A has less energy and lower frequency, while B has greater energy and higher frequency.
Niobium wire with a 2.60 mm diameter has a maximum current capacity of 500 A while still remaining superconducting.
<h3>Describe the present.</h3>
Current is the rate at which charge passes from one point on a circuit to another. In a circuit, a significant current flows when several coulombs or charge pass over the cross section of a wire. When the charge carriers are firmly packed inside the wire, high currents can be generated at low speeds.
<h3>What do current and electron actually mean?</h3>
Electron movement is referred to as electron current. The positive terminal receives electrons that are released by the negative terminal. Traditional current, usually referred to as just current, exhibits behavior consistent with positive charge carriers being the source of current flow. Regular current is received at the positive end and then flows to a negative terminal.
To know more about current visit:
brainly.com/question/15141911
#SPJ4
The moon has a small amount of gravity. Low tides mean the moon is not pulling on the water. High tides mean that the moon is pulling on the water.
<span>1) The differential equation that models the RC circuit is :
(d/dt)V_capacitor </span>+ (V_capacitor/RC) = (V_source/<span>RC)</span>
<span>Where the time constant of the circuit is defined by the product of R*C
Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s
2) C<span>harge of the capacitor 1.57 time constants
1.57*(2.742) = 4.3048 s
The solution of the differential equation is
</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T) + </span>V_capac(∞)
Since the capacitor is initially uncharged V_capac(0) = 0
And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V
This means,
V_capac (t) = (-9V)e ^(-t /T) + 9V
The charge in a capacitor is defined as Q = C*V
Where C is the capacitance and V is the Voltage across
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V
V_capac (4.3048 s) = 7.1275 V
Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C
3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V
Q (∞) = 89.9mF*(9V) = 0.8091 C
