Prestige Institute of Management & Research, Bhopal Sessional Papers PDF

# Prestige Institute of Management & Research, Bhopal ## INDEX | Sr. No. | Experiment Description | Experiment Date | Submission Date | Remarks/Signature | |---|---|---|---|---| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |...

# Prestige Institute of Management & Research, Bhopal ## INDEX | Sr. No. | Experiment Description | Experiment Date | Submission Date | Remarks/Signature | |---|---|---|---|---| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | ## SESSIONAL PAPERS ### Assignment - 1 **Q No 1A) Write statement and find out current across a.c. by using mesh analysis** **Q No 1B) What is the voltage across A and B in the circuit shown in figure below** **Solution: In ADECA mesh I current flows** $10I_1 = -6$ $I_1 = - \frac{6}{10}$ **In mesh FGHBF I2 current flows** $14I_2 = -12$ $I_2 = - \frac{12}{14}$ **for VAB circuit current flow from A to downward - some as mesh current** $-4I_1-12 - 4I_2 = 0$ $-4 \times -\frac{6}{10} - 12 - 4 \times -\frac{12}{14} = 0$ $V_{AB} = + 2.4 - 12 - 8.4$ $V_{AB} = -18V$ **Q No 1C) Write statement of KCL & find current across a.c using nodal analysis.** **Solution: Take VA and VB references circuit KCL equation** for VA : $V_A - 1 + \cfrac{V_A}{2} - \cfrac{V_B}{20} = 0$ for VB : $\cfrac{V_B -20}{10} + \cfrac{V_B}{5} + \cfrac{V_B}{20} - \cfrac{V_A}{5}= 0$ $V_A -5 + \cfrac{V_A}{2} - \cfrac{V_B}{20}= 0$ $\cfrac{2V_A}{2} - \cfrac{10 }{2} + \cfrac{V_A}{2} - \cfrac{V_B}{20} =0$ $\cfrac{3}{2}V_A - 5 - \cfrac{V_B}{20} = 0$ $3V_A - 10 - \cfrac{V_B}{10} = 0$ $30V_A - 100 - V_B = 0$ $V_A -5 + V_A - V_B = 0$ $2V_A - 5 - V_B = 0$ $V_B = 2V_A - 5$ $\cfrac{V_B -20}{10} + \cfrac{4V_B}{20} - \cfrac{4V_A}{20}+ \cfrac{V_B}{20} = 0 $ $\cfrac{V_B}{10} - 2 + \cfrac{V_B}{5} -\cfrac{V_A}{5} +\cfrac{V_B}{20} = 0$ $\cfrac{V_B}{10} - 2 + \cfrac{4V_B}{20} -\cfrac{4V_A}{20} +\cfrac{V_B}{20} = 0$ $\cfrac{V_B}{2} - 2 + \cfrac{4V_B}{20} -\cfrac{4V_A}{20} +\cfrac{V_B}{20} = 0$ $\cfrac{10V_B}{20} - 2 + \cfrac{4V_B}{20} -\cfrac{4V_A}{20} +\cfrac{V_B}{20} = 0$ $ \cfrac{15V_B}{20} - 2 - \cfrac{4V_A}{20} = 0$ $15V_B - 40 - 4V_A = 0 $ $15V_B = 40+ 4V_A$ $V_B = \cfrac{40+4V_A}{15}$ $V_B = 2V_A - 5$ and $V_B = \cfrac{40+4V_A}{15}$ $2V_A - 5 = \cfrac{40+4V_A}{15}$ $30V_A - 75 = 40 + 4V_A$ $26V_A = 115$ $V_A = \cfrac{115}{ 26}$ $I_{load} = \cfrac{V_A}{20} = \cfrac{115}{26 \times 20} = 0.205$ Ampere **Q No 2(a) Discuss about different type of independent and dependent source with suitable sketch.** In this case at a line one source is active and another source replaced by their internal resistance. **Q No 3 (b) Find current across a.c by using Thevenin theorem.** **Step 1:** $R = 10, I = 2$. $R = 2 ohm, I = 5$ $R = 6 ohm, I = 3$ **Step 2:** $R_{AB} = R_{in} - R_{th}$ $R_{in} = 0.4$ **Step 3: ** **Q No 4 (b) Compute the power dissipated in 9 ohm resistor using superposition theorem.** **Step 1:** In mesh ABCTHA $16I_1 - 4I_2 - 32 = 0$ At CTF Branch current equation $I_1 - 4 - I_2 = 0$ And Now for $I_2$ variable we'll use super mesh: - ABCDHGHA $12I_1' + 6I_2' + 9I_3' - 0$ $I_1' = 0.5$ $I_2' =-2$ $I_3' = -2$ **Step 2:** In mesh ABHGA $16I_1' - 4I_2' - 32 = 0$ In mesh BCDEFB $18I_2' - 4I_1' - 32 = 0$ $I_1'' = -1.5$ $I_2'' = -2$ $I_3 = I_2'' + I_3'$ $I_3 = -2 + 2$ $I_3 = 4$ Ampere