Experiement 2

Objectives:-

•

1. To learn how to build a simple resistive network. 2. To learn how to measure the voltage across the resistor and the 3. 4.

5. 6. 7. 8.

9.

current that passes through it. To verify ohm's law, and analysis a circuit using kirchoff's voltage and current law. Calculate power dissipated in DC circuit and show that the power dissipated in a load equal to the power supplied by the source. Calculate the voltage drop across a resistor using voltage divider method. Calculate the current through a resistor by current divider method. plot the I-V characteristic of some linear & nonlinear devices. show the differences between linear and nonlinear devices. Verify the lab experiment with Pspise.

• Equipment:1. Digital multimeter(DMM) 2. Breadboard. 3. DC power supply. 4. Resistors. 5. LED (Light Emitted Diode). 6. Computer software (Pspise). •

contents:1. Ohm's law and DC series circuits. 2. Voltage and current divider. 3. I-V characteristics of tow terminal devices.

• Theory:In the voltage division rule the voltage is divided between tow series resistors in direct to their resistance, while in the current division rule the current is divided between tow parallel resisters in reverse to their resistance.  In the series compination we can use the voltage divider circuit to find the voltage in each resistor 

VR1=(R1/(R1+R2))*V  In the parallel compination we can use the current divider circuit to find the current in each resistor I1=(R2/(R1+R2))*I 

The LED is not emitted until reaching it's barrier potential value.

Calculation &data analyses: 1. Ohm's law & DC series circuits: We connect the circuit shown in figure 2.1 &take fill data in table Quantity Nominated value VR1 5.952 V VR2 4.047 V Idc 5.952 mA PR1 35.42 mW PR2 24.08 mW

Measured value 5.97 4.08 5.99 35.76 24.43

Pspice value 5.952 4.047 5.952 35.42 24.08

Percentage error 1.8% 3.4% 3.8% 1% 1.1%

VR1 = (R1/ (R1+R2))*VS = (1000/1680)*10 = 5.952 V. VR2 = (R2/ (R1+R2))*VS = (680/1680)*10 = 4.047 V. Idc = VS/(R1+R2) = 10/1680 = 5.952 mA. PR1 = I*VR1 = 5.952*5.952 = 35.42 mW. PR2 = I*VR2 = 5.952*4.047 = 24.08 mW. PEV1 = ((NV-MV)/NV)*100% = ((5.952-5.95)/5.952)*100% = 1.8 PEV2 = ((NV-MV)/NV)*100% = ((4.047-4.08)/4.047)*100% = 3.4% PEI = ((NV-MV)/NV)*100% = ((5.952-5.99)/5.952)*100% = 3.8% PEP1 = ((NV-MV)/NV)*100% = ((35.42-35.76)/35.42)*100% = 1% PEP2 = ((NV-MV)/NV)*100% = ((24.08-24.43)/24.08)*100% = 1.1% Voltage ¤t divider: We connect the circuit shown in figure 2.2 & measure the currents & voltages and fill the table .

Quantity IR1 IR2 VR1 VR2 PR1 PR2

Nominated value 4.857 mA 7.143 mA 4.857 V 4.857 V 23.59 mW 34.69 mW

Measured value 4.72 mA 4.8 mA 4.75 V 4.75 V 22.42 mW 33.01 mW

Pspice value 4.857 mA 7.143 mA 4.857 V 4.857 V 23.59 mW 34.69 mW

Percentage error 2.8% 2.7% 2.2% 2.2% 4.96% 4.84%

IR1 = ((R2/(R1+R2))*IT = (680/1680)*0.012 = 4.857 mA IR2= ((R1(R1+R2))*IT = (1000/1680)*0.012 = 7.143 mA VR1 = I1*R1 = 4.857* 1000 = 4.857 V VR1 = VR2 PR1 = I1*V1 = 4.857*4.857 = 23.59 mW. PR2 = I2*V2 = 7.143*4.857 = 34.69 mW. PEI1 = ((NV-MV)/NV)*100% = ((4.857-4.72)/4.857)*100% = 2.8% PEI2 = ((NV-MV)/NV)*100% = ((7.143-6.95)/7.143)*100% = 2.7% PEV1 = ((NV-MV)/NV)*100% = ((4.857-4.75)/4.857)*100% = 2.2% PEV2 = PEV1 = 2.2% PEP1 = ((NV-MV)/NV)*100% = ((23.59-22.42)/23.59)*100% = 4.96% PEP2 = ((NV-MV)/NV)*100% = ((34.69-33.01)/34.69)*100% = 4.84% I-V characteristics of two terminal devices: We plot the diagram of I-V for the LED shown in figure 2.3 . Range -2 -1.5 -1 -0.5 0 0.5 1.5 2 2.5

V LED -2.09 -1.56 -1.05 -0.54 0.01 0.55 1.55 1.64 1.67

ILED = (Vs-VLED)/R1 ILED1.5 = (1.5-1.35)/33 = 4.45 mA ILED2 = (2-1.42)/33 = 17.5 mA ILED2.5= (2.5-1.5)/33 = 30.3 mA

Vv

I-V characteristic 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

5

10

15

20

25

30

35

I mA

Conclusions:1. In the voltage divider we have the same current in the series resistors and different voltage cross them by putting a voltage source with resistors in series with it. 2. We design the current source by putting a power source with resistor in parallel with it, having the same voltage across them and different currents. 3. LED's experiment we see that the relation is nonlinear and there is no current while using negative voltage values.

Circuit connections: R2 680

Vs

R1 1k

10v Figure 2.1

R3 1k R2

R1 1k

mA 12

680

Firure 2.2

R1 33 LED Vs

Figure 2.3