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Chapter 12: Problem 1824

Two electric bulbs marked \(25 \mathrm{~W}-220 \mathrm{~V}\) and $100 \mathrm{~W}-220 \mathrm{~V}\( are connected in series to a \)440 \mathrm{~V}$ supply which of the bulbs will fuse? (A) \(100 \mathrm{~W}\) (B) \(25 \mathrm{~W}\) (C) None of this (D) Both

Short Answer

Expert verified
(B) \(25 \mathrm{~W}\)

Step by step solution

01

Understand the power equation and Ohm's law

For an electric bulb, the power (P) is given by the formula: \[P = V \times I\] where V is the voltage and I is the current passing through the bulb. Ohm's law relates the voltage and current with resistance (R) as: \[V = I \times R\] Thus, we can find the resistance of the bulbs using their power ratings and given voltage.
02

Calculate the resistance of each bulb

We can find the resistance of each bulb using their power ratings and the given voltage. Let's denote the resistance of the 25W bulb as \(R_{25}\) and the resistance of the 100W bulb as \(R_{100}\). The resistance of each bulb can be calculated using: \[R = \frac{V^2}{P}\] Therefore, the resistance of the 25W and 100W bulbs are: \(R_{25} = \frac{(220V)^2}{25W} = 1936\Omega\) \(R_{100} = \frac{(220V)^2}{100W} = 484\Omega\)
03

Calculate the total resistance and the current flowing through the bulbs

As the bulbs are connected in series, their resistances add up: \(R_{total} = R_{25} + R_{100} = 1936\Omega + 484\Omega = 2420\Omega\) The 440V supply is connected across this total resistance: \[\frac{440V}{2420\Omega} = 0.1818(A)\] Hence, the current flowing through the entire circuit, including both bulbs, is \(0.1818A\).
04

Calculate the voltage drop across each bulb

Using Ohm's law, we can now find the voltage drop across each bulb: \(V_{25} = I \times R_{25} = 0.1818A \times 1936\Omega = 352V\) \(V_{100} = I \times R_{100} = 0.1818A \times 484\Omega = 88V\)
05

Determine which bulb will fuse

Now that we have the voltage drops across each bulb, we can compare them to the given voltage ratings to determine which bulb will fuse: (1) For the 25W bulb: The voltage drop across it is 352V, which is higher than its rated voltage of 220V. It will likely fuse. (2) For the 100W bulb: The voltage drop across it is 88V, which is lower than its rated voltage of 220V. It will not fuse due to the given conditions. Therefore, the correct answer is: (B) \(25 \mathrm{~W}\)

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Most popular questions from this chapter

Two electirc bulbs whose resistances are in the ratio of \(1: 2\) are connected in parallel to a constant voltage source the power dissipated in them have the ratio. (A) \(1: 2\) (B) \(1.1\) (C) \(2: 1\) (D) \(1: 4\)

Which of the following has negative temperature coefficient of resistance? (A) \(\mathrm{Fe}\) (B) \(\mathrm{C}\) (C) \(\mathrm{Mn}\) (D) Ag

How many dry cells, each of emf \(1.5 \mathrm{~V}\) and internal resistance $0.5 \Omega\(, much be joined in series with a resistor of \)20 \Omega$ to give a current of \(0.6 \mathrm{~A}\) in the circuit? (A) 2 (B) 8 (C) 10 (D) 12

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) \(1060 \Omega\)

A potentiometer wire of length \(1 \mathrm{~m}\) and resistance \(10 \Omega\) is connected in series with a cell of e.m.f \(2 \mathrm{~V}\) with internal resistance \(1 \Omega\) and a resistance box of a resistance \(R\) if potential difference between ends of the wire is \(1 \mathrm{~V}\) the value of \(R\) is. (A) \(4.5 \Omega\) (B) \(9 \Omega\) (C) \(15 \Omega\) (D) \(20 \Omega\)
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