drop across 3kiloohm resistor will be =current through resistor*resistance
=17.73*3 = 53.29 v.

So the voltage at first junction after 3kiloohm resistor is :- 111-53.29=57.79v.
So the voltage across 8kiloohm resistor will be 57.79-0=57.79 v.

We subtract it with zero because the other end of 8kiloohm resistor is at 0 vas it is connected to -ve end of voltage source.

Now power through 8kiloohm resistor is = V(8kiloohm)*V(8kiloohm)/resistance.
=57.79*57.79/8 =417 mW.

SO OPTION A is correct.
(5)

Durga Prasad said:   4 years ago
Actually mesh analysis is better to get currents and then power.

But it takes so much time.

We have to solve this in the simplest way can anyone tell how to solve this?


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