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 ? $R_1 : R_2 : R_3$
Medium
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(b) Resistors are connected in series. So current through each resistor will be same
$ \Rightarrow i = \frac{{12 - 8}}{{{R_3}}} = \frac{{8 - 4}}{{{R_2}}} = \frac{{4 - 0}}{{{R_1}}}$ $ \Rightarrow \frac{4}{{{R_3}}} = \frac{4}{{{R_2}}} = \frac{4}{{{R_1}}}$
So, ${R_1}:{R_2}:{R_3}::\;1:1:1$.
$ \Rightarrow i = \frac{{12 - 8}}{{{R_3}}} = \frac{{8 - 4}}{{{R_2}}} = \frac{{4 - 0}}{{{R_1}}}$ $ \Rightarrow \frac{4}{{{R_3}}} = \frac{4}{{{R_2}}} = \frac{4}{{{R_1}}}$
So, ${R_1}:{R_2}:{R_3}::\;1:1:1$.
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