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The voltage V in a simple circuit is related to the current I and the resistance R by Ohm’s Law: V=IR. An illustration of a simple square ci
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The voltage V in a simple circuit is related to the current I and the resistance R by Ohm’s Law: V=IR. An illustration of a simple square circuit. Parts of the surface are labeled V for voltage, I for current, and R for resistance, but values for these quantities are not indicated. Find the rate of change dI/dt of the current I if R=600 Ω, I=0.04 A, dR/dt=−0.5 Ω/s, and dV/dt=0.04 V/s.
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3 years
2021-07-20T14:32:02+00:00
2021-07-20T14:32:02+00:00 2 Answers
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Answer:
0.0001 A/s
Explanation:
Since V = IR,
dV/dt = d(IR)/dt = IdR/dt + RdI/dt using product rule
dV/dt = IdR/dt + RdI/dt
dI/dt = (dV/dt – IdR/dt)/R which is the rate of change of current
dV/dt = 0.04 V/s, dR/dt = -0.5 Ω/s, I = 0.04 A and R = 600 Ω
Substituting these values into dI/dt, we have
dI/dt = [0.04 V/s – (0.04 A × -0.5 Ω/s)]/600 Ω
= (0.04 V/s + 0.02 V/s)/600 Ω
= 0.06 V/s/600 Ω
= 0.0001 A/s
Answer:
dI/dt = 0.0004 A/s
Explanation:
R = 600 ohms
I = 0.04 A
dR/dt = -0.5 ohms/s
dV/dt = 0.04 V/s
From Ohm’s law V = IR
Taking the derivative of both sides with respect to t using product rule
dV/dt = I dR/dt + R dI/dt
0.04 = -( 0.04*0.5) + (600) dI/dt
0.04 + 0.2 = 600 dI/dt
0.24 = 600 dI/dt
dI/dt = 0.24/600
dI/dt = 0.0004 A/s