Use simple diagram to explain Thevenin's theorem. For the network shown in fig. Q3b below, calculate the current in the 92 resistor using Thevenin's theorem.

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Asked on July 11, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-07-11T14:50:40.004Z

2a) State superposition principle using simple diagrams. The superposition principle states that in any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit can be calculated as the algebraic sum of the currents or voltages produced by each independent source acting alone. When considering one source, all other independent voltage sources are replaced by short circuits, and all other independent current sources are replaced by open circuits. Diagram 1: Original Circuit (0,0) to[V,v_=V_1] (0,2) to[R=R_1] (2,2) to[R=R_2] (4,2) to[I,i_=I_2] (4,0) -- (0,0); (2,2) -- (2,0) to[R=R_3] (4,0); (Imagine a circuit with a voltage source V_1 and a current source I_2, and resistors R_1, R_2, R_3.) Diagram 2: V_1 acting alone (I_2 open-circuited) (0,0) to[V,v_=V_1] (0,2) to[R=R_1] (2,2) to[R=R_2] (4,2); (2,2) -- (2,0) to[R=R_3] (4,0) -- (0,0); (4,2) -- (4,0); (Imagine the same circuit, but the current source I_2 is replaced by an open circuit.) Diagram 3: I_2 acting alone (V_1 short-circuited) (0,0) -- (0,2) to[R=R_1] (2,2) to[R=R_2] (4,2) to[I,i_=I_2] (4,0) -- (0,0); (2,2) -- (2,0) to[R=R_3] (4,0); (Imagine the same circuit, but the voltage source V_1 is replaced by a short circuit.) 2b) Use superposition theorem to calculate the current and power in the 20 resistor in Fig. Q2a. Step 1: Consider the 20V voltage source acting alone. The 2A current source is open-circuited. The 20 and 4 resistors are in parallel. Their equivalent resistance is: R_p1 = (20 × 4)/(20 + 4) = (80)/(24) = (10)/(3) This equivalent resistance is in series with the 10 resistor. The total resistance seen by the 20V source is: R_total1 = 10 + (10)/(3) = (30+10)/(3) = (40)/(3) The total current from the 20V source is: I_total1 = 20V40/3 = (60)/(40)A = 1.5A This current flows through the 10 resistor. Now, use current division to find the current through the 20 resistor: I'_20 = I_total1 × (4)/(20 + 4) = 1.5A × (4)/(24) = 1.5A × (1)/(6) = 0.25A The direction of I'_20 is downwards. Step 2: Consider the 2A current source acting alone. The 20V voltage source is short-circuited. The 10 resistor is now in parallel with the 20 resistor. Their equivalent resistance is: R_p2 = (10 × 20)/(10 + 20) = (200)/(30) = (20)/(3) The 2A current source splits between the 4 resistor and the parallel combination R_p2. The current flowing into the R_p2 branch (which contains the 20 resistor) is: I_branch = 2A × (4)/(4 + R_p2) = 2A × (4)/(4 + 20/3) = 2A × (4)/((12+20)/3) = 2A × (4)/(32/3) = 2A × (12)/(32) = 2A × (3)/(8) = 0.75A This 0.75A then splits between the 10 and 20 resistors. The current through the 20 resistor is: I''_20 = I_branch × (10)/(10 + 20) = 0.75A × (10)/(30) = 0.75A × (1)/(3) = 0.25A The direction of I''_20 is downwards. Step 3: Calculate the total current and power in the 20 resistor. Since both currents I'_20 and I''_20 flow in the same (downwards) direction, they add up: I_20 = I'_20 + I''_20 = 0.25A + 0.25A = 0.5A The power dissipated in the 20 resistor is: P_20 = I_20^2 × R = (0.5A)^2 × 20 = 0.25 × 20W = 5W 3a) For the network shown in Fig. Q2b, apply Norton's theorem to calculate the potential difference across the 10 resistor. Step 1: Remove the 10 load resistor. Let the terminals where the 10 resistor was connected be A (top) and B (bottom). Step 2: Calculate the Norton current (I_N) by short-circuiting terminals A and B. When A and B are short-circuited, the 16 resistor is in parallel with the short. Therefore, the voltage across the 16 resistor is zero, and it effectively carries no current (or rather, the current takes the path of least resistance through the short). The circuit simplifies to: Left branch: 8V source in series with 4. Right branch: 12V source in series with 5. Both branches are connected across the short circuit. Current from the left branch flowing into the short (from A to B): I_L = 8V4 = 2A. Current from the right branch flowing into the short (from B to A): I_R = 12V5 = 2.4A. The Norton current I_N (from A to B) is the algebraic sum: I_N = I_L - I_R = 2A - 2.4A = -0.4A