### Discussion :: Circuit Theorems and Conversions - General Questions (Q.No.13)

M.V.Krishna/Palvoncha said: (Dec 19, 2011) | |

From super position theorem case 1: using only Vs1 and shorting Vs2. Rt=R1+(R2||R3) Rt=180 ohm; It=Vs1/Rt It=12/180 It=0.067 I(R2)=(R2||R3)*It/R3 I(R2)=0.0333A=33.4mA case 2: using only Vs2 and shorting Vs1. Rt=R3+(R1||R2) Rt=180 ohm It=Vs2/Rt It=6/180=0.0333 I(R2)=(R2||R1)*It/R1 I(R2)=16.7mA Adding case1 I(R2)+ case2 I(R2), because both currents are flowing in same direction. We get 33.3mA+16.7mA=50mA |

Renish said: (Feb 2, 2016) | |

In this question also apply the millman's theorem and it was very easy and fast technique. |

Ritesh Kesharwani said: (Jun 12, 2016) | |

I1 = Vs1/R1. = 12/120 = 0.1. I2 = Vs2/R3 . = 6/120 = 0.05. I = I1 + I2 . = 0.1 + 0.05 = 0.15. I(R2) = I * R2/(R1 + R2 + R2). = 0.15 * 120/360. = 50mA. |

Chhatrapal said: (Feb 1, 2017) | |

Vs1/r1+r2. i.e.12/120 + 120. = 50 * 10^-3. |

Hodavid said: (Jan 29, 2019) | |

Voltage for R2 = Vs1-Vs2. 12-6 = 6v. Try the different options V/R. 6/50mA = 120. |

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