contestada

The voltage,V, (in volts) across a circuit is given by Ohm’s law:V=IR, whereIis thecurrent (in amps) flowing through the circuit andRis the resistance (in ohms). If we placetwo circuits, with resistanceR1andR2, in parallel, then their combined resistance,R, isgiven by1R=1R1+1R2.Suppose the current is 2 amps and increasing at 10−2amp/sec andR1is 3 ohms and incresingat 0.5 ohm/sec, whileR2is 5 ohms and decreasing at 0.1 ohm/sec. Calculate the rate at whichthe voltage is changing.

Respuesta :

Answer:

[tex]\dfrac{dV}{dt}=-0.097\ V/s[/tex]

Explanation:

Given that

Equivalent resistance given as

1/R=1/R₁+1/R₂

V= I R

V=I (1/R₁+1/R₂)

I = 2 A

[tex]\dfrac{dI}{dt}=10^{-2}\ A/s[/tex]

R₁  =3 ohms

[tex]\dfrac{dR_1}{dt}=0.5\ ohms/s[/tex]

R₂=5 ohms

[tex]\dfrac{dR_2}{dt}=-0.1\ ohms/s[/tex]

V=I (1/R₁+1/R₂)

[tex]\dfrac{dV}{dt}=\dfrac{dI}{dt}\left ( \dfrac{1}{R_1}+ \dfrac{1}{R_2}\right )+\left (-\dfrac{1}{R_1^2}\dfrac{dR_1}{dt}- \dfrac{1}{R_2^2}\dfrac{dR_2}{dt}\right )I[/tex]

Now by putting the values

[tex]\dfrac{dV}{dt}=10^{-2}\left ( \dfrac{1}{3}+ \dfrac{1}{5}\right )+\left (-\dfrac{1}{3^2}\times 0.5- \dfrac{1}{5^2}\times (-0.1)\right )2\ V/s[/tex]

[tex]\dfrac{dV}{dt}=-0.097\ V/s[/tex]

Otras preguntas