DC Supply Notes PDF

# UNIT-1-D.C SUPPLY ## 1. Voltage (V) - Defined as work done in forcing IC of charge completely around the circuit. - The unit of voltage is Volt (V). ## 2. Current (I) - Defined as the rate of flow of electric charge. - S.I unit of current is Amperes (A). - $$i = \frac{dQ}{dt}$$ ## 3. Power...

# UNIT-1-D.C SUPPLY ## 1. Voltage (V) - Defined as work done in forcing IC of charge completely around the circuit. - The unit of voltage is Volt (V). ## 2. Current (I) - Defined as the rate of flow of electric charge. - S.I unit of current is Amperes (A). - $$i = \frac{dQ}{dt}$$ ## 3. Power (P) - The rate of doing work is known as power. - S.I unit is watts (W). - $$P = \frac{W}{t}$$ ## 4. Circuit - A circuit consists of voltage source, connecting wires and element forming single closed path. ## 5. Network - The connection of number of circuits together is called as network. - **Active element:** Element which provide energy to the circuit are called active elements. - **Example:** Voltage source or battery. - **Passive element:** Elements which receives energy are called passive element. - **Example:** Resistor, inductor and capacitor. # PARAMETERS OF CIRCUIT ## 1. Resistor - Defined as a property of circuit by which it opposes the flow of current through it. - S.I unit of resistor is ohm (Ω). - **Symbol:** M R (ohm (Ω)) - **Reciprocal of resistance is called conductance (G) = 1/R** - **S.I unit is mho (Ʊ).** - **Specific resistance:** The resistance of conductor depends upon length of conductor, cross-sectional area of conductor and material of conductor. - **$$R = \frac{PL}{A} \Rightarrow P = Resistivity$$** ## 2. Inductor - Inductance is a property of coil by virtue of which it opposes change of current. - S.I unit is Henry (H). - **When coil carries changing current and emf is induced due to change of flux and emf is given by** - **$$e = L \frac{di}{dt}$$** - **When current of constant magnitude flows through inductor there is no shock circuit emf, and it acts as voltage on it.** ## 3. Capacitor - Two conducting surfaces separated by some medium is called as capacitor. Symbol - **$$ \Rightarrow Q \propto V \Rightarrow Q = CV $$** - **Unit is Farads (F).** ## Ohms’s Law: - At constant temperature, voltage is proportional to current. - $$ \Rightarrow V \propto I$$ - $$ \Rightarrow V = IR; $$ - R: Proportionality constant. ## RESISTANCES CONNECTED IN SERIES - $$V = V_1 + V_2 + V_3$$ - $$ = iR_1 + iR_2 + iR_3$$ - $$V = i(R_1 + R_2 + R_3)$$ - $$V = R_1 + R_2 + R_3$$ - $$R_{eq} = R_1 + R_2 + R_3$$ ## RESISTANCES CONNECTED IN PARALLEL - $$i = i_1 + i_2 + i_3$$ - $$ = V(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3})$$ - $$ = V(\frac{1}{R} + \frac{1}{R_2} + \frac{1}{R_3})$$ - $$R_{eq} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ ## INDUCTANCES IN SERIES - $$e = e_1 + e_2 + e_3$$ - $$e= L\frac{di}{dt}$$ - $$e = L \frac{di}{dt} + L_2 \frac{di}{dt} + L_3 \frac{di}{dt}$$ - $$e = \frac{di}{dt}(L_1 + L_2 + L_3)$$ - $$e = \frac{di}{dt} L_{eq} = L_1 + L_2 + L_3$$ ## INDUCTANCES IN PARALLEL - $$\frac{di}{dt} = \frac{di_1}{dt} + \frac{di_2}{dt} + \frac{di_3}{dt}$$ - $$e= \frac{di}{dt} + \frac{di}{dt} + \frac{di}{dt}$$ - $$e = \frac{di}{dt}(\frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3})$$ - $$e = \frac{di}{dt}L_{eq} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}$$ ## CAPACITANCES IN SERIES - $$V = V_1 + V_2 + V_3$$ - $$V = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3}$$ - $$V= Q(\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3})$$ - $$Q = C_{eq}V$$ - $$C_{eq} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$ ## CAPACITANCES IN PARALLEL - $$Q = Q_1 + Q_2 + Q_3$$ - $$Q = C_1V + C_2V + C_3V$$ - $$Q = V(C_1 + C_2 + C_3)$$ - $$C_{eq} = C_1+ C_2 + C_3$$ ### When current is passed through the resistor, power is wasted in the form of heat which is given by - $$P = Vi$$ (or) - $$P = i^2R$$ (or) - $$P = \frac{V^2}{R}$$ ### When current is passed through inductor it stores energy in the form of electromagnetic field - Li². ### When current is passed through capacitor it stores energy in the form of electrostatic field. ½ CV². ### Energy stored by inductor (Derive an exp.) - Consider an inductor of inductance L and current I flows through inductor. - The emf developed across inductor is given by - $$e = L\frac{di}{dt}$$ - $$ dP = dw = eidt $$ - Instantaneous power, $$dP = dw = eidt $$ - $$ \Rightarrow dw = dpdt \Rightarrow P = ei = L \frac{di}{dt}idt$$ - Instantaneous power = $$\int Lidi = \frac{1}{2}Li^2(J)$$ ### Energy stored by capacitor (Derivation) - Consider capacitor of capacitance C, let it be charged through voltage V - $$Q =CV$$ - $$ \Rightarrow P = Vi$$ - $$ \Rightarrow i = \frac{dQ}{dt}$$ - $$ \Rightarrow i = C\frac{dV}{dt}$$ - $$W = \int_{0}^{V}Cvdv$$ - $$W = \frac{1}{2}CV^2$$ - $$ \Rightarrow dpdt = VC\frac{dV}{dt}$$ ### Kirchoff’s Law In order to find current, voltage and power in different branch of the network, Kirchoff’s law is applicable. In this we have 2 laws. #### 1) Kirchoff’s current law (KCL) - The sum of current entering the junction is equal to sum of the current leaving the junction. - At junction O', sum of current entering = sum of current leaving. - $$ i_1 + i_5 + i_6 = i_3+ i_4 + i_7 $$ #### 2) Kirchoff’s voltage law (KVL) - Around any close loop, the algebraic sum of voltage drop around any closed loop is equal to zero. - Loop ABCDA - $$ -iR_1 - iR_2 -iR_3 + V = 0 $$ ## Problem In the circuit, find current & voltage across all branches by Kirchoff’s law. * **Loop ABEFA** - $$ -2.5i - 5i + 10 = 0$$ * **Loop BCDEB** - $$ - 2(i - i_1) - 3(i - i_1) + 5i_1 = 0$$ - $$ -2i + 2i_1 - 3i + 3i_1 + 5i_1 = 0$$ - $$ -5i + 10i_1 = 0$$ - $$i = 2A$$ - **Current