Question

equivalent capacitance between A and B ​

Answer 1

Answer:

Option b is correct i.e 2.5μf.

Explanation:

Here ,

The combination forms a Wheatstone bridge circuit .

In which centre branch can be neglected .

Formula used :

For capacitance in series combination :

$\star \boxed{ \red{ \bold{ \frac{1}{c_{eq}} = \frac{1}{c_1} + \frac{1}{c_2} + \frac{1}{c_3} ...}}}$

For capacitance in parallel combination:

$\star \boxed{ \red{ \bold{ c_{eq} = c_1 + c_2 + c_3....}}}$

Answer 2

2.5 × 10^-6 farad. is the required answer.

Option b is right.

GIVEN:

• C1 = 5 micro farad = 5 × 10^-6 F
• C2 = 5 micro farad = 5 × 10^-6 F
• C3 = 5 micro farad = 5 × 10^-6 F
• C4 = 5 micro farad = 5 × 10^-6 F
• C5 = 5 micro farad = 5 × 10^-6 F
• C6 = 5 micro farad = 5 × 10^-6 F

TO FIND:

Equivalent capacitance between A and B

FORMULA:

• When capacitance are connected in series, net capacitance , 1/C = 1/C1 + 1/C2 + 1/C3 + ...

• When capacitance are connected in parallel, then the net capacitance, C = C1 + C2 + C3 + .....

SOLUTION:

STEP 1:

Consider the circuit C1 ,C2, C3, C4, C5.

It is similar to wheatstone bridge.

Condition for wheatstone bridge:

• In wheatstone bridge , current through the capacitance ,C3 is zero.

• C2 × C4 = C1 × C5

$C_2 \times C_4 = 5 \times {10}^{ - 6} \times 5 \times {10}^{ - 6} \\ \\ C_2 \times C_4 = 25 \times {10}^{ - 12}$

$C_1 \times C_5 = 5 \times {10}^{ - 6} \times 5 \times {10}^{ - 6} \\ \\ C_1 \times C_5 = 25 \times {10}^{ - 12}$

Hence it is a wheatstone bridge.

So capacitance, C3 can be neglected.

STEP 2:

C2 and C5 are in series. Le their effective capacitance be Cb.

$\frac{1}{C_b} = \frac{1}{C_2} + \frac{1}{C_5} \\ \\ \frac{1}{C_b} = \frac{1}{(5 \times {10}^{ - 6} )} + \frac{1}{(5 \times {10}^{ - 6} )} \\ \\ \frac{1}{C_b} = \frac{2}{5 \times {10}^{ - 6} } \\ \\ C_b = \frac{5 \times {10}^{ - 6} }{2} \\ \\ C_b = 2.5 \times {10}^{ - 6} \: farad$

STEP 3:

C1 and C4 are in series. Le their effective capacitance be Ca.

$\frac{1}{C_a} = \frac{1}{C_1} + \frac{1}{C_4} \\ \\ \frac{1}{C_a} = \frac{1}{(5 \times {10}^{ - 6} )} + \frac{1}{(5 \times {10}^{ - 6} )} \\ \\ \frac{1}{C_a} = \frac{2}{5 \times {10}^{ - 6} } \\ \\ C_a = \frac{5 \times {10}^{ - 6} }{2} \\ \\ C_a = 2.5 \times {10}^{ - 6} \: farad$

STEP 4:

Ca and Cb are in parallel. Le their effective capacitance be Cx.

$C_x = C_a + C_b \\ \\ C_x = (2.5 \times {10}^{ - 6} ) + (2.5 \times {10}^{ - 6} ) \\ \\ cx = (2.5 + 2.5) \times {10}^{ - 6} \\ \\ C_x = 5 \times {10}^{ - 6} farad$

STEP 5 :

Cx and C6 are in series. Le their effective capacitance be C.

$\frac{1}{C} = \frac{1}{C_x} + \frac{1}{C_6} \\ \\ \frac{1}{C} = \frac{1}{(5 \times {10}^{ - 6} )} + \frac{1}{(5 \times {10}^{ - 6} )} \\ \\ \frac{1}{C} = \frac{2}{5 \times {10}^{ - 6} } \\ \\ C = \frac{5 \times {10}^{ - 6} }{2} \\ \\ C = 2.5 \times {10}^{ - 6} \: farad$

ANSWER:

The effective capacitance between A and B is 2.5 × 10^-6 farad.