All the elements in the circuits are ideal. The power delivered by the 10 V source in watts is

The correct option is 0.

Step-by-Step Explanation:

To find the power delivered by the $10 V$ voltage source, we can apply Kirchhoff's Current Law (KCL) at the central node of the circuit.

Let the voltage at the central node (relative to the reference/ground node at the bottom) be denoted as $V_n$.

Based on the provided circuit diagram:
1. The left branch contains a $10 V$ ideal voltage source in series with a resistor of value $\alpha \Omega$.
2. The middle branch consists of a $1 \Omega$ resistor connected to ground.
3. The right branch contains a $10 A$ current source pointing upwards in series with a $2 \Omega$ resistor, which injects a constant current of $10 A$ into the central node.

Applying Kirchhoff's Current Law (KCL) at the central node (setting the sum of currents leaving the node equal to zero):

$\frac{V_n-10}{\alpha }+\frac{V_n}{1}-10=0$

We can rewrite the equation by grouping the terms containing $(V_n-10)$:

$\frac{V_n-10}{\alpha }+(V_n-10)=0$

Factoring out $(V_n-10)$ from the terms:

Since the resistance value $\alpha$ is positive ($\alpha >0$), the term $[\frac{1}{\alpha }+1]$ is strictly positive and non-zero. Therefore, we must have:

$V_n-10=0\Rightarrow V_n=10 V$

Now, let's find the current $I_s$ leaving the positive terminal of the $10 V$ voltage source:

$I_s=\frac{10-V_n}{\alpha }$

Substituting $V_n=10 V$ into the equation:

$I_s=\frac{10-10}{\alpha }=0 A$

The power $P$ delivered by the $10 V$ source is:

$P=V\times I_s=10 V\times 0 A=0 W$