Question

The equivalent capacitance of the combination of the capacitors is–

• A. (a) 3.20 mF
• A. (b) 7.80 mF
• A. (c) 3.90 mF
• A. (d) 2.16 mF

Answer 1

(a)

C_eq = $\frac { 16 \times 4 } { 16 + 4 }$ = 3.20 mF