The negative sign indicates that the current flows from B to A. So, I_N = 0.4A from B to A. Step 3: Calculate the Norton resistance (R_N) by turning off independent sources. Replace the 8V voltage source with a short circuit. Replace the 12V voltage source with a short circuit. Looking into terminals A and B: The 4 resistor is connected from A to ground. The 5 resistor is connected from B to ground. The 16 resistor is connected between A and B. This forms a parallel combination of the 16 resistor with the series combination of the 4 and 5 resistors. R_series = 4 + 5 = 9 R_N = (16 × 9)/(16 + 9) = (144)/(25) = 5.76 Step 4: Connect the 10 load resistor (R_L) to the Norton equivalent circuit. The Norton equivalent circuit consists of a current source I_N = 0.4A (flowing from B to A) in parallel with R_N = 5.76. The load resistor R_L = 10 is connected in parallel with this combination. The current through the 10 resistor (I_10) can be found using current division: I_10 = I_N × (R_N)/(R_N + R_L) = 0.4A × (5.76)/(5.76 + 10) = 0.4A × (5.76)/(15.76) I_10 ≈ 0.4A × 0.36548 ≈ 0.14619A This current flows from B to A. Step 5: Calculate the potential difference across the 10 resistor. V_10 = I_10 × R_L = 0.14619A × 10 = 1.46V Since the current flows from B to A, terminal B is at a higher potential than terminal A. 5a) Use simple diagram to explain Thevenin's theorem. Thevenin's theorem states that any linear electrical network containing voltage sources, current sources, and resistors can be replaced by an equivalent circuit consisting of a single voltage source (V_TH) in series with a single resistor (R_TH) connected across the load terminals. Diagram 1: Original Complex Circuit (0,0) to[V,v_=V_s] (0,2) to[R=R_1] (2,2) to[R=R_2] (2,0) -- (0,0); (2,2) to[R=R_3] (4,2) node[right]A; (2,0) to[R=R_4] (4,0) node[right]B; (4,2) to[R=R_L] (4,0); (Imagine a complex circuit with sources and resistors connected to a load resistor R_L at terminals A and B.) Diagram 2: Thevenin Equivalent Circuit (0,0) to[V,v_=V_TH] (0,2) to[R=R_TH] (2,2) node[right]A; (0,0) -- (2,0) node[right]B; (2,2) to[R=R_L] (2,0); (Imagine the complex circuit replaced by a voltage source V_TH in series with a resistor R_TH, connected to the same load resistor R_L at terminals A and B.) 5b) For the network shown in Fig. Q3b, calculate the current in the 9 resistor using Thevenin's theorem. Assumption: The question refers to Fig. Q5b (as Q3b is already used) and the 9 resistor is the load resistor (R_L) connected between the top-right and bottom-right terminals of the circuit. Step 1: Remove the 9 load resistor (R_L). Let the terminals be A (top-right, after 14) and B (bottom-right, after 6). Step 2: Calculate Thevenin voltage (V_TH = V_AB). With terminals A and B open, no current flows through the 14 and 6 resistors. The circuit consists of the 15A current source in parallel with the 12 resistor, connected to the series combination of 3 and 8. The equivalent resistance of the series branch is R_series = 3 + 8 = 11. The 15A current splits between the 12 resistor and the 11 series branch. The current flowing through the 11 branch (I_11) is: I_11 = 15A × (12)/(12 + 11) = 15A × (12)/(23) = (180)/(23)A ≈ 7.826A The voltage across the 8 resistor is V_8 = I_11 × 8 = (180)/(23)A × 8 = (1440)/(23)V. The voltage at the node between 3 and 8 (let's call it V_X) relative to ground is V_X = V_8. The voltage at the node where the 15A source, 12, and 3 connect (let's call it V_Y) is V_Y = I_11 × (3 + 8) = (180)/(23)A × 11 = (1980)/(23)V. Since terminals A and B are open, V_A is the voltage at the node before the 14 resistor, and V_B is the voltage at the node before the 6 resistor. So, V_A = V_Y and V_B = V_X. V_TH = V_A - V_B = V_Y - V_X = (1980)/(23)V - (1440)/(23)V = (540)/(23)V ≈ 23.478V Step 3: Calculate Thevenin resistance (R_TH) by turning off independent sources. Replace the 15A current source with an open circuit. Looking into terminals A and B: The 12 resistor is in series with the 3 resistor. This combination is in parallel with the 8 resistor. R_series' = 12 + 3 = 15 R_parallel' = (15 × 8)/(15 + 8) = (120)/(23) This R_parallel' is in series with the 14 and 6 resistors. R_TH = R_parallel' + 14 + 6 = (120)/(23) + 20 = (120 + 460)/(23) = (580)/(23) ≈ 25 ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._

