Use the equation dV/dt = (dV/dI) (dI/dt) + (dV/dR) (dR/dt) to find how the current is changing at the instant when R = 600 ohms, I = 0.04 amp, dR/dt = 0.5 ohm/sec, and dV/dt = -0.0... Use the equation dV/dt = (dV/dI) (dI/dt) + (dV/dR) (dR/dt) to find how the current is changing at the instant when R = 600 ohms, I = 0.04 amp, dR/dt = 0.5 ohm/sec, and dV/dt = -0.01 volt/sec.
Understand the Problem
The question is asking for the calculation of how the current (I) is changing in an electrical circuit at a specific instant, given the variable conditions of voltage (V) and resistance (R). It requires the use of the provided differential equation along with the specified values to compute the rate of change of current.
Answer
The rate of change of current is \( \frac{dI}{dt} = -0.00005 \) amp/sec.
Answer for screen readers
The rate of change of current is ( \frac{dI}{dt} = -0.00005 ) amp/sec.
Steps to Solve
- Identify the given values We are given the following values to plug into the equation:
- Resistance, ( R = 600 ) ohms
- Current, ( I = 0.04 ) amp
- Rate of change of resistance, ( \frac{dR}{dt} = 0.5 ) ohm/sec
- Rate of change of voltage, ( \frac{dV}{dt} = -0.01 ) volt/sec
- Determine derivatives of voltage with respect to current and resistance Using the relationship ( V = IR ), we can derive the partial derivatives:
- The derivative of voltage with respect to current:
$$ \frac{dV}{dI} = R $$
- The derivative of voltage with respect to resistance:
$$ \frac{dV}{dR} = I $$
At the given instance, we have:
$$ \frac{dV}{dI} = 600 \text{ ohms} $$ $$ \frac{dV}{dR} = 0.04 \text{ amp} $$
- Plug values into the differential equation Now substitute all known values into the given differential equation:
$$ \frac{dV}{dt} = \left( \frac{dV}{dI} \right) \left( \frac{dI}{dt} \right) + \left( \frac{dV}{dR} \right) \left( \frac{dR}{dt} \right) $$
Substituting:
$$ -0.01 = (600) \left( \frac{dI}{dt} \right) + (0.04)(0.5) $$
- Calculate the contribution from ( \frac{dR}{dt} ) Calculate ( (0.04)(0.5) ):
$$ 0.04 \times 0.5 = 0.02 $$
- Rearranging to solve for ( \frac{dI}{dt} ) Now substitute back into the equation:
$$ -0.01 = 600 \left( \frac{dI}{dt} \right) + 0.02 $$
Rearranging gives:
$$ 600 \left( \frac{dI}{dt} \right) = -0.01 - 0.02 $$
- Calculating ( \frac{dI}{dt} ) Combine the values on the right:
$$ 600 \left( \frac{dI}{dt} \right) = -0.03 $$
Now, divide by 600:
$$ \frac{dI}{dt} = \frac{-0.03}{600} $$
- Final calculation of ( \frac{dI}{dt} ) Calculate:
$$ \frac{dI}{dt} = -0.00005 \text{ amp/sec} $$
The rate of change of current is ( \frac{dI}{dt} = -0.00005 ) amp/sec.
More Information
This result indicates that the current in the circuit is decreasing at a rate of ( 50 ) microamperes per second at the specified instant. This aligns with the observation that the voltage is decreasing while the resistance is increasing.
Tips
- Neglecting to correctly calculate the partial derivatives from the basic relationships in the circuit, such as ( V = IR ).
- Forgetting to account for the signs when substituting values—especially relevant for rates of change like ( \frac{dV}{dt} ).
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