Electronics - Series-Parallel Circuits - Discussion

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.

More description anybody?

By using superposition theorem.

We will find out current at resistors, so by using i*i*r we find power.

Explain it clearly.

I can't understand, can someone please elaborate further and use terminologies we all understand kindly?

Please tell, how to get current in each resistor?

Using KVL on three loops, your equations will be:

Equation 1: 11kI1-8kI2-111 = 0.

Equation 2: -8kI1+17kI2-7kI3 = 0.

Equation 3: -7kI2+14kI3 = 0.

Solving the equations will give.

I1 = 17.73mA, I2 = 10.5mA, I3 = 5.25mA.

Go back to circuit you will find that,

P2 = (I1-I2)^2*R2 = 417mW.

P4 = (I2-I3)^2*R4 = 193mW.

P6 = I3^2*R6 = 166mW.

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)?

Please explain this clearly.

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