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Chapter 12: Problem 1732

An electric kettle has two coils. When one of them is switched on, the water in the kettle boils in 6 minutes. When the other coil is switched on, the water boils in 3 minutes If the two coils are connected in series the time taken to boil water in the kettle is: (A) 3 minutes (B) 6 minutes (C) 2 minutes (D) 9 minutes

Short Answer

Expert verified
The time taken to boil water when the coils are connected in series is 18 minutes. However, the given options in the question might have a typo or provide wrong information, as none of the options match the calculated time.

Step by step solution

01

Find the power of each coil

Firstly, we need to find the power of each coil using the time taken to boil water. The relationship between power and time is given by: \(P = \frac{Q}{t}\), where Q is the heat required to boil water, P is the power and t is the time taken. Since the amount of heat required to boil water is the same in both cases, we can say that the power of coil 1 is twice the power of coil 2: \(P_{1} = 2P_{2}\)
02

Use the power formula to find the equivalent resistance when the coils are connected in series

When the coils are connected in series, their equivalent resistance is given by the sum of their individual resistances: \(R_{eq} = R_{1} + R_{2}\) We can use the power formula: \(P = \frac{V^2}{R}\) for each coil, where P is the power, V is the voltage across the coil and R is its resistance. We know that the power of coil 1 is twice the power of coil 2, so we can write the equation: \(\frac{V^2}{R_{1}} = 2\frac{V^2}{R_{2}}\) Dividing both sides by \(V^2\) and multiplying both sides by \(R_{1} * R_{2}\), we get: \(R_{2} = 2R_{1}\) Now, we have the relationship between the resistances of both coils.
03

Substitute the relationship between resistances into the equivalent resistant equation and find the weightage factor

Substitute the relationship between resistances of coils into the equivalent resistance equation: \(R_{eq} = R_{1} + 2R_{1}\) Thus, the equivalent resistance is 3 times the resistance of coil 1: \(R_{eq} = 3R_{1}\) Since \(R_{eq}\) is 3 times \(R_{1}\), and the voltage across the series connection is the same as that across each coil individually, the power is divided by a factor of 3 when the coils are connected in series.
04

Calculate the time taken to boil water when coils are connected in series

We know the relationship between power and time is given by: \(P = \frac{Q}{t}\) Since the power has been divided by a factor of 3 when both coils are connected in series, the time required to boil water will increase by a factor of 3. Hence, the time taken to boil water when the coils are connected in series is: \(t_{eq} = 3*t_{1} = 3*6\) \t_eq = 18 minutes. However, none of the options given in the exercise matches the calculated time taken to boil water in the kettle. So, there might be a typo in the exercise options or some wrong information provided in the question.

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Most popular questions from this chapter

The tungsten filament of bulb has resistance equal to \(18 \Omega\) at \(27^{\circ} \mathrm{C}\) temperature \(0.25 \mathrm{~A}\) of current flows, when \(45 \mathrm{~V}\) is connected to it If $\alpha=4.5 \times 10^{-3} \mathrm{~K}^{-1}$ for a tungsten then find the temperature of the filament. (A) \(2160 \mathrm{~K}\) (B) \(1800 \mathrm{~K}\) (C) \(2070 \mathrm{~K}\) (D) \(2300 \mathrm{~K}\)

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at $500 \mathrm{~cm}$ on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The resistance of the voltmeter is (A) \(500 \Omega\) (B) \(290 \Omega\) (C) \(49 \Omega\) (D) \(20 \Omega\)

A wire \(50 \mathrm{~cm}\) long and \(1 \mathrm{~mm}^{2}\) in cross-section carries a current of \(4 \mathrm{~A}\) when connected to a \(2 \mathrm{~V}\) battery. The resistivity of the wire is: (A) \(2 \times 10^{-7} \Omega \mathrm{m}\) (B) \(5 \times 10^{-7} \Omega \mathrm{m}\) (C) \(4 \times 10^{-6} \Omega \mathrm{m}\) (D) \(1 \times 10^{-6} \Omega \mathrm{m}\)

An infinite sequence of resistances is shown in the figure. The resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\) will be, when \(\mathrm{R}_{1}=1 \mathrm{ohm}\) and \(\mathrm{R}_{2}=2 \mathrm{ohm}\) (A) \(3 \Omega\) (B) \(2 \Omega\) (C) \(1 \Omega\) (D) \(1.5 \Omega\)

Length of a wire of resistance \(R \Omega\) is increased to 10 times, so its resistance becomes \(1000 \Omega\), therefore \(R=\ldots .\) (The volume of the wire remains same during increase in length) (A) \(0.01 \Omega\) (B) \(0.1 \Omega\) (C) \(1 \Omega\) (D) \(10 \Omega\)
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