Accepted Answer · 2026-07-11T14:50:40.004Z

2a) State superposition principle using simple diagrams. The superposition principle states that in any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit can be calculated as the algebraic sum of the currents or voltages produced by each independent source acting alone. When considering one source, all other independent voltage sources are replaced by short circuits, and all other independent current sources are replaced by open circuits. Diagram 1: Original Circuit (0,0) to[V,v_=V_1] (0,2) to[R=R_1] (2,2) to[R=R_2] (4,2) to[I,i_=I_2] (4,0) -- (0,0); (2,2) -- (2,0) to[R=R_3] (4,0); (Imagine a circuit with a voltage source V_1 and a current source I_2, and resistors R_1, R_2, R_3.) Diagram 2: V_1 acting alone (I_2 open-circuited) (0,0) to[V,v_=V_1] (0,2) to[R=R_1] (2,2) to[R=R_2] (4,2); (2,2) -- (2,0) to[R=R_3] (4,0) -- (0,0); (4,2) -- (4,0); (Imagine the same circuit, but the current source I_2 is replaced by an open circuit.) Diagram 3: I_2 acting alone (V_1 short-circuited) (0,0) -- (0,2) to[R=R_1] (2,2) to[R=R_2] (4,2) to[I,i_=I_2] (4,0) -- (0,0); (2,2) -- (2,0) to[R=R_3] (4,0); (Imagine the same circuit, but the voltage source V_1 is replaced by a short circuit.) 2b) Use superposition theorem to calculate the current and power in the 20 resistor in Fig. Q2a. Step 1: Consider the 20V voltage source acting alone. The 2A current source is open-circuited. The 20 and 4 resistors are in parallel. Their equivalent resistance is: R_p1 = (20 × 4)/(20 + 4) = (80)/(24) = (10)/(3) This equivalent resistance is in series with the 10 resistor. The total resistance seen by the 20V source is: R_total1 = 10 + (10)/(3) = (30+10)/(3) = (40)/(3) The total current from the 20V source is: I_total1 = 20V40/3 = (60)/(40)A = 1.5A This current flows through the 10 resistor. Now, use current division to find the current through the 20 resistor: I'_20 = I_total1 × (4)/(20 + 4) = 1.5A × (4)/(24) = 1.5A × (1)/(6) = 0.25A The direction of I'_20 is downwards. Step 2: Consider the 2A current source acting alone. The 20V voltage source is short-circuited. The 10 resistor is now in parallel with the 20 resistor. Their equivalent resistance is: R_p2 = (10 × 20)/(10 + 20) = (200)/(30) = (20)/(3) The 2A current source splits between the 4 resistor and the parallel combination R_p2. The current flowing into the R_p2 branch (which contains the 20 resistor) is: I_branch = 2A × (4)/(4 + R_p2) = 2A × (4)/(4 + 20/3) = 2A × (4)/((12+20)/3) = 2A × (4)/(32/3) = 2A × (12)/(32) = 2A × (3)/(8) = 0.75A This 0.75A then splits between the 10 and 20 resistors. The current through the 20 resistor is: I''_20 = I_branch × (10)/(10 + 20) = 0.75A × (10)/(30) = 0.75A × (1)/(3) = 0.25A The direction of I''_20 is downwards. Step 3: Calculate the total current and power in the 20 resistor. Since both currents I'_20 and I''_20 flow in the same (downwards) direction, they add up: I_20 = I'_20 + I''_20 = 0.25A + 0.25A = 0.5A The power dissipated in the 20 resistor is: P_20 = I_20^2 × R = (0.5A)^2 × 20 = 0.25 × 20W = 5W 3a) For the network shown in Fig. Q2b, apply Norton's theorem to calculate the potential difference across the 10 resistor. Step 1: Remove the 10 load resistor. Let the terminals where the 10 resistor was connected be A (top) and B (bottom). Step 2: Calculate the Norton current (I_N) by short-circuiting terminals A and B. When A and B are short-circuited, the 16 resistor is in parallel with the short. Therefore, the voltage across the 16 resistor is zero, and it effectively carries no current (or rather, the current takes the path of least resistance through the short). The circuit simplifies to: Left branch: 8V source in series with 4. Right branch: 12V source in series with 5. Both branches are connected across the short circuit. Current from the left branch flowing into the short (from A to B): I_L = 8V4 = 2A. Current from the right branch flowing into the short (from B to A): I_R = 12V5 = 2.4A. The Norton current I_N (from A to B) is the algebraic sum: I_N = I_L - I_R = 2A - 2.4A = -0.4A The negative sign indicates that the current flows from B to A. So, I_N = 0.4A from B to A. Step 3: Calculate the Norton resistance (R_N) by turning off independent sources. Replace the 8V voltage source with a short circuit. Replace the 12V voltage source with a short circuit. Looking into terminals A and B: The 4 resistor is connected