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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

The circuit shown in fig consists of the following \(E_{1}=6, E_{2}=2, E_{3}=3\) Volt \(\mathrm{R}_{1}=6, \mathrm{R}_{4}=3 \mathrm{Ohm}\) \(\mathrm{R}_{3}=4, \mathrm{R}_{2}=2 \mathrm{Ohm}\) $\mathrm{C}=5 \mu \mathrm{F}$$E_{1}=6 \mathrm{~V} \quad E_{2}=2 \mathrm{~V} \quad E_{3}=3 \mathrm{~V} \quad \mathrm{R}_{1}=6 \Omega$ \(R_{2}=2 \Omega \quad R_{3}=4 \Omega \quad R_{4}=3 \Omega\) The energy stored in the capacitor is. (A) \(4.8 \times 10^{-6} \mathrm{~J}\) (B) \(9.6 \times 10^{-6} \mathrm{~J}\) (C) \(1.44 \times 10^{-5} \mathrm{~J}\) (D) \(1.92 \times 10^{-5} \mathrm{~J}\)

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 resistance of the series combination of two resistances is \(\mathrm{S}\), when they are joined in parallel the total resistance is \(\mathrm{P}\) If $S=n P\(, then the minimum possible value of \)n$ is.... (A) 4 (B) 3 (C) 2 (D) 1

Two wires of resistances \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have temperature coefficient of resistances \(\alpha_{1}\) and \(\alpha_{2}\) respectively they are joined in series the effective temperature coefficient of resistance is .... (A) \(\left[\left(\alpha_{1}+\alpha_{2}\right) / 2\right]\) (B) \(\sqrt{\alpha_{1} \alpha_{2}}\) (C) $\left[\left(\alpha_{1} R_{1}+\alpha_{2} R_{2}\right) /\left(R_{1}+R_{2}\right)\right]$ (D) $\left[\sqrt{\left(R_{1}\right.} R_{2} \alpha_{1} \alpha_{2}\right) / \sqrt{ \left.\left(R_{1}^{2}+R_{2}^{2}\right)\right]}$

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