### Discussion :: RL Circuits - General Questions (Q.No.5)

Joseph said: (Aug 1, 2011) | |

Current through inductor = 15/125=0.12 = 120 mA. |

Elavendhan said: (Jan 3, 2012) | |

According Ohms law V = IR I = v/R V = 15 v I = ? R1 = 470 R2 = 125 Both the Resistance in parallel so, R = (R1*R2)/(R1+R2) = (470*125)/(470+125) = 98.739 I = 15/98.739 Ans = 152 mA |

Biswajit Oec said: (Feb 2, 2012) | |

After calculate 152MA PUT votage divider rule it is i*R1/R1+R2. |

Paban Borah(Assam) said: (Jan 20, 2013) | |

R1=470. R2=125. Circuit current(i)=152mA. Current through inductor(i2)=(i*R1)/(R1+R2)=120mA. |

Darji Divyang said: (Mar 2, 2014) | |

Here circuit current is 152mA but dear friends in question said that find the current through an inductor so that answer is 120mA. 15/125=120mA. |

Priyaphilemon said: (Feb 4, 2015) | |

In an AC circuit containing pure inductance the following formula applies: Current through ac inductance = voltage/opposition to current flow. I = V/XL. I = 15/125. I = 0.12a. I = 120ma. |

Vijay Bairoji said: (Aug 23, 2015) | |

According Ohms law. V = IR. I = v/R. V = 15 v I = ? R1 = 470. R2 = 125. Both the Resistance in parallel so, R = (R1*R2)/(R1+R2). = (470*125)/(470+125). = 98.739. I = 15/98.739. Answer = 152 mA. But current through inductor is: IL= (i*R1)/(R1+R2). |

Sushant said: (Sep 11, 2015) | |

Current through conductor = (152*470)/595. |

Prasanta Bcet said: (Dec 27, 2016) | |

Here, we have to find the source current then put current division rule. |

Supriyo Kumar said: (Feb 3, 2017) | |

Since it is voltage source is parallel to both resistances. So, In parallel circuit v1=v2=v3=v. Current in inductor, I = V/R =15/125 = 120 mA. |

Ratna said: (Apr 28, 2017) | |

Since the voltage source is parllel to both resistances. |

Mohsin said: (Jan 18, 2019) | |

You can't just simply calculate Req like 2 resistors in parallel, proper impedance have to be calculated for parallel RL circuit. I = V/Z. Z= 1/*(1/R)^2+(1/XL)^2. |

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