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insens350 [35]
4 years ago
9

To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to

the car's battery, which has an emf E = 10.0 V and an internal resistance rint = 0.03 Ω . The heating element that is immersed in the cup of water is a resistive coil with resistance R. David wants to experiment with the device, so he connects an ammeter into the circuit and measures 10.0 A when the device is connected to the car's battery. From this, he calculates the time to boil a cup of water using the device. If the energy required is 100 kJ , how long does it take to boil a cup of water?
Physics
1 answer:
mr_godi [17]4 years ago
6 0

Answer:

 t = 1030 s

Explanation:

Let's start by calculating the resistance of the coil,

    V = I (R + ri)

    R = V / I - ri

    R = 10/10 -0.03

    R = 0.97  Ω

Now we can calculate the power supplied to the water

   P = I2 R

   P = 10 2 0.97

   P = 97 w

Work energy is power for time

   E = W = P t

   t = W / P

   t = 100 103/97

   t = 1030 s

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Answer:

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Solving the differential equation,

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where k = RC

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The Current through the capacitor is given as the time rate of change of charge on the capacitor.

I(t) = -dQ/dt

But, the charge on a capacitor is given as

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(dQ/dt) = (d/dt) (CV)

Since C is constant,

(dQ/dt) = (CdV/dt)

V(t) = I(t) × R

V(t) = -(CdV/dt) × R

V = -RC (dV/dt)

(dV/dt) = -(RC/V)

(dV/V) = -RC dt

∫ (dV/V) = ∫ -RC dt

Let k = RC

∫ (dV/V) = ∫ -k dt

Integrating the the left hand side from V₀ (the initial voltage of the capacitor) to V (the voltage of the resistor at any time) and the right hand side from 0 to t.

In V - In V₀ = -kt

In(V/V₀) = - kt

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V = V₀ e⁻ᵏᵗ

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Hope this Helps!!!

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