The equivalent capacitance is 20 $ \mu $F by joining two capacitors of capacitances $ C_1 $ $ \mu $F and $ C_2 $ $ \mu $F in parallel. If they are joined in series, the equivalent capacitance is 4.8 $ \mu $F. Find the ratio of the capacitances $ C_1 $ and $ C_2 $ ( $ C_1C_2 $).

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The equivalent capacitance is 20 $ \mu $F by joining two capacitors of capacitances $ C_1 $ $ \mu $F and $ C_2 $ $ \mu $F in parallel. If they are joined in series, the equivalent capacitance is 4.8 $ \mu $F. Find the ratio of the capacitances $ C_1 $ and $ C_2 $ ( $ C_1>C_2 $).

cdquestions Admin · Accepted Answer

Using the formulas for parallel and series combinations: $$ C_1 + C_2 = 20 $$ $$ \frac{C_1 C_2}{C_1 + C_2} = 4.8 $$ Solving for the ratio: $$ C_1 = 16 \, \mu\text{F}, \quad C_2 = 4 \, \mu\text{F} $$ $$ \boxed{C_1 : C_2 = 4 : 1} $$

The equivalent capacitance is 20 \( \mu \)F by joining two capacitors of capacitances \( C_1 \) \( \mu \)F and \( C_2 \) \( \mu \)F in parallel. If they are joined in series, the equivalent capacitance is 4.8 \( \mu \)F. Find the ratio of the capacitances \( C_1 \) and \( C_2 \) ( \( C_1>C_2 \)).

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