### Discussion :: Parallel Circuits - General Questions (Q.No.13)

Satish Kumar said: (Dec 7, 2010) | |

Since 1KOhm||2KOhm therefore Req=2/3 KOhm. now using Ohm's law I=V/R we get I=18A. So current in 2KOhm = (1*18)/(1+2)=6mA. |

Raju said: (Mar 26, 2011) | |

This 18mA current how? Please explain. |

Brajesh said: (Jun 17, 2011) | |

@Raju R = r1*r2/r1+r2 = 2/3 kOhm v = 12v So current I = V/R = 12*3/2=18mA. |

Mohan said: (Jun 24, 2011) | |

@sathish Explain the last step (ie., current in 2kohm = (1*18)/(1+2)=6mA.) What formula you have use for substitution ? |

Lakshmi said: (Jun 25, 2011) | |

I1=v/R1 =12/1k =12mA I2=v/R2 =12/2k =6mA total current=12+6=18mA |

Vineet said: (Jul 3, 2011) | |

@Mohan The formula in last step is voltage divider rule. |

Rajnish said: (Oct 30, 2011) | |

Not voltage divider but current divider. |

Seema Ingole said: (Nov 26, 2011) | |

Voltage accros parallel ckt is same so v=12v & we want to find out current accros 2kohm resistor so R=2kohm as , v=ir i=v/r i=12/2 =6 Hence option d is right. |

M.V.Krishna/Palvoncha said: (Dec 14, 2011) | |

Given R1=1k & R2=2k; current(I1) at R1=(V/R1); current(I2) at R2=(V/R2) I1=12/1K; I2=12/2K; I1=12mA; I2=6mA; from the question, we need I2. so, option b is right. |

Clara said: (Jan 4, 2012) | |

V=IR 12 = I(2K) I=12/2K I= 6mA |

Javed said: (Jan 19, 2012) | |

Both resistors are in parallel so voltage remains same. I1 at 2k=V/R i.e. 12/2K=6mA |

Jaggubai said: (Apr 7, 2012) | |

V = IR I = V/R So V = 12V ,R = 2000 ohms I = 12/2000 = 0.006 = 6mA |

Suchitra said: (May 8, 2012) | |

Voltage division formula r2=r1*I/r1+r2;r1=r2*I/r1+r2 |

Harendra Kumar said: (Sep 16, 2012) | |

In parallel circuit voltage remains same. So, I1=V/R1=12/1=12mA. I2=V/R1=12/2=6mA. This is right answer. So for this question option B is true. |

Jaideep said: (Nov 14, 2013) | |

Can we use current divider rule here? |

Vedanti said: (Feb 4, 2014) | |

I1 = V/R1 = 12mA, I2 = V/R2 = 6mA, These are the currents flowing through individual branches. But if the resistances are connected in parallel. Total current = 18mA. So in across 1st resistance 12 mA will get drop so 6 mA will flow through 2nd resistor. |

Senthil said: (Feb 22, 2014) | |

My simple way to calculate the current flowing through 2kohm resistance is, If we are exchange 1KOhm resistance by 2 parallel connected 2KOhm resistance, the total circuit will become 3 parallel connected 2KOhm resistances which are having voltage source of 12v. Total current flowing through the circuit is 18A. So 6mA current will flow through 3 branches of the circuit. |

Prasanna Kumar said: (May 10, 2014) | |

There is formula for this: i2 = (r1*I/r1+r2). I = total current. |

Suraj said: (Jan 14, 2015) | |

In || circuit the voltage drop is same so v = 12 answer resistance is 2. So effective I = v/R, 12/2 = 6. So answer B. |

S.Balakumar said: (Jun 8, 2015) | |

We know the parallel circuit voltage is same. The voltage across the 2 K ohm resistor is 12 v. Apply I = V/R. I = 12/2 kohm = 6 ma. |

Rashmi said: (Jun 24, 2015) | |

The voltage through both the resistors are same but the current divides. According to KCL (the algebraic sum of all the individual current is 0). Therefore the current through R2=6mA. |

Armandwish said: (Sep 11, 2015) | |

If 2k resistor connected 1st i.e. beside of voltage source then answer will be 18. |

Kiran.Kk said: (Jan 17, 2016) | |

I = V/R. R1 = 1 kohm. R2 = 2 kohm. Their four r2 = 12/2 = 6 ma. |

Lakshmi Prasanna said: (Jan 21, 2016) | |

Voltage is same in parallel circuit so directly 12/2k = 6 mA. |

Yogash Joshi said: (Aug 20, 2016) | |

I = I1 + I2. I = V/R. 1/R = 1/R1 + 1/R2. 1/R = 1 + 1/2 = 3/2, So R = 2/3. NOW I = V/R SO, I = 12/ (2/3) = 18. AND I1= V/ R1 = 12/ 1= 12. SO IN EQUATION 1. I = I1 + I2. 18 = 12 + I2. SO, I2 = 18 - 12 = 6mA. |

Souvik said: (Jan 17, 2018) | |

Circuit current = V/R = 12/(2/3) = 18mA Current through the 2k resistor = total current*opposite site resistor/ sum of the two resistors. that is equal to 18*1/(2+1) = 6mA. |

Faizan said: (Nov 25, 2018) | |

Here using current divider you can easily find. Since 1KOhm||2KOhm. therefore Req=2/3 KOhm. now using Ohm's law. I=V/R. we get I=m18A. So current in 2KOhm = (1*18)/(1+2)=6mA. |

Arjun said: (Jan 20, 2019) | |

I = V/R. I = 12/1+12/2, In this 12/2 stands for the 2KOhm resistor. So it is 6mA. |

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