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(B) The circuit consists of three resistors $R_1, R_2, R_3$ connected in series across a $12\,V$ source.Let the current flowing through the circuit be

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$1 : 1 : 1$

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(B) The circuit consists of three resistors $R_1, R_2, R_3$ connected in series across a $12\,V$ source.Let the current flowing through the circuit be $I$.The voltage drops across the resistors are proportional to their resistances since $V = IR$.The potential at the bottom is $0\,V$.The potential at the junction of $R_1$ and $R_2$ is $4\,V$. Thus,the voltage drop across $R_1$ is $V_1 = 4\,V - 0\,V = 4\,V$.The potential at the junction of $R_2$ and $R_3$ is $8\,V$. Thus,the voltage drop across $R_2$ is $V_2 = 8\,V - 4\,V = 4\,V$.The potential at the top of $R_3$ is $12\,V$. Thus,the voltage drop across $R_3$ is $V_3 = 12\,V - 8\,V = 4\,V$.Since the current $I$ is the same for all resistors in series,and the voltage drops are equal $(V_1 = V_2 = V_3 = 4\,V)$,the resistances must be equal.Therefore,$R_1 = R_2 = R_3$,which gives 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?

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