The two condensers of capacitance's $4 μ F$ and $6 μ F$ are connected in parallel. This combination is now connected in series to a third condenser. If the resultant capacitance becomes $3 \frac{1}{3}$ up, the capacitance of the third condenser is:

2 μ F

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3 μ F

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5 μ F

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4 μ F

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Solution

The correct option is C $5 μ F$
The two condensers of capacitance's $= 4 μ F$ and $6 μ F$
When they are in parallel $= (4 + 6) = 10 μ F$
Now, let the third condenser $x μ F$
As per question
$\frac{10 \times x}{10 + x} = 3 \frac{1}{3} \frac{10 x}{10 + x} = \frac{10}{3} 30 x = 100 + 10 x 20 x = 100 x = 5 μ F$

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The two condensers of capacitance's 4μF and 6μF are connected in parallel. This combination is now connected in series to a third condenser. If the resultant capacitance becomes 313 up, the capacitance of the third condenser is:

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The two condensers of capacitance's $4 μ F$ and $6 μ F$ are connected in parallel. This combination is now connected in series to a third condenser. If the resultant capacitance becomes $3 \frac{1}{3}$ up, the capacitance of the third condenser is: