Slove in English

In the given circuit, it is a Wheatstone bridge. It is stated that no current flows through the $4\Omega$ resistor (between B and D). This means the bridge is balanced.
The condition for a balanced Wheatstone bridge is:

$$\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}}$$

From the diagram, the resistances are:
$R_{AB} = 1\Omega$
$R_{BC} = R$
$R_{AD} = 1\Omega$
$R_{DC} = 1\Omega$

Substituting these values into the balance condition:

$$\frac{1\Omega}{R} = \frac{1\Omega}{1\Omega}$$ $$1 = \frac{R}{1\Omega}$$ $$R = \boxed{\text{1}\Omega}$$

Since the bridge is balanced, no current flows through the $4\Omega$ resistor. Therefore, we can ignore this resistor for current calculations in the main branches.
The total potential difference across A and C is $V_{AC} = 2V$.

The resistances in the lower branch (ADC), $R_{AD}$ and $R_{DC}$, are in series.
$R_{AD} = 1\Omega$
$R_{DC} = 1\Omega$
The total resistance of the lower branch is $R_{lower} = R_{AD} + R_{DC} = 1\Omega + 1\Omega = 2\Omega$.

The current flowing through the lower branch (ADC), $I_{ADC}$, is:

$$I_{ADC} = \frac{V_{AC}}{R_{lower}} = \frac{2V}{2\Omega} = 1A$$

The potential difference across A and D ($V_{AD}$) can be found using Ohm's law:

$$V_{AD} = I_{ADC} \times R_{AD}$$ $$V_{AD} = 1A \times 1\Omega$$ $$V_{AD} = \boxed{\text{1V}}$$