Ohm's law relates the volate V across a resistor, the electrical current I passing through the resistor, and the resistance R. In particular: V=IR.

Question

Ohm's law relates the volate V across a resistor, the electrical current I passing through the resistor, and the resistance R. In particular: V=IR. Currently, the resistance is constant, 4 ohms. The current is increasing at a rate of 0.3 amps/sec. How quickly is the voltage changing?

Answer 1

V = IR

V' = I'R + IR'

R' = 0

V' = I'R

I' = 0.3A/s, R = 4Ω

The derivative of the voltage is 0.3A/s * 4Ω = 1.2 V/s (volts / second)

Answer 2

1. Ohm's Law Formula

Ohm's Law states that:

V = I * R

where:

- V is the voltage across the resistor (in volts, V),

- I is the current through the resistor (in amperes, A),

- R is the resistance of the resistor (in ohms, Ω).

2. Given Information

- Resistance, R = 4 Ω (constant)

- Rate of change of current, dI/dt = 0.3 A/s

- We need to find the rate of change of voltage, dV/dt.

3. Calculating Rate of Change of Voltage

To determine how quickly the voltage is changing, we need to differentiate the voltage with respect to time. Using the formula V = I * R, we take the derivative of both sides with respect to time:

dV/dt = R * (dI/dt)

Since the resistance R is constant, it comes out of the derivative.

4. Calculation

Substituting the given values:

dV/dt = 4 Ω * 0.3 A/s

dV/dt = 1.2 V/s

Therefore, the voltage is increasing at a rate of 1.2 volts per second.

5. Interpretation

The positive rate of change of voltage (1.2 V/s) indicates that the voltage is increasing at this rate as the current increases. This is consistent with Ohm's Law, as a higher current through a constant resistance results in a proportional increase in voltage.

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