PHYSICS HIGH SCHOOL

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.

Answers

Answer 1

Answer:

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

Read more at; brainly.com/question/13539417

Answer 2

Answer:

Answer:

$(dV)/(dt) = 0.453 Volts/s$

Explanation:

As we know that two resistors are in parallel

so we have

$V = i R$

where we know that

$(1)/(R) = (1)/(R_1) + (1)/(R_2)$

so we have

$(1)/(V) = (1)/(i)((1)/(R_1) + (1)/(R_2))$

now to find the rate of change we have

$(1)/(V^2)(dV)/(dt) = (1)/(i^2)(di)/(dt)((1)/(R_1) + (1)/(R_2)) + (1)/(i)((1)/(R_1^2)((dR_1)/(dt)) + (1)/(R_2^2)((dR_2)/(dt)))$

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

$(1)/(V) = 0.194$

now from above equation we have

$(0.194)^2(dV)/(dt) = (1)/(3^2)(0.02)((1)/(4) + (1)/(3)) + (1)/(3)((1)/(4^2)(0.4) + (1)/(3^2)(0.2))$

$0.0376(dV)/(dt) = 1.296* 10^(-3) + 0.0157$

$(dV)/(dt) = 0.453 Volts/s$

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Answers

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Answers

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Answer:

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Explanation:

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Answers

Answer:

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