Using thevenin s theorem through direct formula clearly step by step please

To find $R_{Th}$, we first remove the load resistor (the 10 $\Omega$ resistor between terminals A and B). Then, we turn off all independent voltage sources (replace them with short circuits) and independent current sources (replace them with open circuits).

• Replace the 6V voltage source with a short circuit. This connects node C directly to node D.
• Replace the 9V voltage source with a short circuit. This connects node F directly to node D.

Now, nodes C, D, and F are all electrically connected to the same point (let's call it ground for simplicity).
Looking into terminals A and B:

• The 4 $\Omega$ resistor is connected between terminal A and node C (which is now ground).
• The 6 $\Omega$ resistor is connected between terminal B and node D (which is now ground).
• The 5 $\Omega$ resistor is connected between node C and node F. Since C and F are both connected to D (ground), the 5 $\Omega$ resistor is connected between two points that are at the same potential. This means the 5 $\Omega$ resistor is shorted out and does not contribute to $R_{Th}$ between A and B.

The effective path between A and B is through the 4 $\Omega$ resistor to ground (C/D/F) and then from ground (C/D/F) through the 6 $\Omega$ resistor to B. Therefore, the 4 $\Omega$ and 6 $\Omega$ resistors are in series when viewed from terminals A and B.

$$R_{Th} = R_{4\Omega} + R_{6\Omega}$$ $$R_{Th} = 4 \, \Omega + 6 \, \Omega$$ $$R_{Th} = 10 \, \Omega$$

To find $V_{Th}$, we determine the open-circuit voltage across the terminals where the load resistor was removed (terminals A and B). This is $V_{AB}$.

• With the 10 $\Omega$ load removed, no current flows through the 4 $\Omega$ resistor (connected to A) or the 6 $\Omega$ resistor (connected to B) because the circuit is open at A-B.
• Let's set node D as the reference node ($V_D = 0V$).
• The 6V source is connected between C and D, with C being positive. So, $V_C = 6V$.
• The 9V source is connected between F and D, with F being positive. So, $V_F = 9V$.
• The current flowing through the 5 $\Omega$ resistor (between C and F) is $I_{CF}$: $$I_{CF} = \frac{V_C - V_F}{5 \, \Omega} = \frac{6V - 9V}{5 \, \Omega} = \frac{-3V}{5 \, \Omega} = -0.6A$$ This means $0.6A$ flows from F to C.
• Since no current flows through the 4 $\Omega$ resistor, there is no voltage drop across it. Thus, the voltage at terminal A is the same as the voltage at node C: $$V_A = V_C = 6V$$
• Since no current flows through the 6 $\Omega$ resistor, there is no voltage drop across it. Thus, the voltage at terminal B is the same as the voltage at node D: $$V_B = V_D = 0V$$
• The Thevenin voltage $V_{Th}$ is the potential difference between A and B: $$V_{Th} = V_A - V_B = 6V - 0V$$ $$V_{Th} = 6V$$

The Thevenin equivalent circuit consists of $V_{Th}$ in series with $R_{Th}$. Now, reconnect the 10 $\Omega$ load resistor ($R_L$) to this equivalent circuit.
The total resistance in the series circuit is $R_{total} = R_{Th} + R_L$.

$$R_{total} = 10 \, \Omega + 10 \, \Omega = 20 \, \Omega$$

The current flowing through the load resistor is given by Ohm's Law:

$$I_L = \frac{V_{Th}}{R_{total}}$$ $$I_L = \frac{6V}{20 \, \Omega}$$ $$I_L = 0.3A$$

The current across the 10 $\Omega$ resistance is $\boxed{\text{0.3 A}}$.

Using Ohm's Law for the load resistor:

$$V_L = I_L \times R_L$$ $$V_L = 0.3A \times 10 \, \Omega$$ $$V_L = 3V$$

The p.d across the 10 $\Omega$ resistance is $\boxed{\text{3 V}}$.

Using the formula for power:

$$P_L = I_L^2 \times R_L$$ $$P_L = (0.3A)^2 \times 10 \, \Omega$$ $$P_L = 0.09 A^2 \times 10 \, \Omega$$ $$P_L = 0.9W$$

The power consumed by the 10 $\Omega$ resistance is $\boxed{\text{0.9 W}}$.