An initially uncharged 3.67 μF capacitor and a 8.01 k Ω resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes?

The initial current in the circuit is 0.1873 mA

Explanation:

Given;

capacitance is given as 3.67 x 10⁻⁶ F

resistance is given as 8010 Ω

voltage across the circuit is 1.5 V

Since the capacitor is initially uncharged, the capacitive reactance is zero.

From ohms law;

Voltage across the circuit is directly proportional to the opposition to the flow of current.

V = IR

I = V/R

I = 1.5/8010

I = 0.0001873 A = 0.1873 mA

Therefore, the initial current in the circuit is 0.1873 mA

answer: the answer is a

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: