Question

Two capacities of 4 μf and 12 μf are joined in series the total capacitance is

Answer 1

$1 \div cne t = 1 \div c1 + 1 \div c2$

$cnet = c1 \times c2 \div c1 + c2$

4×12=48

and, 4 +12 =16

now 48÷16 =3

Answer 2

The total capacitance of the combination is $3\ \mu F$.

Explanation:

It is given that,

Capacitor 1, $C_1=4\ \mu F$

Capacitor 2, $C_2=12\ \mu F$

It is mentioned that two capacitors are connected in series. We know that the total capacitance in series combination of capacitor is given by :

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$

$\dfrac{1}{C}=\dfrac{1}{4}+\dfrac{1}{12}$

$C=3\ \mu F$

So, the total capacitance of the combination is $3\ \mu F$. Hence, this is the require solution.

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