Two capacitors, each having capacitance 40 μ F are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became 24 μ F. The value of K will be :

Question

Two capacitors, each having capacitance 40 μ F are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became 24 μ F. The value of K will be : (A) 1.5 (B) 2.5 (C) 1.2 (D) 3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a4ab9481db2fa2f8e6650ac815817516-.png" /> C eq =( C ( KC )/ C + KC )=( KC / K +1) 24=( K 40/ K +1) [ K =1 ⋅ 5]

Q. Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24 μ F$ . The value of $K$ will be :

$1.5$

$2.5$

$1.2$

3

$C_{e q} = \frac{C ( K C )}{C + K C} = \frac{K C}{K + 1}$
$24 = \frac{K 40}{K + 1}$
$[K = 1 \cdot 5]$

Q. Two capacitors, each having capacitance 40μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became 24μF. The value of K will be :

1.5

2.5

1.2

3

Ceq​=C+KCC(KC)​=K+1KC​ 24=K+1K40​ [K=1⋅5]

Q. Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24 μ F$ . The value of $K$ will be :

$1.5$

$2.5$

$1.2$

$C_{e q} = \frac{C ( K C )}{C + K C} = \frac{K C}{K + 1}$
$24 = \frac{K 40}{K + 1}$
$[K = 1 \cdot 5]$