In the circuit shown, find the current through $4\Omega$ resistor using Superposition theorem.
Show Hint
While Superposition is explicitly requested, Nodal Analysis serves as a fantastic verification tool! If you set the bottom wire to ground ($0\text{V}$) and write KCL at the top middle node $V_x$:
$$\frac{V_x - 20}{5} + \frac{V_x}{4} - 5 = 0$$
Multiplying by 20: $4(V_x - 20) + 5V_x - 100 = 0 \Rightarrow 9V_x = 180 \Rightarrow V_x = 20\text{V}$.
Then Current $= \frac{V_x}{4} = \frac{20}{4} = 5\text{ A}$. Extremely fast check!
Understanding the Concept:
The Superposition Theorem states that the total current flowing through any branch of a linear bilateral network containing multiple independent energy sources equals the algebraic sum of the individual currents produced by each source acting independently, with all other independent sources deactivated.
• To deactivate an independent voltage source, replace it with a short circuit ($0\text{V}$).
• To deactivate an independent current source, replace it with an open circuit ($0\text{A}$).
Let the total current passing down through the $4\Omega$ resistor ($R_3$) be denoted as $I_{R3} = I^\prime + I^{\prime\prime}$.
Case 1: Activating only the $20\text{V$ voltage source ($5\text{A}$ source is open-circuited).}
• Open circuit the $5\text{A}$ current source on the right. This leaves the $10\Omega$ resistor disconnected at one end, meaning zero current flows through it.
• The circuit simplifies to a simple series loop consisting of the $20\text{V}$ source, the $5\Omega$ resistor ($R_1$), and the $4\Omega$ resistor ($R_3$).
• Calculate the current $I^\prime$ using Ohm's Law:
$$I^\prime = \frac{V_1}{R_1 + R_3} = \frac{20}{5 + 4} = \frac{20}{9}\text{ A}$$
Case 2: Activating only the $5\text{A$ current source ($20\text{V}$ source is short-circuited).}
• Short circuit the $20\text{V}$ voltage source on the left side.
• The $5\Omega$ resistor ($R_1$) is now connected directly in parallel with the $4\Omega$ resistor ($R_3$). Let's find their parallel combination equivalent:
$$R_{13} = R_1 \parallel R_3 = \frac{5 \times 4}{5 + 4} = \frac{20}{9}\Omega$$
• This parallel group ($R_{13}$) is in series with the $10\Omega$ resistor ($R_2$). The total current injected into this parallel network by the independent current source is $5\text{A}$.
• Use the Current Division Rule to find the fraction of the $5\text{A}$ current that flows down through the $4\Omega$ branch ($I^{\prime\prime}$):
$$I^{\prime\prime} = I_1 \times \frac{R_1}{R_1 + R_3} = 5 \times \frac{5}{5 + 4} = 5 \times \frac{5}{9} = \frac{25}{9}\text{ A}$$
Step 3: Calculating Total Current via Superposition Algebraic Summation.
Both partial currents $I^\prime$ and $I^{\prime\prime}$ flow downwards through the $4\Omega$ resistor. Summing them yields:
$$I_{R3} = I^\prime + I^{\prime\prime} = \frac{20}{9} + \frac{25}{9} = \frac{20 + 25}{9} = \frac{45}{9} = 5\text{ A}$$
Thus, the total current through the $4\Omega$ resistor is exactly $5\text{ A}$, corresponding to Option (B).
