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A complex circuit with two voltage sources and three resistive loads can be analyzed using Kirchhoff’s Voltage Law (KVL). KVL states that the sum of all voltages in a closed loop must be equal to zero. In this circuit, we can apply KVL to the outer loop to find the current flowing through the circuit.

Let’s assume that the current flows in the direction shown in the circuit diagram. Applying KVL to the outer loop, we get:

5V – i(1?) – i(9?) – 10V – i(10?) = 0
implifying this equation, we get:

5V – 10V – i(1?) – i(9?) – i(10?) = 0

-5V – i(20?) = 0

i = -0.25A

Therefore, the current flowing through the circuit is 0.25A in the opposite direction to the one shown in the circuit diagram.

Using Ohm’s Law, we can calculate the voltage across the 1? resistor:

Vab = i(1?) = -0.25V

The voltage at node B is greater than the voltage at node A:

Vb = 5V + Vab = 5V – 0.25V = 4.75V

Va = 5V

Thus, the voltage across the 1? resistor is -0.25V, and the voltage at node B is 4.75V.

In conclusion, we can analyze a complex circuit with two voltage sources and three resistive loads using Kirchhoff’s Voltage Law. By applying KVL to the outer loop, we can find the current flowing through the circuit and the voltage across the resistive loads. This method is useful for analyzing circuits with multiple voltage sources and resistive loads.

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