C. Forces have mass and take up space
We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:

where

is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance
In our problem, I=4.00 A and

, so the potential difference is
Answer:
at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all
Explanation:
we know that for series RLC circuit impedance is given by

but we know that at resonance
putting
in impedance formula , impedance will become
Z=R so at resonance impedance of series RLC is equal to resistance only
now quality factor of series resonance is given by
so from given expression it is clear that quality factor depends on R L and C
For the answer to the question above
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Very (VERY) basic CRM for small businesses
I hope my answer helped you.
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s