in 2.5Ω = i2.5 = i = 2A** - **Current in 5Ω = i_1 = i = 1A** - **i_2 = (i - i_1) = (2 - 1) = 1A** - **i_3 = (i - i_1) = (2 - 1) = 1A** * **Voltage drop ** - V2.5 = i (2.5) = 2 (2.5) = 5V - V5Ω = i_1 (5) = 1(5) = 5V - V2Ω = (i - i_1) 2 = 1(2) = 2V - V3Ω = (i - i_1) 3 = 1(3) = 3V ## Find current in 25Ω resistor by Kirchoff’s law ### Loop ABEFA - $$ -100i - 25i + 12 = 0$$ ### Loop BCDEB - $$18 - 75(i-1_1) + 40 - 20(i - i_1) + 25i_1 = 0$$ - $$ i = 0.201 $$ ### Find current in all the resistors by Kirchoff’s law ### Loop ABEFA - $$ -220i - 470i_1 + 5 = 0$$ ### Loop BCDEB - $$ -680(i - i_1) - 470i_1 = 0$$ - $$ i = 0.024 $$ ## Mesh analysis / Loop analysis (KVL) - In this method, branched currents are replaced by loop current / mesh current. - A loop current is a current which is assumed to flow around a closed loop that is through all elements of the loop. - loop ABEFA - $$ -i_1R_1 - (i_1 - i_2)R_2 + V = 0 $$ - loop BCDEB - $$ -i_2R_3-(i_2- i_1)R_2 = 0$$ - loop ABEFA - $$ -i_1R_1 - (i_1 + i_2)R_2 + V = 0$$ - loop BCDEB - $$ -i_2R_3+(i_1 + i_2)R_2 = 0$$ ### In the circuit, find current in all elements by mesh analysis. - loop ABEFA - $$ -i_12.5 - (i_1 - i_2)5 + 10 = 0$$ - loop BCDEB - $$ -2i_2 - 3i_2 - ( i_2 - i_1) 5 = 0$$ - $$ \Rightarrow -5i_2 - 5i_2 + 5i_1 = 0 $$ - $$\Rightarrow -10i_2 + 5i_1 = 0$$ - $$\Rightarrow i_1 = 2i_2$$ - $$ \Rightarrow -2i_2(2.5) - (2i_2 - i_2)5 + 10 = 0 $$ - $$ \Rightarrow -5i_2 - 5i_2 + 10 = 0$$ - $$ \Rightarrow - i_2 = 1 $$ - $$ i_1 = 2$$ ### In the network shown, find current in all elements and power in 5Ω by mesh analysis. - loop ABEFA - $$ -i_13-(i_1 - i_2)5- 6i_1+50=0$$ - loop BCDEB - $$ -2i_2 - 25- 8i_2 - ( i_2 - i_1)5 = 0$$ - $$ \Rightarrow -10i_2 - 25- 5i_2 15i_1 = 0$$ - $$ \Rightarrow 5i_1 - 15i_2 -25 = 0$$ - From ① - $$ - 9i_1 - 5i_1 + 5i_2 + 50 = 0 $$ - $$ - 14i_1 + 5i_2 + 50 = 0 $$ - From ③ - $$ (i_1 = 3.37A, i_2 = 0.54 A )$$ - $$ i_1 - i_2 = 3.37 + 0.54 = 3.91 A $$ - $$ P = i^2R = (3.91)^2 * 5 = 15.28 * 5 = 76.44 W$$ ### Find current in 4Ω by mesh analysis. - loop ABCA - $$ -5i_1 - 10 - (i_1 - i_2)4 = 0 $$ - $$ -5i_1 - 10(i_1 - i_2) 4= 0 $$ - loop ACDA - $$ (i_2 - i_i)4 - 3(i_2 - i_3) 2i_2 = 0$$ - $$ (i_2 - i_1)4 - 3(i_2 - i_3) 2i_2 = 0$$ - loop BDCB - $$ -i_2- 3(i_3-i_2) 10(i_3-i_1) = 0$$ - $$ -i_2- 3(i_3-i_2) 10(i_3-i_1) = 0$$ - $$ -5i_1 - 10i_1 +10i_2 + 4i_2 + 4i_2 = 0$$ - $$= -15i_1 + 10i_2 + 4i_2 = 0$$ - $$ \Rightarrow -15i_1 + 14i_2 = 0$$ - $$ -4i_2 + 4i_1 - 3i_2 + 3i_3- 2i_2 = 0 $$ - $$ \Rightarrow 4i_1 - 9i_2 + 3i_3 = 0$$ - $$ \Rightarrow 10i_1 + 3i_2 - 13i_3 = 12 $$ - $$ i_1 = -1.738A, i_2 = -1.653A, i_3 = -2.642A $$ - Current in 4Ω = $$ ( i_1 - i_2 ) = -0.08A$$ ### Find current in 4kΩ by mesh analysis. - loop ABEFA - $$ -i_1 2000 -(i_1 - i_2) 2000 + 12 = 0$$ - $$ -6000i_1 -2000i_2 + 2000i_2 + 12 = 0$$ - $$ - 8000i_1 + 2000i_2 + 12 = 0 $$ - loop BCDEB - $$ - 