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 r1 and r2, in parallel, then their combined resistance, r, is given by 1r=1r1+1r2. suppose the current is 3 amps and increasing at 0.02 amps/sec and r1 is 4 ohms and increasing at 0.4 ohms/sec, while r2 is 3 ohms and decreasing at 0.2 ohms/sec. calculate the rate at which the voltage is changing.

The rate at which the voltage of the given circuit is changing is gotten to be;

dV/dt = 0.452 V/s

We are given;

Current; I = 3 A

Resistance 1; R1 = 4Ω

Resistance 2; R2 = 3Ω

dR1/dt = 0.4 Ω/s

dR2/dt = 0.2 Ω/s

dI/dt = 0.02 A/s

Now, formula for voltage with resistors in parallel is;

1/V = (1/I)(1/R1 + 1/R2)

Plugging in the relevant values, we can find V;

1/V = (1/3)(1/4 + 1/3)

Simplifying this gives;

1/V = 0.194

Now, we want to find the rate at which the voltage is charging, we need to find dV/dt.

Thus, let us differentiate 1/V = (1/I)(1/R1 + 1/R2) with respect to t to get;

(1/V)²(dV/dt) = [(1/i²)(di/dt)(1/R1 + 1/R2)] + (1/I)[(1/R1²)(dR1/dt) + (1/R2²)(dR2/dt)]

Plugging in the relevant vies gives us;

0.194²(dV/dt) = [(1/3²)(0.02)(¼ + ⅓)] + (⅓)[(1/3²)(0.4) + (1/4²)(0.3)]

>> 0.037636(dV/dt) = 0.001296 + 0.0157

>> dV/dt = 0.016996/0.037636

>> dV/dt = 0.452 V/s

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Explanation:

As we know that two resistors are in parallel

so we have

where we know that

so we have

now to find the rate of change we have

now from above equation we have