Question

The voltage, V (in volts), across a circuit is given by Ohm's law: V=IR, where I is the current (in amps) flowing through the circuit and R is the resistance (in ohms). If we place two circuits, with resistance R₁and R₂, in parallel, then their combined resistance, R₁, is given by 1/R = 1/R₁ + 1/R₂ . Suppose the current is 5amps and increasing at 0.04 amps/sec and R₁ is 4 ohms and increasing at 0.4 ohms/sec, while R₂ is 3 ohms and decreasing at 0.3ohms/sec. Calculate the rate at which the voltage is changing.

_____volts/sec

Answer 1

To calculate the rate at which the voltage is changing, we can use calculus and apply the product rule of differentiation to Ohm's law. By substituting the given values for current, resistance, and their rates of change, we can find the rate at which the voltage is changing.

Step-by-step explanation:

The rate at which the voltage is changing can be calculated using calculus.

We have V = IR, and we want to find dV/dt, the rate of change of voltage with respect to time.

Differentiating the equation with respect to time, we get dV/dt = d(IR)/dt.

By applying the product rule of differentiation, we have dV/dt = I dR/dt + R dI/dt.

Given that I = 5 A and dI/dt = 0.04 A/sec, and R = 4 Ω and dR/dt = 0.4 Ω/sec, we can substitute these values into the equation to find dV/dt.