In the given figure, the capacitors C₁, C₃, C₄, C₅ have a capacitance 4 F each. If the capacitor C₂ has a capacitance 10 F , then effective capacitance between A and B will be

Step-by-step solution

When a battery is applied across A and B , then the points b and c will be at the same potential ( C₁=C₄=C₃=C₅=4 F ) Therefore, no charge flows through C ₂ . As , C₁ and C₅ are in series. Their equivalent capacitance, C^ = C₁ C₅ C₁+C₅ = 4 4 4+4 =2 F Similarly, C₄ and C₃ are in series. Therefore, their equivalent capacitance C^ = C₃ C₄ C₃+C₄ = 4 4 4+4 =2 F Now, C^ and C^ are in parallel. Therefore, effective capacitance between A and B=C^ +C^ =2+2=4 F