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Chapter 14: Problem 159

An electric bulb is rated \(220 \mathrm{~V}-100 \mathrm{~W}\). The power consumed by it when operated on \(110 \mathrm{~V}\) will be (A) \(75 \mathrm{~W}\) (B) \(40 \mathrm{~W}\) (C) \(25 \mathrm{~W}\) (D) \(50 \mathrm{~W}\)

Short Answer

Expert verified
The power consumed by the electric bulb when operated at 110V is 25W (option C). This is found by first calculating the bulb's resistance at its rated voltage and power (484 ohms), and then using this resistance to find the power consumed at the lower voltage.

Step by step solution

01

Find the resistance at the rated voltage and power

We are given the rated voltage (220V) and power (100W) of the bulb, and we need to find the resistance. We can use the formula for electrical power: Power = Voltage^2 / Resistance Rearranging the formula to find the resistance, we get: Resistance = Voltage^2 / Power Substituting the given values into the formula, we find the resistance of the bulb at its rated voltage: Resistance = (220V)^2 / 100W Resistance = 48400 / 100 Resistance = 484 ohms
02

Find the power consumed at the lower voltage

Now that we have the resistance of the bulb, we can find the power consumed when the bulb is operated at a lower voltage (110V). We can use the formula for electrical power again: Power = Voltage^2 / Resistance Substituting the lower voltage (110V) and the resistance (484 ohms) into the formula, we find the power consumed at the lower voltage: Power = (110V)^2 / 484 ohms Power = 12100 / 484 Power = 25 W So the power consumed by the bulb when operated at 110V is 25W, which corresponds to option (C).

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Most popular questions from this chapter

Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The balance length when key \(K_{2}\) is open and \(K_{1}\) is closed is given by (A) \(10.5 \mathrm{~cm}\) (B) \(11.5 \mathrm{~cm}\) (C) \(12.5 \mathrm{~cm}\) (D) \(13.5 \mathrm{~cm}\)

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