3000i_2 - 4000(i_2) - 2000(i_2 - i_1) = 0$$ - $$ -3000i_2 - 4000i_2 - 2000 i_2 + 2000 i_1 = 0$$ - $$ -9000 i_2 + 2000 i_1 = 0 $$ - $$ i_1 = 1.58 * 10^{-3} A, i_2 = 3.52 * 10^{-4} A$$ - $$ i_1 = 1.58 mA, i_2 = 0.352 mA$$ ### Source transformation - A current source in parallel with resistance can be converted into voltage source in series with resistance and vice versa - $$V = iR$$ - In series, voltages are added up. - In parallel, currents are added. ### Reduce network into single voltage source in series with resistance between A & B - $$ i_1 + i_2 = 3A $$ ### Find thevenin’s equivalent by finding current through the element in 4Ω. * **Step 1:** Remove 4Ω, find Vth/Voc - $$ -2 i - 2i - 12 + 8 = 0$$ - $$ i = -1A$$ - $$ -2i - Vth + 8 = 0$$ - $$ -2(-1) + 8 = Vth$$ - $$ Vth = 10 $$ * **Step 2:** Find Rth, by S.C the voltage source - $$ Rth = (2//2) $$ - $$= \frac{12}{2} $$ - $$ = 1 $$ * **Step 3:** Draw equivalent circuit and connect 4Ω to find current - $$I_L = \frac{Vth}{Rth + R_L}$$ - $$ = \frac{10}{1 + 4} $$ - $$ = 2A $$ ### Find current in 7.5Ω by thevenin's theorem. * **Step 1:** Remove 7.5Ω, find Vth/Voc - $$ -100i - 25i + 12 = 0$$ - $$ \Rightarrow +125i = 12$$ - $$ i = 0.096 A$$ - $$ 18 - Vth + 40 +25 = 0 $$ - $$ Vth = 60.4V $$ * **Step 2:** Find Rth, by S.C voltage source - $$ Rth = (100//25) + 20 $$ - $$ = \frac{100 * 25}{100+25} + 20$$ - $$= 20+20 $$ - $$ = 40Ω $$ * **Step 3:** Draw equivalent circuit and connect 7.5Ω to find current - $$ IL = \frac{Vth}{Rth + R_L}$$ - $$ = \frac{60.4}{40+75}$ $ - $$ = 0.525A$$ ### Find current in 25Ω by thevenin's theorem. * **Step 1:** Remove 25Ω, find Vth/Voc - $$ -100i + 18 - 75i + 40 - 20i +12 = 0$$ - $$ -195i +70 = 0$$ - $$ i = 0.35A$$ - $$ -100i - Vth + 12 = 0$$ - $$Vth = -23$$ * **Step 2:** Find Rth, by S.C voltage source - $$ Rth = 100 // (20+75) $$ - $$= \frac{100 * 100}{100+25} $$ - $$ = 48.71Ω $$ * **Step 3:** Draw equivalent circuit and connect 25Ω to find current - $$ IL = \frac{Vth}{Rth + R_L}$$ - $$ = \frac{-23.8}{48.153+ 25} $$ - $$ = -23.8 $$ - $$ = -0,322A$$ ### Norton's theorem - Any active linear network with 2 terminals can be replaced by an equivalent circuit comprising of current source in // with resistance, which would bring equal to current, the current being flow if terminals are short circuitted and resistance being equal to resistance between terminals by replacing all sources with internal resistance. - $$ I_L = \frac{I_N * R_N}{R_N + R_L} $$ ### Example: - To find current in R_L by Norton's theorem. - **Step 1:** Remove R_L - $$ -i_1R_1 - i_2R_2 + V_1 = 0 $$ - $$ - (i_1 - i_2)R_3 - V_2 + i_2R_2= 0$$ - **Step 2:** $$ i_2 = I_{sc} = i_m$$ - **Step 3:** Remove current sources, Volt (R_L = 0) $$i = 0.0$$ - $$ RN = (R_1 || R_2) + R_3 $$ - **Step 4:** Draw equivalent circuit - $$ I = \frac{I_NR_N}{R_N + R_L} $$ ### Find current in 30Ω by Norton's