The voltage, V (in volts), across a circuit is given by Ohm\'s law: V=IR, where

Question

The voltage, V (in volts), across a circuit is given by Ohm's law: V=IR, where II is the current (in amps) flowing through the circuit and R is the resistance (in ohms). If we place two circuits, with resistance R1 and R2, in parallel, then their combined resistance, R, is given by

1/R=1/R1+1/R2. (R1 and R2 are simply to indicate different variable, not multiplied by one and two)

Suppose the current is 3 amps and increasing at 0.04 amps/sec and R1 is 4 ohms and increasing at 0.5 ohms/sec, while R2 is 5 ohms and decreasing at 0.2 ohms/sec. Calculate the rate at which the voltage is changing.

Explanation / Answer

Rewrite 1/R = 1/R + 1/R in terms of R:
R = [1/R + 1/R]¹.

When R = 4 and R = 5, R = 20/9 ohm.

Differentiate both sides with respect to time t:
dR/dt = -[1/R + 1/R]² * [(-1/R²) dR/dt + (-1/R²) dR/dt]
dR/dt = [1/R + 1/R]² [(1/R²) dR/dt + (1/R²) dR/dt]

Using the given information, we can solve for dR/dt:
dR/dt = [1/4 + 1/5]² [(1/4²) (0.5) + (1/5²) (-0.2)] = 0.11339

Now, differentiate both sides of V = IR with respect to time:
dV/dt = I dR/dt + R dI/dt

So, dV/dt = 3 * (0.11339) + (20/9) * 0.04 = 0.429058 volts/second.

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