Two identical parallel plate capacitors are connected in series and then joined in series with a battery of 100 V. A slab of dielectric constant K=3 is inserted between the plates of the first capacitor. Then, the potential difference across the capacitors will be, respectively,

Question

Two identical parallel plate capacitors are connected in series and then joined in series with a battery of 100 V. A slab of dielectric constant K=3 is inserted between the plates of the first capacitor. Then, the potential difference across the capacitors will be, respectively, (A) 25 V , 75 V (B) 75 V , 25 V (C) 20 V , 80 V (D) 50 V , 50 V

Tardigrade Admin · Accepted Answer

V1=(C × 100/3 C+C)=25 V , V2=100-25=75 V

Q. Two identical parallel plate capacitors are connected in series and then joined in series with a battery of $100 V$ . A slab of dielectric constant $K = 3$ is inserted between the plates of the first capacitor. Then, the potential difference across the capacitors will be, respectively,

$25 V, 75 V$

$75 V, 25 V$

$20 V, 80 V$

$50 V, 50 V$

$V_{1} = \frac{C \times 100}{3 C + C} = 25 V$ ,
$V_{2} = 100 - 25 = 75 V$

Q. Two identical parallel plate capacitors are connected in series and then joined in series with a battery of 100V. A slab of dielectric constant K=3 is inserted between the plates of the first capacitor. Then, the potential difference across the capacitors will be, respectively,

25V,75V

75V,25V

20V,80V

50V,50V