Given that the three resistors $R_1 = 10 \ \Omega$, $R_2 = 50 \ \Omega$, and $R_3 = 190 \ \Omega$ are connected in series, and a potential difference of $V = 250 \ \text{V}$ is applied across the combination. The equivalent resistance $R_s$ of the series circuit is: $$ \begin{aligned} R_s &= R_1 + R_2 + R_3 \\ &= (10 + 50 + 190) \ \Omega \\ &= 250 \ \Omega \end{aligned} $$ The total current $I$ flowing through the circuit is: $$ \begin{aligned} I &= \frac{V}{R_s} \\ &= \frac{250 \ \text{V}}{250 \ \Omega} \\ &= 1 \ \text{A} \end{aligned} $$ Since the resistors are in series, the same current $I$ flows through each of them. The potential difference across each resistor is calculated as follows: For the first resistor $R_1$: $$ \begin{aligned} V_1 &= I \times R_1 \\ &= 1 \ \text{A} \times 10 \ \Omega \\ &= 10 \ \text{V} \end{aligned} $$ For the second resistor $R_2$: $$ \begin{aligned} V_2 &= I \times R_2 \\ &= 1 \ \text{A} \times 50 \ \Omega \\ &= 50 \ \text{V} \end{aligned} $$ For the third resistor $R_3$: $$ \begin{aligned} V_3 &= I \times R_3 \\ &= 1 \ \text{A} \times 190 \ \Omega \\ &= 190 \ \text{V} \end{aligned} $$ Therefore, the potential differences across the three resistors are $10 \ \text{V}$, $50 \ \text{V}$, and $190 \ \text{V}$ respectively.