Question

Kindly solve both the questions.

Thanks.​

Answer 1

1st Question:

Given:

3 capacitors are connected at a junction as provided in the figure.

To find:

Potential at point O (Junction)

Concept:

We shall apply Extension of Kirchoff's Current Law , which applies for charge conservation ; As per the law Charge will not be accumulated at the junction.

Calculation:

Let potential at point O be x :

$\therefore \: \sf{incoming \: charge = outgoing \: charge}$

$= > \: (10 - x)1 + (4 - x)2 = (x - 0)7$

$= > 10 - x + 8 - 2x = 7x$

$= > 18 = 10x$

$= > x = 1.8 \: volt$

So final answer is 1.8 volt.

2nd question:

Given:

A circuit consisting of capacitors have been provided.

To find:

Equivalent capacitance of the circuit.

Concept:

For difficult circuits , we need to assign potential drops and then simplify the circuit.

Calculation:

First refer to the diagram to understand the simplified circuit.

Net capacitance in upper segment will be :

$\therefore \: C1 = \dfrac{2C \times 3C}{2C + 3C }$

$= > \: C1 = \dfrac{6 {C}^{2} }{5C }$

$= > \: C1 = \dfrac{6 C }{5 }$

Net capacitance of lower segment is C .

So equivalent capacitance of whole circuit will be :

$C \: eq. = C + \dfrac{6C}{5}$

$= > C \: eq. = \dfrac{11C}{5}$

So final answer is (11C)/5.

Answer 2

1).

FORMULA USED:

• Q = CV

SOLUTION:

=> Q + Q' + Q'' = 0

=> 1(V-10) + 2(V-4) + 7(V) = 0

=> V - 10 + 2V - 8 + 7V = 0

Collecting like terms...

=> v + 2v + 7V - 10 - 8

=> 10V - 18 = 0

=> 10V = 18

=> V = 18/10

=> V = 1.8 volt

option (4) is correct ✔

2).

Here, the equivalent capacitance between point A and B = total equivalent capacitance in the whole circuit

=> C eq.= (2c × 3c)/(2c + 3c) + c

=> C eq.= 6c²/5c + c

cancelling C from numerator and denominator

=>C eq. = 6c/5 + c

=> C eq. = 11c/5

option 4 is correct ✔