In the circuit shown below C1=10 μ F , C2=C3=20 μ F, and C4=40 μ F. If the charge on C1 is 20 μ C then potential difference between X and Y is

Question

In the circuit shown below C1=10 μ F , C2=C3=20 μ F, and C4=40 μ F. If the charge on C1 is 20 μ C then potential difference between X and Y is (A) 2 V (B) 3 V (C) 6 V (D) 3.5 V

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b7599b912ddd3f81c3a438b6d5647a51-.png" /> q3=20 μ C q4=q4=40 μ C Δ V=(20/10)+(20/20)=3 V

Q. In the circuit shown below $C_{1} = 10 μ F, C_{2} = C_{3} = 20 μ F$ , and $C_{4} = 40 μ F$ . If the charge on $C_{1}$ is $20 μ C$ then potential difference between $X$ and $Y$ is

2 V

3 V

6 V

3.5 V

$q_{3} = 20 μ C$
$q_{4} = q_{4} = 40 μ C$
$Δ V = \frac{20}{10} + \frac{20}{20} = 3 V$

Q. In the circuit shown below C1​=10μF,C2​=C3​=20μF, and C4​=40μF. If the charge on C1​ is 20μC then potential difference between X and Y is

2 V

3 V

6 V

3.5 V

q3​=20μC q4​=q4​=40μC ΔV=1020​+2020​=3V

Q. In the circuit shown below $C_{1} = 10 μ F, C_{2} = C_{3} = 20 μ F$ , and $C_{4} = 40 μ F$ . If the charge on $C_{1}$ is $20 μ C$ then potential difference between $X$ and $Y$ is

$q_{3} = 20 μ C$
$q_{4} = q_{4} = 40 μ C$
$Δ V = \frac{20}{10} + \frac{20}{20} = 3 V$