### Discussion :: Series-Parallel Circuits - General Questions (Q.No.3)

Yuvaraj said: (Feb 25, 2011) | |

I need explanation for this problem. |

Nagamma said: (Mar 15, 2011) | |

Which formula used to solve the problem? |

Priti said: (Mar 17, 2011) | |

Please explain. Thank you. |

Ishwar said: (Mar 27, 2011) | |

Voltage division principle is used. This way First determine the parallel combination of R2 and RL and then find the voltage division between the resultant and R1. |

Madhusudan said: (Jul 23, 2011) | |

RL||R2=7.5k and vltg drop across 7.5k s (7.5/(7.5+2))*20=15.79... |

Karthik said: (Aug 20, 2011) | |

In the Given Circuit RL=120K So RL and R2 in parallel, R2=8K 1/R=1/R2+1/RL 1/R=1/8+1/120 1/R=0.13333 R=7.5K THEN R AND R1 IN SERIES SO TOTAL R =9.5K CURRENT I =0.0021A SO VL = I*R VL=0.0021*7.5K VL =15.75.. Thanks, Karthik |

Amit said: (Sep 21, 2011) | |

RL is given as 120k So 1st calculate the total resistance by 120k||8K=7500 then 2k+7500=9500 N v=ir 20=i*9500 i=20/9500=2.1052exp-3 Now by current division rule current through RL will be say I2=i*8k/(8k+120k)=1.315exp-4 again v=ir So say voltage across RL be v2 v2=1.315exp-4*120k =15.79v same can be solved using voltage division rule |

Rajesh Madavi said: (Jul 18, 2012) | |

8*120/8+120 =7.5k v=20*7.5k/7.5k+2k =20*15/19 =300/19 =15.79v |

Rekha said: (Nov 10, 2012) | |

Very simple procedure, Step 1:Find parallel combination of 8K & 120K = 7.5K Step 2:Now 2k &7.5k are in series so total resistance=9.5k Step 3:Find ct I=20v/9.5k= 2.105mA, Step 4:find drop across 2K is 2.105m*2K=4.2V Step 5:Finally Vl=20-4.2=15.8v. |

Manoj said: (Jul 11, 2016) | |

Thanks @Rekha. |

Yoosuf said: (Oct 28, 2016) | |

Thanks for explaining the solution. |

Aravind said: (May 30, 2017) | |

8k parallel 10k= (8*120)/(8+120)= 7.5k. Now, Voltage across 7.5k = Vol.across 120k (bz 8k parallel 120k) So vol across 7.5k =(20V * 7.5k)/2k + 7.5k, = 15.79V. (Vol.divider equation) |

Vaishnavi said: (Jul 13, 2017) | |

Thanks for explaining it. |

Pragadee said: (Jul 27, 2017) | |

Thanks for explaining this. |

Shik Shansha Vali said: (Sep 15, 2017) | |

RL = 120 anyone can explain? |

Vc_Davv said: (Nov 19, 2017) | |

Don't be so complex. Use thevenin theorem. So vth = 16 v and Rth = 1.6 kohm. Voltage across Rl (load 120kohm) by VDL = (16*120K/120k+1.6k)=15.7v. |

Thiru said: (Feb 16, 2018) | |

Can we choose this method? Apply *voltage Divider Rule * V=((Total voltage *opposite Resistance) /Total Resistance) 20*8/8+2=16V. Whatever the Voltage across parallel is same as the other side That's why its 16v. |

Alex said: (Dec 14, 2018) | |

Is this series, parallel, or an open circuit? Please tell me. |

Akhil Va said: (Feb 27, 2019) | |

resistor 120k is parallel with 8k. So r=1/120+1/8 = 7.5k. vout=r2/(r1+r2) = 7.5/((2+7.5). = 15.78947368421053 V, = 15.79 V. |

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