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Chapter 12: Problem 1729

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) \(1: 4\)

Short Answer

Expert verified
The ratio of the power dissipated in the two electric bulbs connected in parallel to a constant voltage source is \(\frac{P_1}{P_2} = \frac{2}{1}\), or \(2:1\). So, the correct answer is (C).

Step by step solution

01

Write down the given information.

We are given the ratio of resistances of the electric bulbs as \(1:2\). Let's denote the resistances of the bulbs as \(R_1\) and \(R_2\) such that \(\frac{R_1}{R_2} = \frac{1}{2}\). Since the bulbs are connected in parallel, we know that the voltage across them is the same and denoted as V.
02

Use Ohm's Law formula.

Ohm's Law states that Voltage (V) equals the product of current (I) and resistance (R). We can write this formula for each bulb, where \(I_1\) and \(I_2\) are the currents flowing through bulb 1 and bulb 2 respectively. For Bulb 1: \(V = R_1 I_1\) For Bulb 2: \(V = R_2 I_2\)
03

Calculate the power dissipated in each bulb.

The formula for power (P) is the product of Voltage (V) and current (I). We can write this formula for each bulb as well. For Bulb 1: \(P_1 = V I_1\) For Bulb 2: \(P_2 = V I_2\)
04

Use Ohm's Law to substitute the current values in the power formulas.

We can use Ohm's Law to rewrite the Power formulas in terms of resistances. For Bulb 1: \(P_1 = V \cdot \frac{V}{R_1} = \frac{V^2}{R_1}\) For Bulb 2: \(P_2 = V \cdot \frac{V}{R_2} = \frac{V^2}{R_2}\)
05

Calculate the ratio of the power dissipated in both bulbs.

Now, we simply need to find the ratio of the power dissipated in each bulb. Using the formulas from Step 4, we can write the ratio as: \(\frac{P_1}{P_2} = \frac{\frac{V^2}{R_1}}{\frac{V^2}{R_2}}\) To solve this, we can cancel out the \(V^2\) terms and then substitute the ratio of the resistances given in the problem: \(\frac{P_1}{P_2} = \frac{R_2}{R_1} = \frac{2}{1}\) So, the ratio of the power dissipated in the bulbs is \(2:1\). This corresponds to the correct answer (C).

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Most popular questions from this chapter

For a cell of e.m.f \(2 \mathrm{~V}\), a balance is obtained for $50 \mathrm{~cm}\( of the potentiometer wire If the cell is shunted by a \)2 \Omega$ resistor and the balance is obtained across \(40 \mathrm{~cm}\) of the wire, then the internal resistance of the cell is. (A) \(1 \Omega\) (B) \(0.5 \Omega\) (C) \(1.2 \Omega\) (D) \(2.5 \Omega\)

The reading of ammeter shown in figure is.... (A) \(2.18 \mathrm{~A}\) (B) \(3.28 \mathrm{~A}\) (C) \(6.56 \mathrm{~A}\) (D) \(1.09 \mathrm{~A}\)

A wire of resistor \(R\) is bent into a circular ring a circular ring of radius \(\mathrm{r}\) Equivalent resistance between two points \(\mathrm{X}\) and \(\mathrm{Y}\) on its circumference, when angle xoy is \(\alpha\), can be given by (A) $\left\\{(\mathrm{R} \alpha) /\left(4 \pi^{2}\right)\right\\}(2 \pi-\alpha)$ (B) \((\mathrm{R} / 2 \pi)(2 \pi-\alpha)\) (C) \(\mathrm{R}(2 \pi-\alpha)\) (D) \((4 \pi / \mathrm{R} \alpha)(2 \pi-\alpha)\)

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 voltmeter reading will be. (A) \(1.96 \mathrm{~V}\) (B) \(1.8 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(0.96 \mathrm{~V}\)

In each of the following questions, match column \(\mathrm{I}\) and column II and select the correct match out of the four given choicesColumn I Column II (a) The series combination of cells is for (p) More current (b) The parallel combination of cell is for (q) More voltage (c) In series combination of n cells, each of (r) \(\varepsilon\) \(\mathrm{emf}_{\mathrm{e}}\) the effective voltage is (d) In parallel combination of n cells, each of emf \(\varepsilon\) (s) ne the effective voltage is (A) \(a-p, b-q, c-r, d-s\) (B) a - q, b-p, c-r, d - s (C) \(a-q, b-p, c-s, d-r\) (D) \(a-p, b-q, c-s, d-r\)
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