Question

Evaluate the time constant of the circuit: τ = RC = (7.1 105 Ω)(4.00 µF) = 1 s Evaluate the maximum charge on the capacitor: Q = C = (4.00 µF)(12.0 V) = 2 µC Evaluate the maximum current in the circuit:

Answer 1

Answer:

maximum current is 1.68 A

Explanation:

Maximum current is given as:

$I_{max}=\frac{\epsilon}{R}$

ε = 12 V

R=7.1105Ω

$I_{max}=\frac{12}{7.1105}\\\\I_{max}=1.68 A$