The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
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Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.
Answer:
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Answer:a. Magnetic dipole moment is 0.3412Am²
b. Torque is zero(0)N.m
Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A
That is,
U = n*I*A
But Area A is given as pi*radius² since it is a circular coil
Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is
A = 3.142*(0.05)² =7.86*EXP {-3} m²
Current I is 2 A
Number of turns is 20
So magnetic dipole moment U is
U = 20*2*7.86*EXP {-3}=0.3142A.m²
b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U
Torque = B x U =B*U*Sine(theta)
But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m
See the attached picture for answers
Answer:
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