Three sub-parts to this question. The potential, P, in a circuit, measured in vo

Question

Three sub-parts to this question.

The potential, P, in a circuit, measured in volts, is a function of time, t in seconds. The rate of change of the potential in the circuit on the time interval [0, 4] is graphed below. For each group of four statements below, write down the letter of all true statement(s). The potential is decreasing between 0 seconds and 1 second. The potential is increasing between 0 seconds and 1 second. The potential is constant between 0 seconds and 1 second. There is not enough information to determine whether the potential is increasing or decreasing between 0 seconds and 1 second. The potential reaches its highest point between 2 seconds and 3 seconds. The potential reaches its lowest point between 2 seconds and 3 seconds. The potential reaches its highest point at 4 seconds. The rate of change of potential reaches its highest point at 4 seconds. The potential in the circuit is lower at 2 seconds than it is at 0.25 seconds. The potential in the circuit is higher at 2 seconds than it is at 0.25 seconds. The potential in the circuit is the same at 2 seconds as it is at 0.25 seconds. There is not enough information to compare the potential at 2 seconds to the potential at 0.25 seconds.

Explanation / Answer

1)
Between 0 to 1 dp/dt is negative , potential function will decreasing during this interval
Choice (a) is correct.
2)
Since derivative (dp/dt) is changing sign from -ve to +ve between 2 to 3, hence potential function value will be minimum at this point
choice (b) is correct answer.
3)
Last choice is correct answer.
From the given information we can say about local minimum value , we cannot compare them together.

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