Respuesta :
Answer:
[tex]\frac{dV}{dt} = 0.453 Volts/s[/tex]
Explanation:
As we know that two resistors are in parallel
so we have
[tex]V = i R[/tex]
where we know that
[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]
so we have
[tex]\frac{1}{V} = \frac{1}{i}(\frac{1}{R_1} + \frac{1}{R_2})[/tex]
now to find the rate of change we have
[tex]\frac{1}{V^2}\frac{dV}{dt} = \frac{1}{i^2}\frac{di}{dt}(\frac{1}{R_1} + \frac{1}{R_2}) + \frac{1}{i}(\frac{1}{R_1^2}(\frac{dR_1}{dt}) + \frac{1}{R_2^2}(\frac{dR_2}{dt}))[/tex]
[tex]\frac{1}{V} = \frac{1}{3}(\frac{1}{4} + \frac{1}{3})[/tex]
[tex]\frac{1}{V} = 0.194[/tex]
now from above equation we have
[tex](0.194)^2\frac{dV}{dt} = \frac{1}{3^2}(0.02)(\frac{1}{4} + \frac{1}{3}) + \frac{1}{3}(\frac{1}{4^2}(0.4) + \frac{1}{3^2}(0.2))[/tex]
[tex]0.0376\frac{dV}{dt} = 1.296\times 10^{-3} + 0.0157[/tex]
[tex]\frac{dV}{dt} = 0.453 Volts/s[/tex]
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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