Three capacitors each of 4 μ F are to be connected in such a way that the effective capacitance is 6 μ F .…

1. all in series
2. all in parallel
3. two in parallel and one in series
4. two in series and one in parallel

Solution

For top branch,

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{4}$

$\Rightarrow {\mathrm{C}}_{\mathrm{eq}}=2 \mathrm{\Omega }$

For parallel combination, ${\mathrm{C}}_{\mathrm{net}}=2 \mathrm{\Omega }+4 \mathrm{\Omega }=6 \mathrm{\Omega }$ .

Asked in: JEE Mains - Capacitance - Test 1

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