* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

# Download KIRCHHOF`S LAW KIRCHOOF`S CURRENT LAW

Survey

Document related concepts

Integrating ADC wikipedia , lookup

Valve RF amplifier wikipedia , lookup

Operational amplifier wikipedia , lookup

Schmitt trigger wikipedia , lookup

Power electronics wikipedia , lookup

Resistive opto-isolator wikipedia , lookup

Opto-isolator wikipedia , lookup

Josephson voltage standard wikipedia , lookup

Voltage regulator wikipedia , lookup

Current mirror wikipedia , lookup

Switched-mode power supply wikipedia , lookup

Surge protector wikipedia , lookup

Current source wikipedia , lookup

Power MOSFET wikipedia , lookup

Transcript

Example We want to calculate the voltage Vo .We will the same four steps. First Step: We have removed RL to calculate Vth . Second Step: Vth We want to calculate Vth . Apply KVL in two loops to calculate the individual loop currents. Here I1 = 2mA I2 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. 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 c A I1 9kI2 -3kI1 + 3 =0 9kI2 – 6 +3 =0 I2 = 1/3 mA I2 = 0.33mA I2 Vth c A I1 Now VAB = 4k(I1) = 4k( 2m) = 8 volts. I2 Vth c A I1 VBC = 6k(I2) = 6k x 0.33m = 2 volts. I2 Vth c A I1 So Vth = VAB + VBC = 8 +2 = 10 volts. I2 Third Step: To calculate Rth .Short circuiting the voltage source and open circuiting the current source. V Rth Rth th c A I1 I2 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 I1 6k Vth I2 Putting in equation for loop1 4kI2 +1 –1kI2 =2 3kI2 =1 I2 =0.33mA I1 6k Vth I2 from equation1 I1 = (I2 +2)/2 =0.33+2/2 =1.166mA I1 6k Vth I2 VR6k = 6kI1 = 6k(1.166m) =7volts. I1 6k 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 . Second Step: ++ VV A A -- ++- 4V A 4V A VV 0 th Voltage across 4k resistor V4k = (4k/6k) 12 =8volts Voltage across 2k resistor V2k= (2k/6k) 12 = 4volts. ++ VV A A -- ++- 4V A 4V A Vth = 8 – 4VA = 8 – 4(4) = 8 – 16 = -8 volts. VV 0 th Third Step: We will first find Isc to calculate Rth. Here VA = 2kI1 - + 4VA +- VA Isc I1 Applying KVL to loop 1 -12 +2kI1 +(I1 – Isc)4k=0 -12 +2kI1 +4kI1 – 4kIsc=0 6kI1 -4kIsc =12 Isc - + 4VA +- VA Isc I1 For other loop (Isc –I1)4k +4VA = 0 4kIsc -4I1 +4(2kI1)=0 4kIsc -4I1 +8I1 =0 I1 = -Isc Isc - + 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 So Rth = Vth/Isc = -8/(-6/5 m) =6.67k Isc Fourth Step: So V0 = (6k/6k +6.67k) x8 =6k/12.67k x8 =3.78 volts