In the circuit shown in figure, potential of points A and B are respectively

Question

In the circuit shown in figure, potential of points A and B are respectively (A) -8V,2V (B) (10/3)V, (20/3)V  (C) -6V ,4V (D) - (10/3)V, (20/3)V

Tardigrade Admin · Accepted Answer

Effective capacitance = (2C/5) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f2d6cdb236b4564a711358c02affd247-.png" /> Charge on system = (20C/5) = 4C According to K.V.L. VA + (3q/2C) = 0 VA = - (3× q/2C)× 4C = -6V and VB= (q/C)=0⇒ VB=4

Q. In the circuit shown in figure, potential of points A and B are respectively

$- 8 V, 2 V$

$\frac{10}{3} V, \frac{20}{3} V$

$- 6 V, 4 V$

$- \frac{10}{3} V, \frac{20}{3} V$

Effective capacitance = $\frac{2 C}{5}$

Charge on system = $\frac{20 C}{5} = 4 C$ According to $K . V . L .$ $V_{A} + \frac{3 q}{2 C} = 0$
$V_{A} = - \frac{3 \times q}{2 C} \times 4 C = - 6 V$
$an d V_{B} = \frac{q}{C} = 0 \Rightarrow V_{B} = 4$

Q. In the circuit shown in figure, potential of points A and B are respectively

−8V,2V

310​V,320​V

−6V,4V

−310​V,320​V

Effective capacitance = 52C​ Charge on system = 520C​=4C According to K.V.L. VA​+2C3q​=0 VA​=−2C3×q​×4C=−6V andVB​=Cq​=0⇒VB​=4

$- 8 V, 2 V$

$\frac{10}{3} V, \frac{20}{3} V$

$- 6 V, 4 V$

$- \frac{10}{3} V, \frac{20}{3} V$

Effective capacitance = $\frac{2 C}{5}$

Charge on system = $\frac{20 C}{5} = 4 C$ According to $K . V . L .$ $V_{A} + \frac{3 q}{2 C} = 0$
$V_{A} = - \frac{3 \times q}{2 C} \times 4 C = - 6 V$
$an d V_{B} = \frac{q}{C} = 0 \Rightarrow V_{B} = 4$