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(B) The resistors $R_1, R_2,$ and $R_3$ are connected in series across a $12\, V$ source.Since they are in series,the same current $I$ flows through e

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$1 : 1 : 1$

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(B) The resistors $R_1, R_2,$ and $R_3$ are connected in series across a $12\, V$ source.Since they are in series,the same current $I$ flows through each resistor.Using Ohm's law,the voltage drop across each resistor is proportional to its resistance.The voltage across $R_1$ is $V_1 = 4\, V - 0\, V = 4\, V$.The voltage across $R_2$ is $V_2 = 8\, V - 4\, V = 4\, V$.The voltage across $R_3$ is $V_3 = 12\, V - 8\, V = 4\, V$.Since the current $I$ is the same for all resistors,we have $I = \frac{V_1}{R_1} = \frac{V_2}{R_2} = \frac{V_3}{R_3}$.Substituting the voltage values: $\frac{4}{R_1} = \frac{4}{R_2} = \frac{4}{R_3}$.This implies $R_1 = R_2 = R_3$.Therefore,the ratio $R_1 : R_2 : R_3 = 1 : 1 : 1$.

$A$ potential divider is used to give outputs of $4\, V$ and $8\, V$ from a $12\, V$ source. Which combination of resistances $(R_1, R_2, R_3)$ gives the correct voltages? Find the ratio $R_1 : R_2 : R_3$.

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