Answer:
Connect the test light in series with the negative post, and start pulling feed wires. The first to check is the heavy charging wire from the alternator. A bad or leaky diode in an alternator is a very common source of overnight battery drain. Connect wires one at a time to see what lead is drawing current.
Answer:
ω=314.15 rad/s.
0.02 s.
Explanation:
Given that
Motor speed ,N= 3000 revolutions per minute
N= 3000 RPM
The speed of the motor in rad/s given as

Now by putting the values in the above equation

ω=314.15 rad/s
Therefore the speed in rad/s will be 314.15 rad/s.
The speed in rev/sec given as

ω= 50 rev/s
It take 1 sec to cover 50 revolutions
That is why to cover 1 revolution it take

Answer:
a) 2,945 mC
b) P(t) = -720*e^(-4t) uW
c) -180 uJ
Explanation:
Given:
i (t) = 6*e^(-2*t)
v (t) = 10*di / dt
Find:
( a) Find the charge delivered to the device between t=0 and t=2 s.
( b) Calculate the power absorbed.
( c) Determine the energy absorbed in 3 s.
Solution:
- The amount of charge Q delivered can be determined by:
dQ = i(t) . dt

- Integrate and evaluate the on the interval:

- The power can be calculated by using v(t) and i(t) as follows:
v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt
v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV
P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)
P(t) = -720*e^(-4t) uW
- The amount of energy W absorbed can be evaluated using P(t) as follows:

- Integrate and evaluate the on the interval:
