Question

In the following circuit, the equivalent capacitance between terminal A and terminal B is:

पर्याय

• 0.5 μF

• 4 μF

• 2 μF

• 1 μF

Answer 1

2 μF

Explanation:

As the bridge is balanced, the capacitor 3 can be neglected.

Hence,

C₁ and C₂ are in series:

`C_(s_1)` = 1µF

Again, C₄ and C₅ are in series:

`C_(s_2)` = 1µF

Now `C_(s_1)` and `C_(s_2)` are in parallel. So,

C = `C_(s_1) + C_(s_1)`

= 1 + 1

= 2µF