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 = 0.08 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?

Answer :

Answer:

t=17 min

Explanation:

Given that

E= 10 V

I=10 A

Internal resistance Rin=  0.08 Ω

E= I.R(net)

E=I(R+r)

10 =  10 (R+0.08)

R= 1 - 0.0 8 = 0.02 Ω

R= 0.98 Ω

So the heat given as

Q=I²Rt

Q=10² x 0.98 x t

Q= 98 t J

Given that Q= 100 KJ

98 t = 100 x 1000 J

t= 100 x 1000/98 sec

t=17 min

The time to boil a cup of water using the device by David If the energy required is 100 kJ is 17 minutes.

What is the Ohm's law?

Ohm's law states that for a flowing current, the potential difference of the circuit is directly proportional to the current flowing in it. Thus,

[tex]V\propto I[/tex]

Here, (V) is the potential difference and (I) is the current.

It can be written as,

[tex]V=IR[/tex]

Here, (R) is the resistance of the circuit.

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 = 0.08 Ω .

The heating element that is immersed in the cup of water is a resistive coil with resistance R. The total resistance is,

[tex]R_n=R+0.08\rm \; ohm[/tex]

The value of current in ammeter into the circuit is 10.0 A when the device is connected to the car's battery. Put these values in the ohm's law.

[tex]10=10(R+0.08)\\R=0.98\rm\; ohm[/tex]

The heat generate in a circuit in time (t) can be given as,

[tex]Q=I^2Rt[/tex]

The energy required is 100 kJ. Thus, by the above expression,

[tex]100\times 10^3=(10)^2(0.92)t\\t=1020.41\rm\; sec\\[/tex]

[tex]t=1020.41\times\dfrac{1}{60}\rm\; min\\t=17\;min[/tex]

Thus, the time to boil a cup of water using the device by David If the energy required is 100 kJ is 17 minutes.

Learn more about the Ohm's law here;

https://brainly.com/question/25822112

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