EXPERMENT NO. 4

Componints: 1 * resistor, 1 *5V1 400mW Zener diode, 1 *Diode 1N4007.

Vs= 10 & 15V, R= 1Kohms

                                             10 Volts                                                      15 Volts
Volt drop V1:                        4.63V                                                    4.80V
Volt drop V2:                        0.63V                                                    0.66V
Volt drop V3:                        5.26V                                                    5.47V
Volt drop V4:                        4.73V                                                    9.54V
 
Calculate current A:     4.74A                                                     9.54A
I= Vs - V1(ZdV)+V3(DV) / R.
I= 10V - 4.6V + 0.63V / 1 Kohms=4.74 A                       I=15V - 4.80+ 0.66V / 1Kohms=9.54A
I=4.54A                                                                             I= 9.54A

There are some pictures of my experement.

Discribe what is happining and why you are geting these reading?

In this circuit i put 1* Zener diode on reverse biase and on diode on forward biase. The zener diode on forward biase on circuit working same normal diode , but onreverse biase depend on diode voltge. This zener diode is 5v in this reading we can see wich (V1) is zener diiode on reverse the voltage drop is 4.63V, that way for the safe region voltage throgh on diode in reverse biase up to 5V.
The other diiode wich is 1N400 in forward biase in this circuit the votge drop is normally 0.6V to 0.7V. In my reading is on 10Vs voltge drop 0.63V & 15Vs is Vd 0.66V. If we see between 15Vs & 10V diffrence voltge drop acrross the diodes noot toomuch littile bit , And also i had current calculation from my reading. I had got correct reading in this experemint.