Answer:
Vdc=10V
Explanation:
in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,
if the loop remains closed for a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.
Answer:
Current = 0.063 Amperes
Explanation:
Let the three resistors be R1, R2, and R3 respectively.
Given the following data;
R1 = 25.0Ω,
R2 = 30.0Ω
R3 = 40.0Ω
Voltage = 6 Volts
First of all, we would determine the equivalent or total resistance;
Total resistance (in series) = R1 + R2 + R3
Total resistance = 25.0Ω + 30.0Ω + 40.0Ω
Total resistance = 95 Ω
Next, we find the current flowing through the circuit;
Voltage = current * resistance
Substituting into the formula, we have;
6 = current * 95
Current = 6/95
Current = 0.063 Amperes
The potential energy will be 1.46*10^-4J.
To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.
<h3>How to find the potential energy of the loop?</h3>
- We have the expression for torque acting on a current loop in a uniform magnetic field as,
where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.
- As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
- Thus, the potential energy will be,
Thus, we can conclude that, the potential energy will be 1.46*10^-4J.
Learn more about the torque here:
brainly.com/question/27949876
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<span>The
answer is Local Area Network (LAN). Examples of LANs, are those in workplaces,
libraries, and universities. The nodes are
connected and share common resources over
a WiFi, Ethernet network, Token rings, and
etcetera. The opposite of LAN is a Wide Area Network (WAN) that covers a
wide geographical area.</span>
