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.

First determine the current as I1,I2 and I3.
I1=0.01778A
I2=0.01058A
I3=0.00529A
Now current across R2=0.0072A
" " " R4=0.00529A
" " " R6=0.00529A
By using the formula for power as P=I2R
We will get power across R2=414mW
"""""""""""""""""""""""""R4=196mW
"""""""""""""""""""""""""R6=168mW

So answer will be [A].

By mesh analysis inspection method:

11 I1 - 8 I2 = 111 --> EQN 1.
-8 I1 + 17 I2 - 7 I3= 0 --> EQN 2.
-7 I2 + 14 I3 = 0 --> EQN 3.

BY RESOLVING THESE EQUATIONS WE GET,

I1 = 17.72 mA
I2 = 10.50 mA
I3 = 5.25 mA

Current through R2 is I1-I2 = 7.22 mA.

And by P = (I^2)R formula we can calculate power across the resistor R2

P = 7.22*7.22*10^-6*8*10^3 = 417 mW.

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.

Make the circuit simple first such away that all will be in parallel and now cdr only apply for parallel those which will be in series are not consider.

First r5 and r6 make series this make big resistance parallel with r4 so new resistance is r456 this is in series with in r3 so r3456 which is parallel with r2 now apply but omit r1.

Apply kvl to the loop and find current
Which is,

I1=0.01778
I2=0.01058
I3=0.00529

Now current across R2=0.0072
Now current across R4=0.00529
Now current across R6=0.00529

We Know that P=I2R
SO P across R2=(0.0072)2*R2
=(0.0072)2*8000
=417mW.

P across R4=193
P across R6=163

Node connecting r1, r2, r3-X say X volt also,

.............................. r3, r4, (r5+r6) - Y.
Apply nodal analysis.
(x-111)/3+(x-y) /2+(x/8)=0.

(y-x)/2+y/7+y/(1+6)=0,
Solve X and Y.
Use power=V*V/R.
For r2 p2=(x-y) ^2/2,
P4=y^2/7,
Don't have to solve p6.

Redraw the circuit to it's simplest series form by getting the total resistance and current for the entire circuit then work your way backwards by applying ohm's law to determined the other values.

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?

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.