Example We want to calculate the voltage Vo .We will the same four steps.

First Step: We have removed RL to calculate Vth .

We want to calculate Vth . Apply KVL in two Second Step: Vth We want to calculate Vth . Apply KVL in two loops to calculate the individual loop currents.

Here I1 = 2mA

KVL for the super mesh -6 +6kI2 +12k(I1 +I2) + 12k(I1 +I2) =0 Vth I1 KVL for the super mesh -6 +6kI2 +12k(I1 +I2) + 12k(I1 +I2) =0 -6 +6kI2 +12kI2 + 12kI2 +24kI1 =0 30kI2 – 6 +24k(2m)=0 30KI2 -6 +48 =0

I2 Vth I1 I2 = -42/30mA = -1.4mA

I2 Vth I1 Now V6k = 6kI2 =6k(-1.4m) = -8.4volts

I2 Vth I1 V12k = 12k(I1 +I2) = 12k( 2m -1.4m) =7.2volts

I2 Vth I1 So Vth = V6k + V12k = -8.4 +7.2 = -1.2volts.

Third Step: 6k is in series with 12k .The resultant of these two is in parallel with 12k. (6k+12k)||12k = 18k x12k/(12k+18k) =7.2k

Fourth Step: V0 = -1.2 x4k/(7.2k + 4k) = - 0 .4285 volts.

Example We want to calculate the value of RL and the maximum power dissipation across it by Thevenin’s Theorem.

First Step: To remove RL to calculate Vth .In this case our Rth is our RL.

We want to calculate Vth . Apply KVL in two Second Step: Vth A c We want to calculate Vth . Apply KVL in two loops to calculate the individual loop currents.

Here I1 = 2mA Applying KVL to loop 2 6kI2 + 3k(I2 – I1) +3 =0

Vth A c I2 I1 9kI2 -3kI1 + 3 =0 9kI2 – 6 +3 =0 I2 = 1/3 mA I2 = 0.33mA

Vth A c I2 I1 Now VAB = 4k(I1) = 4k( 2m) = 8 volts.

Vth A c I2 I1 VBC = 6k(I2) = 6k x 0.33m = 2 volts.

Vth A c I2 I1 So Vth = VAB + VBC = 8 +2 = 10 volts.

Third Step: To calculate Rth .Short circuiting the voltage source and open circuiting the current source.

3k is in parallel with 6k and 4k is series with these two so Vth Rth Rth A c I2 I1 3k is in parallel with 6k and 4k is series with these two so 3k||6k + 4k = 3k x 6k/(3k + 6k) +4k =2k +4k = 6k = Rth =RL

Fourth Step: For maximum power dissipation RL =Rth = 6k PL = I2R =(10/12k)2 (6k)

Rth PL = 25/6 mW = 4.1 mW

Example We want to calculate the value of RL and the maximum power dissipation across it by Thevenin’s Theorem.

First Step: To remove RL to calculate Vth .In this case our Rth is our RL.

Second Step: KVL for loop1 -12 +12I1 – 6I2 = 0 2I1 – I2 =2

6k I1 Vth I2 KVL for loop2 12I2 – 6I1 +3 =0 4I2 – 2I1 +1 =0 2I1 =4I2

Putting in equation for loop1 4kI2 +1 –1kI2 =2 3kI2 =1 I2 =0.33mA Vth I2 Putting in equation for loop1 4kI2 +1 –1kI2 =2 3kI2 =1 I2 =0.33mA

6k I1 Vth I2 from equation1 I1 = (I2 +2)/2 =0.33+2/2 =1.166mA

6k I1 Vth I2 VR6k = 6kI1 = 6k(1.166m) =7volts.

6k I1 Vth I2 Vth = 7V +3V =10 volts.

Third Step: We want to calculate Rth 6k is in parallel with 6k .The resultant is again parallel with third 6k resistor.

6k Rth 6k||6k = (6k x 6k)/6k + 6k =3k 3k||6k = 6k x 3k/(6k + 3k) =2k

6k Rth So Rth = 2k

Fourth Step: VRL = 10 x2k/4k =5 volts PL = V2/R =25/2k =12.5mW

THEVENIN’S THEOREM and DEPENDENT SOURCES. Working with dependent sources is different from working with independent sources while applying Thevenin’s theorem . While calculating Rth we can simply open circuit current sources and short circuit voltage sources.

THEVENIN’S THEOREM and DEPENDENT SOURCES. Because the voltage or current of the dependent sources is dependent on the voltage or current of these independent spources.

THEVENIN’S THEOREM and DEPENDENT SOURCES. While calculating Rth we will short circuit the open terminals of the Thevenin circuit and will calculate the Isc and then divide Vth with Isc to calculate Rth.

Example We want to calculate the voltage Vo by Thevenin’s theorem.

First Step: To remove RL to calculate Vth .

Voltage across 4k resistor V4k = (4k/6k) 12 =8volts Second Step: + - + - + - + - VA 4VA VA 4VA V0 Vth Voltage across 4k resistor V4k = (4k/6k) 12 =8volts Voltage across 2k resistor V2k= (2k/6k) 12 = 4volts.

Vth = 8 – 4VA = 8 – 4(4) = 8 – 16 = -8 volts. + - + - + - + - VA 4VA

Third Step: We will first find Isc to calculate Rth. Here VA = 2kI1

Applying KVL to loop 1 -12 +2kI1 +(I1 – Isc)4k=0 4VA - + + - VA Isc I1 Isc Applying KVL to loop 1 -12 +2kI1 +(I1 – Isc)4k=0 -12 +2kI1 +4kI1 – 4kIsc=0 6kI1 -4kIsc =12

For other loop (Isc –I1)4k +4VA = 0 4kIsc -4I1 +4(2kI1)=0 4kIsc -4I1 +8I1 =0 I1 = -Isc

Putting in equation for loop1 3I1 -2Isc =6 3(-Isc) -2Isc =6 4VA - + + - VA Isc I1 Isc Putting in equation for loop1 3I1 -2Isc =6 3(-Isc) -2Isc =6 Isc = -6/5 mA

4VA - + + - VA Isc I1 Isc So Rth = Vth/Isc = -8/(-6/5 m) =6.67k

Fourth Step: So V0 = (6k/6k +6.67k) x8 =6k/12.67k x8 =3.78 volts