# Diodes question (I Got the answer but I just need more clarification)

Problem : finding I2 and V

at the end of solving I got I2 = 0 by making D1 open and D2 is short circuit.

I just don't understand how the answer for V at the end is -1.7

I tried open D1 and therefore I2 = 0

then I thought it is all together now so I1 = I3 =( 5 - (-5))/15 = 0.666666666mA

then I got lost on how to find V, since I read the solution to be V = -1.7

I tried doing the .6666 * 5 or .66666 * 10 which didn't get me toward the answer

Any explanation that can walk me through would be much appreciated

• Post your solution, what you did. Sep 10, 2015 at 3:04

R1, D2, and R3 are dropping a total of 10V. Since D2 is ideal, it drops 0V. This means that the current flowing must be:

${10\text{V} \over {10\text{k}\Omega + 5\text{k}\Omega}} \approx 667\mu\text{A}$

The voltage dropped by R3 must therefore be:

$667\mu\text{A} \cdot 5\text{k}\Omega \approx 3.34\text{V}$

The voltage at V must then be:

$3.34\text{V} + - 5\text{V} = -1.66\text{V}$

It looks like you are assuming that the diodes are perfect.

Since (as you found out) D1 is not conducting you can ignore D1 and the ground.

I1 = I3.

Do the calculations relative to the -5V rail.

Total voltage is 5V -(-5) = 10V

V1 + V2 = 10

Since the same current is flowing through both resistors the voltage across them is:

Upper resistor : I1 * 10K (call that V1) Lower resistor : I1 * 5k (call that V2)

(I1*10000) + (I1*5000) = 10

I1*15000 = 10

I1 = 10/15000

Voltage across lower resistor = 5000 * 10/15000 = 10/3 = 3.33

Voltage at V relative to ground = -5 + 3.3 = -1.67