from A to ground. The 5 resistor is connected from B to ground. The 16 resistor is connected between A and B. This forms a parallel combination of the 16 resistor with the series combination of the 4 and 5 resistors. R_series = 4 + 5 = 9 R_N = (16 × 9)/(16 + 9) = (144)/(25) = 5.76 Step 4: Connect the 10 load resistor (R_L) to the Norton equivalent circuit. The Norton equivalent circuit consists of a current source I_N = 0.4A (flowing from B to A) in parallel with R_N = 5.76. The load resistor R_L = 10 is connected in parallel with this combination. The current through the 10 resistor (I_10) can be found using current division: I_10 = I_N × (R_N)/(R_N + R_L) = 0.4A × (5.76)/(5.76 + 10) = 0.4A × (5.76)/(15.76) I_10 ≈ 0.4A × 0.36548 ≈ 0.14619A This current flows from B to A. Step 5: Calculate the potential difference across the 10 resistor. V_10 = I_10 × R_L = 0.14619A × 10 = 1.46V Since the current flows from B to A, terminal B is at a higher potential than terminal A. 5a) Use simple diagram to explain Thevenin's theorem. Thevenin's theorem states that any linear electrical network containing voltage sources, current sources, and resistors can be replaced by an equivalent circuit consisting of a single voltage source (V_TH) in series with a single resistor (R_TH) connected across the load terminals. Diagram 1: Original Complex Circuit (0,0) to[V,v_=V_s] (0,2) to[R=R_1] (2,2) to[R=R_2] (2,0) -- (0,0); (2,2) to[R=R_3] (4,2) node[right]A; (2,0) to[R=R_4] (4,0) node[right]B; (4,2) to[R=R_L] (4,0); (Imagine a complex circuit with sources and resistors connected to a load resistor R_L at terminals A and B.) Diagram 2: Thevenin Equivalent Circuit (0,0) to[V,v_=V_TH] (0,2) to[R=R_TH] (2,2) node[right]A; (0,0) -- (2,0) node[right]B; (2,2) to[R=R_L] (2,0); (Imagine the complex circuit replaced by a voltage source V_TH in series with a resistor R_TH, connected to the same load resistor R_L at terminals A and B.) 5b) For the network shown in Fig. Q3b, calculate the current in the 9 resistor using Thevenin's theorem. Assumption: The question refers to Fig. Q5b (as Q3b is already used) and the 9 resistor is the load resistor (R_L) connected between the top-right and bottom-right terminals of the circuit. Step 1: Remove the 9 load resistor (R_L). Let the terminals be A (top-right, after 14) and B (bottom-right, after 6). Step 2: Calculate Thevenin voltage (V_TH = V_AB). With terminals A and B open, no current flows through the 14 and 6 resistors. The circuit consists of the 15A current source in parallel with the 12 resistor, connected to the series combination of 3 and 8. The equivalent resistance of the series branch is R_series = 3 + 8 = 11. The 15A current splits between the 12 resistor and the 11 series branch. The current flowing through the 11 branch (I_11) is: I_11 = 15A × (12)/(12 + 11) = 15A × (12)/(23) = (180)/(23)A ≈ 7.826A The voltage across the 8 resistor is V_8 = I_11 × 8 = (180)/(23)A × 8 = (1440)/(23)V. The voltage at the node between 3 and 8 (let's call it V_X) relative to ground is V_X = V_8. The voltage at the node where the 15A source, 12, and 3 connect (let's call it V_Y) is V_Y = I_11 × (3 + 8) = (180)/(23)A × 11 = (1980)/(23)V. Since terminals A and B are open, V_A is the voltage at the node before the 14 resistor, and V_B is the voltage at the node before the 6 resistor. So, V_A = V_Y and V_B = V_X. V_TH = V_A - V_B = V_Y - V_X = (1980)/(23)V - (1440)/(23)V = (540)/(23)V ≈ 23.478V Step 3: Calculate Thevenin resistance (R_TH) by turning off independent sources. Replace the 15A current source with an open circuit. Looking into terminals A and B: The 12 resistor is in series with the 3 resistor. This combination is in parallel with the 8 resistor. R_series' = 12 + 3 = 15 R_parallel' = (15 × 8)/(15 + 8) = (120)/(23) This R_parallel' is in series with the 14 and 6 resistors. R_TH = R_parallel' + 14 + 6 = (120)/(23) + 20 = (120 + 460)/(23) = (580)/(23) ≈ 25 ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._

Use simple diagram to explain Thevenin's theorem. For the network shown in fig. Q3b below, calculate the current in the 92 resistor using Thevenin's theorem.

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Use simple diagram to explain Thevenin's theorem. For the network shown in fig. Q3b below, calculate the current in the 92 resistor using Thevenin's theorem.

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