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Chapter 12: Problem 1733

An electric kettle has two coils when one of these is switched on. the water in the kettle boils in 6 minutes. When the other coil is switched on, boils in 3 minutes If the two coils are connected in parallel, 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 both coils are connected in parallel is 2 minutes. The correct answer is (C) 2 minutes.

Step by step solution

01

1. Calculate the Power Associated with Each Coil Individually

First, let's calculate the power of each coil when it heats up the water on its own. We can use the formula: Power = Work / Time Since we don't know the specific work done by each coil, we can use a constant C to represent the work done in boiling the water. We know that: Power_1 = C / 6 (for the first coil that takes 6 minutes) Power_2 = C / 3 (for the second coil that takes 3 minutes)
02

2. Determine the Equivalent Power for Parallel Connection

Next, we need to find the equivalent power of the two coils connected in parallel. In a parallel connection, the total power is the sum of the individual powers of the coils. So, we can write: Power_total = Power_1 + Power_2
03

3. Calculate the Total Power and Express in Terms of C

Plug in the values of Power_1 and Power_2 from step 1 in the equation above, we get: Power_total = (C / 6) + (C / 3) By solving for Power_total, we get: Power_total = C / 2
04

4. Determine the Time Taken to Boil Water with the Equivalent Power

Now we need to find the time taken to boil water with the equivalent power when both coils are connected in parallel. We can use the same formula as in step 1. Therefore, Time_total = Work / Power_total Plugging in the total power and the work done constant C, we get: Time_total = C / (C / 2)
05

5. Calculate the Time_total for Coils Connected in Parallel

By solving the equation in step 4, we get: Time_total = 2 So, when both coils are connected in parallel, the time taken to boil water is 2 minutes. The correct answer is (C) 2 minutes.

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Most popular questions from this chapter

Two resistances \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have effective resistance \(\mathrm{R}_{\mathrm{s}}\) when connected in sires combination and \(R_{p}\) when connected in parallel combination if $\mathrm{R}_{8} \mathrm{R}_{\mathrm{p}}=16\( and \)\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)=4\( the values of \)\mathrm{R}_{1}\( and \)\mathrm{R}_{2}$ are (A) \(2 \Omega\) and \(0.5 \Omega\) (B) \(1 \Omega\) and \(0.25 \Omega\) (C) \(8 \Omega\) and \(2 \Omega\) (D) \(4 \Omega\) and \(1 \Omega\)

A Potentiometer wire, \(10 \mathrm{~m}\) long, has a resistance of \(40 \Omega\). It is connected in series with a resistance box and a \(2 \mathrm{v}\) storage cell If the potential gradient along the wire is $0.1 \mathrm{~m} \mathrm{v} / \mathrm{cm}$, the resistance unplugged in the box is. (A) \(260 \Omega\) (B) \(760 \Omega\) (C) \(960 \Omega\) (D) \(1060 \Omega\)

What maximum power can be obtained from a battery of \(\mathrm{emf} \mathrm{E}\) and internal resistance \(\mathrm{r}\) connected with an external resistance \(\mathrm{R}\) ? (A) \(\left(E^{2} / 4 r\right)\) (B) \(\left(E^{2} / 3 r\right)\) (C) \(\left(E^{2} / 2 r\right)\) (D) \(\left(\mathrm{E}^{2} / \mathrm{r}\right)\)

Two batteries each of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series to a resistor \(R\). Maximum Possible power consumed by the resistor \(=\ldots .\) (A) \(3.2 \mathrm{~W}\) (B) \((16 / 9) \mathrm{W}\) (C) \((8 / 9) \mathrm{W}\) (D) \(2 \mathrm{~W}\)

Three identical resistors connected in series with a battery, together dissipate \(10 \mathrm{~W}\) of power. What will be the power dissipated, if the same resistors are connected in paralle1 across the same battery? (A) \(60 \mathrm{~W}\) (B) \(30 \mathrm{~W}\) (C) \(90 \mathrm{~W}\) (D) \(120 \mathrm{~W}\)
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