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

Hemanth said: (Apr 11, 2011) | |

Can you give description anybody? |

Ranjan Singh said: (May 24, 2011) | |

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]. |

Avi said: (Aug 27, 2011) | |

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

Priya said: (Mar 14, 2012) | |

In this we use mesh analysis method for current flowing. where have to find I1=delta1/delta; I2=delta2/delta; I3=delta3/delta |

Ron said: (Jul 24, 2012) | |

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. |

Sonu Pradhan said: (Dec 17, 2012) | |

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 |

Yeyo said: (Aug 2, 2013) | |

More description anybody? |

Jorge said: (Aug 10, 2013) | |

Hello if is possible I would like to see the answer procedure step by step of this exercise. Thank you. 2. What is the power dissipated by R2, R4, and R6? |

Jorge said: (Aug 13, 2013) | |

Hi does anyone can tell me this example step by step? |

Arun said: (Oct 21, 2013) | |

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. |

Vinay said: (Nov 24, 2013) | |

Can anybody please tell me what is total current in the circuit and the voltage drop at R2 step by step. |

Megha said: (Jul 31, 2014) | |

Since the voltage drop is same in parallel circuits why can't we take. P = V^2/R formula. |

Abdul Haleem said: (Dec 4, 2014) | |

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. |

Jehangir Khan Khaksar said: (May 25, 2015) | |

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. |

Mahesh Reddy said: (Jun 24, 2015) | |

We can use this also I think. Please correct me if I am wrong. P=V2/R. We know that voltage will be same across all the branches i.e. 111. |

Kim said: (Sep 22, 2015) | |

Guys can somebody explain this step by step like finding required, voltage drops, currents of each resistors and also the power of each things. |

Kim said: (Sep 22, 2015) | |

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. |

Reetika said: (Oct 22, 2015) | |

Please solve this circuit by using ohm's law. |

Lara said: (Oct 27, 2015) | |

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

Jorda said: (Nov 16, 2015) | |

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). |

Steve said: (Dec 24, 2015) | |

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

Amit Kumar said: (Mar 9, 2016) | |

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

S.L.Kulayappa said: (Aug 22, 2016) | |

Please explain this clearly. |

Srikanta Roy Chowdhury said: (Sep 9, 2016) | |

Please explain this clearly. |

Happy said: (Sep 16, 2016) | |

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)? |

Richard Msesi said: (Sep 24, 2016) | |

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

Aseel said: (Nov 5, 2016) | |

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

Sue said: (Nov 16, 2016) | |

By using the formula p = I^2/R. WE CAN CALCULATE POWER IN R2, R4 AND R6. |

Nagesh said: (Nov 22, 2016) | |

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

Saipriya said: (Nov 25, 2016) | |

Please tell, how to get current in each resistor? |

Emediong said: (Jan 20, 2017) | |

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

Sana said: (Apr 3, 2017) | |

How do we get the loops? Describe it in detail. |

D. Mk said: (May 23, 2017) | |

Easy to follow your solution. Thanks @Arun. |

Baby said: (Sep 9, 2017) | |

I agree @Arun. |

Mayank said: (Sep 29, 2017) | |

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. |

Aarti said: (Nov 4, 2017) | |

Very nice answer and question because answer. |

Abhinav said: (Dec 10, 2017) | |

Can we not apply p = v^2/R? |

Christine said: (Apr 24, 2018) | |

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

Gopala R said: (May 28, 2018) | |

Piease explain clearly. |

Gopala R said: (May 28, 2018) | |

Explain it clearly. |

Sankeerthana said: (Jul 9, 2018) | |

Why the answer is different when we apply v^2/R? |

Hanis said: (Nov 21, 2018) | |

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

Ganesh said: (Sep 1, 2019) | |

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

Rahul said: (Feb 19, 2020) | |

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. |

Durga Prasad said: (May 18, 2021) | |

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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