The current source is 5 mA. This current will flow through two paths, one path has a 1.5 k ohm resistor and the other has two resistors in series their total resistance is 3 k ohm ( = 1 k + 2 k)

The 1.5 k ohm resistor will draw twice the current of the 3 k ohm path (let I be the current in the 3 k ohm resistor). The voltage drop across the 1.5 k ohm ( 1.5k*(X) must equal the voltage drop across the 3 k ohm resistance formed by the 1 k and 2 k resistors (3k *I)

1.5 k*X = 3k*(I) from this X = 2I, so the 1.5 k ohm resistor has twice the current of the 3 k resistance in the second branch. The two currents equal 5 ma.

5ma = 2I + I ========> I = 5/3 ma = 1.66 ma

The current in 1 k ohm resistor is 1.66 ma, this is also the current in the 2 k ohm resistor

The 1.5 k ohm resistor will have a current of 3.33 ma

The 1 k ohm resistor will have a voltage of 1.66 volts ( = 1.66 ma* 1 k ohm)

Convert the current source and the 1.5k resistor to a Thevenin equivlaent circuit of V=5mA*1.5=7.5 volts and R=1.5k. Now the circuit current is 7.5/(1.5k+2k+1k)=1.6667 mA. The voltage v is 1.66667 volts.

Rtotal = ( 2 +1 ) = 3 k(ohm)



Rtotal = 1/1.5 + 1/3 = 1 k(ohm) = 1000 ohm



Vtot = iRtot = 1000 * 0.005 = 5



i1 = V/R = 5/1.5 = 5/1500= 0.0033



i = i1+ i2 = 0.005 - 0.0033



i2 = 0.0017



V = 1000 * 0.0017 = 1.7 => 1.66666

I had a look in the circuit diagram, Thus:



2k resistance is in series with 1 k resistance



Hence,Their total = 3k.



Now the a/m resistance, 3k is parallel with 1.5k resistance[See cct, diagram given in the above link].



Now applying the formula for parallel connection of the resistances, then,(R1*R2)/(R1+R2)



where R1 & R2 are resistances connected in parallel.



Hence total equivalent Resistance, RT =(3*1.5)/((3+1.5) =4.5/4.5 =1 k?



And we know, v =IR where v is the voltage, I is the current and R is the reresistanceThen, v= (5/1000) (1000) = 5volts.



Hence, v= 5volts is the answer. Thank you.

V of 1k resistor = (I of 1k resistor)*(1k resistor)



V of 1k resistor = {[(1.5)/(1.5 +2 +1)]*(5mA)}* (1k Ohm) = (1.667mA)*(1 k Ohm) = 1.667 Volts

V= I*1E3





by current divider:



I = 1.5K/(1.5K + 3K)*5E-3



I = (1.5/4.5)*5E-3



I = 0.333333*5E-3



I = 0.001667





V = 0.001667*1E3



V = 1.667 V .................. Answer





1/Req = 1/1.5K + 1/3K



1/Req = 0.001



Req = 1E3



Req = 1K ..................... Answer

It says to find v in the figure, given that i=5 mA. Here is the figure:





http://s11.postimage.org/6720sxpkj/circuit.jpg





I keep trying to find the equivalent resistor and then using V=IR, but this is wrong.