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Arlecino [84]
3 years ago
5

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric

al power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is:_______. (a) What is the value of P in the limit that R is very small?
(b) What is the value of P in the limit that R is very large?(c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has [Math Processing Error] and [Math Processing Error]. What is the power output of this battery when it is connected to a resistor R, for [Math Processing Error], [Math Processing Error], and [Math Processing Error]? Are your results consistent with the general result that you derived in part (b)?
Physics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

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3 years ago
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Sam receives the kicked football on the 3 yd line and runs straight ahead toward the goal line before cutting to the right at th
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Answer:

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Explanation:

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A positively charged body exerts an electric field that causes a positively charged particle to move away from it. If work is do
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b. electric potential energy.

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3 years ago
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
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Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

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   P = ρ g h

   h = \dfrac{P}{\rho\ g}

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b) when air density decreased linearly to zero.

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now, Pressure at depth x

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dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

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height of the atmosphere is equal to 15902.67 m.

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