Question

The equivalent capacitance of two capacitors in series is 1.5microfarad.The capacitance of one of them is 4microfarad.The value of the other capacitance is

Answer 1

Answer:

For series. ...... c' = c^1 ×c^2 / c^1 +c^2

but c^1=4microfarad

c'=1.5microfarad.

1.5=4×c^2. /4+ c^2

c^2 =2.4microfarad.

Explanation:

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Answer 2

Given:

• The equivalent capacitance of two capacitors in series is 1.5 microfarad.
• The capacitance of one of them is 4 microfarad.

To find:

• Value of the other capacitor?

Calculation:

The general expression for capacitors in series combination is :

$\dfrac{1}{C_{eq.}} = \dfrac{1}{C1}+ \dfrac{1}{C2}$

Now, C1 = 4 micro-farad and C_(eq) = 1.5 micro-farad.

So, putting the necessary values:

$\implies\dfrac{1}{1.5} = \dfrac{1}{4}+ \dfrac{1}{C2}$

$\implies\dfrac{2}{3} = \dfrac{1}{4}+ \dfrac{1}{C2}$

$\implies \dfrac{1}{C2} = \dfrac{2}{3}-\dfrac{1}{4}$

$\implies \dfrac{1}{C2} = \dfrac{8-3}{12}$

$\implies \dfrac{1}{C2} = \dfrac{5}{12}$

$\implies C_{eq}=2.4 \:\mu F$

So, the value of the other capacitor is 2.4 micro-farad.