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PHY301 – Circuit Theory Virtual University PHY 301 LECTURE 26 Example: Calculate the voltage V by using Thevenin’s theorem . o Solution: We want to calculate V by using Thevenin’s theorem .we will follow the 0 following steps . First step: Removing R L Here R is 4k resistor at which we want to calculate the voltage V L o. Second step: Calculating V th We want to calculate V . Apply KVL in two loops to calculate the individual loop th currents. Here I = 2mA 1 KVL for the super mesh -6 +6kI +12k(I +I ) + 12k(I +I ) =0 2 1 2 1 2 -6 +6kI +12kI + 12kI +24kI =0 2 2 2 1 30kI – 6 +24k(2m)=0 2 30KI -6 +48 =0 2 I = -42/30mA 2 I = -1.4mA 2 Now we will calculate the voltage across 6k and 12 k to find the value of V th Therefore, V = 6kI 6k 2 = 6k(-1.4m) V = -8.4volts 6k V = 12k(I1 +I2) 12k Virtual University of Pakistan Page 107 PHY301 – Circuit Theory = 12k( 2m -1.4m) V = 7.2volts 12k So, V = V6k + V12k th = -8.4 +7.2 V = -1.2volts th Third step: Calculating R th 6k is in series with 12k .The resultant of these two is in parallel with 12k. (6k+12k)||12k = 18k x12k/(12k+18k) R = 7.2k th Fourth step: Calculating the unknown quantity. After calculating V and R , re-inserting the load resistance R in the circuit in series with th th L R and considering the V as a battery in series with these two resistances. th th V = -1.2 x4k/(7.2k + 4k) 0 V = - 0 .4285 volts 0 Example: Calculate the value of RL and the maximum power dissipation across it by Thevenin’s Theorem. Solution: We want to calculate R and the maximum power across it by using Thevenin’s theorem .we will follow L the steps given earlier. First step: Removing R L To remove R to calculate V .In this case our R is our R . L th th L Second step: Calculating V th We want to calculate V . Apply KVL in two loops to calculate the individual loop th currents. Virtual University of Pakistan Page 108 PHY301 – Circuit Theory Here I = 2mA 1 Applying KVL to loop 2 6kI + 3k(I – I ) +3 =0 2 2 1 9kI -3kI + 3 =0 2 1 9kI – 6 +3 =0 2 I = 1/3 mA 2 I = 0.33mA 2 Now we will calculate the voltage across 4k and 6k to find the value of V th Therefore, V = 4k(I ) AB 1 = 4k( 2m) V = 8 volts AB V = 6k(I ) BC 2 = 6k x 0.33m V = 2 volts BC So V =V +V th AB BC = 8 +2 V = 10 volts th Third step: Calculating R th To calculate R Short circuiting the voltage source and open circuiting the current source. th . 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 = R =R th L R =R = 6k th L Fourth step: Calculating unknown quantity. After calculating V and R , re-inserting the load resistance R in the circuit in series with th th L R and considering the V as a battery in series with these two resistances. th th For maximum power dissipation R =R = 6k L th 2 P = I R ------------- (A) L From Ohm’s law V=IR and I = V/R so I =10/12 put in A we have 2 P = (10/12k) (6k) L 2 = (0.83) X 6 P = 4.1 mW L Virtual University of Pakistan Page 109 PHY301 – Circuit Theory Example: Calculate the value of RL and the maximum power dissipation across it by Thevenin’s Theorem. Solution: We want to calculate R and the maximum power across it by using Thevenin’s theorem .we will follow L the following steps . First step: Removing R L To remove R to calculate V .In this case our R is our R . L th th L Second step: Calculating V th Apply KVL to the circuit For loop1 -12 +6I + 6(I - I ) = 0 1 1 2 -12 +12I – 6I = 0 1 2 2I – I = 2 --------------- (A) 1 2 For loop 2 12I – 6I +3 =0 2 1 4I – 2I +1 =0 2 1 2I = 4I 1 2 put in equation (A) we have 4kI +1 –1kI =2 2 2 3kI =1 2 I = 0.33mA 2 from equation (A) I = (I2 +2)/2 1 = 0.33+2/2 I = 1.166mA 1 Now voltage across 6k resistor V = 6kI R6k 1 = 6k(1.166m) V = 7volts R6k V = 7V +3V th V =10 volts th Virtual University of Pakistan Page 110 PHY301 – Circuit Theory Third step: Calculating R th We want to calculate R 6k is in parallel with 6k .The resultant is again parallel with third 6k resistor. th 6k||6k = (6k x 6k)/6k + 6k = 3k 3k||6k = 6k x 3k/(6k + 3k) = 2k So R =R = 2k th L Fourth step: Calculating unknown quantity. After calculating V and R , re-inserting the load resistance R in the circuit in series with th th L R and considering the V as a battery in series with these resistances. th th V = 10 x2k/4k RL = 5 volts To calculate the power dissipation we have 2 P = V /R = 25/2k L P = 12.5mW L 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. Because the voltage or current of the dependent sources is dependent on the voltage or current of these independent spources. While calculating R we will short circuit the open terminals of the Thevenin circuit and will calculate the I th sc and then divide V with I to calculate R . th sc th Example: Calculate the voltage V by using Thevenin’s theorem . o Solution: We want to calculate V by using Thevenin’s theorem .we will follow the following steps . 0 First step: Removing R L Here R is 6k resistor at which we want to calculate the voltage V L o. Virtual University of Pakistan Page 111 PHY301 – Circuit Theory Second step: Calculating V th Voltage across 4k resistor V = (4k/6k) 12 4k V = 8volts 4k Voltage across 2k resistor V = (2k/6k) 12 2k V = V = 4volts A 2k V = 8 – 4V th A = 8 – 4(4) = 8 – 16 V = - 8 volts th Third step: When we use dependent sources we will use the following technique . In the above circuit output which is opened after removing the load resistance now we will replace this open circuit with short circuit. As in the circuit below. We will first find I to calculate R . sc th Here V = 2kI A 1 Applying KVL to loop 1 -12 +2kI +(I – I )4k= 0 1 1 sc -12 +2kI +4kI – 4kI = 0 1 1 sc 6kI -4kI = 12 1 sc For other loop (I –I )4k +4V = 0 sc 1 A 4kI -4I +4(2kI )=0 sc 1 1 4kI -4I +8I =0 sc 1 1 I = -I 1 sc Putting in equation for loop1 3I -2I = 6 1 sc 3(-I ) -2I = 6 sc sc I = - 6/5 mA sc So Rth = Vth/Isc = -8/(-6/5 m) Rth = 6.67k Fourth step: After calculating V and R , re-inserting the load resistance R in the circuit in series with th th L R and considering the V as a battery in series with these resistances. th th So V = (6k/6k +6.67k) x8 0 = 6k/12.67k x8 V = 3.78 volts 0 Virtual University of Pakistan Page 112

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