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Chapter 12: Problem 1782

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) \(120 \mathrm{~W}\)

Short Answer

Expert verified
The power dissipated in the parallel configuration is \(30 W\). The correct answer is (B) \(30 \mathrm{~W}\).

Step by step solution

01

Find total resistance in the series configuration

Given, three identical resistors connected in series dissipate 10W of power. Let the resistance of each resistor be R. In a series connection, total resistance will be the sum of the individual resistances:\ \[R_{total\_series} = R + R + R = 3R\]
02

Calculate current in the series configuration

We are given that the total power dissipated in the series configuration is 10W. We can use Ohm's law and the power formula to find the current (I) flowing through the series resistors.\ \[P = I^2 R_{total\_series}\]\ \[10 = I^2 (3R)\]\ Solving for I, we get:\ \[I = \sqrt{\frac{10}{3R}}\]
03

Calculate the battery voltage

Using Ohm's law, we can now find the battery voltage (V):\ \[V = IR_{total\_series}\]\ Plugging in the values found in step 2, we get:\ \[V = \sqrt{\frac{10}{3R}} \cdot 3R\]\ Simplifying, we get:\ \[V = \sqrt{10R}\]
04

Find total resistance in the parallel configuration

Now, let's find the total resistance in the parallel configuration:\ \[\frac{1}{R_{total\_parallel}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R}\]\ \[R_{total\_parallel} = \frac{R}{3}\]
05

Calculate current in the parallel configuration

Using Ohm's law, we can find the current in the parallel configuration:\ \[I_{total\_parallel} = \frac{V}{R_{total\_parallel}}\]\ Plugging in the values found in step 3 and 4, we get:\ \[I_{total\_parallel} = \frac{\sqrt{10R}}{\frac{R}{3}}\]\ Simplifying, we get:\ \[I_{total\_parallel} = 3\sqrt{\frac{10}{R}}\]
06

Calculate power in the parallel configuration

Now, we can calculate the total power dissipated in the parallel configuration using the power formula:\ \[P_{total\_parallel} = I_{total\_parallel}^2 R_{total\_parallel}\]\ Plugging in the values found in step 4 and 5, we get:\ \[P_{total\_parallel} = (3\sqrt{\frac{10}{R}})^2 \cdot \frac{R}{3}\]\ Simplifying, we get:\ \[P_{total\_parallel} = 3 \cdot 10 = 30 W\] So, the power dissipated in the parallel configuration is \(30 W\). The correct answer is (B) \(30 \mathrm{~W}\).

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Most popular questions from this chapter

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) \(20 \Omega\)

Two wires of equal diameters of resistivity's \(\rho_{1}\) and \(\rho_{2}\) are joined in series. The equivalent resistivity of the combination is.... (A) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (B) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$ (C) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (D) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$

The potential difference through the \(3 \Omega\) resistor shown in fig is.... (A) Zero (B) \(1 \mathrm{~V}\) (C) \(3.5 \mathrm{~V}\) (D) \(7 \mathrm{~V}\)

A carbon resistor has a set of coaxial colored rings in the order brown, violet brown and silver. The value of resistance (in ohms) is. (A) \((27 \times 10) \pm 5 \%\) (B) \((27 \times 10)=10 \%\) (C) \((17 \times 10)=5 \%\) (D) \((17 \times 10) \pm 10 \%\)

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) \(1: 4\)
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