theorem - **Step 1:** Remove 30Ω, S.C terminal - $$ -5i - 10(i - i_2) + 10 = 0 $$ - $$ -5i - 10i + 10i_2 + 10 = 0 $$ - $$ -15i + 10i_2 + 10 = 0 $$ - $$ - 20i_2 - 10(i_2 - i_1) = 0 $$ - $$ -20i_2 - 10i_2 + 10i_1 = 0 $$ - $$ +10i_1 - 30 i_2 = 0 $$ - $$ i_1 = - 0.85, i_2 = -0.285 A $$ - **Step 2:** Find RN, by S.C voltage source. - $$ RN = (5||10) + 20 $$ - $$ = \frac{5 * 10}{5 + 10} + 20 $$ - $$ = 23.3Ω $$ - **Step 3:** Draw equivalent circuit and connect 30Ω to find current. $$ I = \frac{I_NR_N}{R_N + R_L} $$ - $$= \frac{0.285* 23.3}{23.3+30}$$ - $$ = 0.1241 $$ ## Superposition theorem - In any linear network containing two or more source. The current in any part of the circuit is equal to algebraic sum of current in that branch by each of the source with all the other sources equal to zero. ### NOTE: - Any voltage source present in the circuit can be removed by short circuiting. - If any current source is present, it can be removed by open circuiting. ### Example: **Step 1:** Consider voltages V_1 and V_2 and find the value of i_2. - $$ -i_1R_1 - i_2R_2 + V_1 = 0 $$ - $$ -(i_1 - i_2)R_3 - V_2 + i_2R_2 = 0$$ - $$ i_2 = A $$ **Step 2:** Consider V_1, remove V_2 by short circuiting the terminals. - $$ -i_1R_1 - i_2R_2 + V_1 =_ 0$$ - $$ -R_3(i_1 - i_2) + i_2R_2 = 0$$ - $$i_2 = A$$ **Step 3:** Consider V_2, remove V_1 by short circuiting the terminal - $$ -iR_1 - i_2R_2 = 0$$ - $$ - (i_1 - i_2)R_3 - V_2 + i_2R_2 = 0 $$ - $$ i_2 = A $$ **Step 4:** step 1 + step 2 + step 3 ### Find current in 1Ω by superposition theorem - **Step 1:** Keep 100V, Remove 50V. - $$ -15i_1 - i_2 + 100 = 0$$ - $$ -6i_1 + 7i_2 = 0$$ - $$ i_1 = 6.34, i_2 = 5.4A $$ - **Step 2:** Keep 50V, Remove 100V. - $$ -15i_1 - i_2 + 50 = 0$$ - $$ -6i_1 + 7i_2 = 0$$ - $$ i_1 = 0.45A, i_2 = -6.75A $$ - **Step 3:** Current in 1Ω = step 1 + step 2. - $$ 5.4 + (-6.75) = -1.35 A $$ ### Limitation of Superposition theorem: - Superposition theorem is applicable if a circuit consists of 2 sources ### Find current in 10Ω by superposition theorem. - **Step 1:** Keep 4V, remove 2V. - $$ -25i_1-10(i_1 + i_2) + 4 = 0$$ - $$ -100i_1 +10(i_1+ i_2 ) = 0 $$ - **Step 2:** Keep 2V, Remove 4V. - $$ -25i_1 - 10(i_1 +i_2) + 2 = 0 $$ - $$ -10i_2 + 100i_1 = 0 $$ - **Step 3:** Current in 10Ω = step 1 + step 2 - $$ = -0.026 + 0.05$$ - $$ = 0.024 A$$ ### Thevenin’s theorem - Thevenin’s theorem states that any active network with 2 terminals can be replaced by an equivalent circuit consists of voltage source in series with resistance. The source voltage being equal to open circuit voltage and resistance being equal to resistance of the network by replacing sources. - Voltage source → S.C - Current source → 0.C ### Equivalent circuit - $$ I = \frac{Vth}{Rth + R_L}$$ ### Steps to solve Thevenin’s theorem - Find current in Ru by thevenin’s theorem. - **Step 1:** Remove Ru, find Vth/Voc