Given the circuit in Fig. 4.75, calculate ix and the power dissipated by the 10-Ω resistor using superposition.
The Correct Answer and Explanation is :
To solve this using superposition, we need to consider each independent source (the current source and the voltage source) separately while turning off the other sources. In this case, “turning off” a voltage source means replacing it with a short circuit, and “turning off” a current source means replacing it with an open circuit.
Step 1: Consider the Voltage Source Only
- Turn off the current source: This means replacing the 2 A current source with an open circuit.
- Analyze the circuit: Now the circuit only has the 24 V voltage source with three resistors. The two 5-Ω resistors are in series, and this combination is in parallel with the 10-Ω resistor. The equivalent resistance ( R_{eq} ) of the two 5-Ω resistors in series is ( 5 \Omega + 5 \Omega = 10 \Omega ). The total resistance seen by the 24 V source is:
[
R_{total} = \frac{10 \Omega \times 10 \Omega}{10 \Omega + 10 \Omega} = 5 \Omega
] The total current from the voltage source ( I ) is:
[
I = \frac{24 V}{5 \Omega} = 4.8 A
] The voltage across each parallel branch (10-Ω and the combined 10-Ω from series resistors) is 24 V. The current through the 10-Ω resistor ( I_{10\Omega} ) is:
[
I_{10\Omega} = \frac{24 V}{10 \Omega} = 2.4 A
]
Step 2: Consider the Current Source Only
- Turn off the voltage source: This means short-circuiting the 24 V source.
- Analyze the circuit: Now, the 2 A current source feeds directly into the parallel combination of the three resistors. Since the path through the shorted voltage source does not impede current flow, all the 2 A will flow through the short, contributing nothing through the 10-Ω resistor. The current through the 10-Ω resistor ( I_{10\Omega} ) is 0 A in this scenario.
Step 3: Superposition
Add the currents from each scenario at the 10-Ω resistor:
[
I_{x} = 2.4 A + 0 A = 2.4 A
]
Power Dissipated by the 10-Ω Resistor
Using the current from superposition:
[
P = I_{x}^2 \times R = (2.4 A)^2 \times 10 \Omega = 57.6 W
]
Thus, the current ( I_x ) through the 10-Ω resistor is 2.4 A, and the power dissipated by it is 57.6 W. This approach of superposition allows us to isolate the impact of each source, simplifying complex circuits into more manageable problems.