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Chapter 12: Problem 1715

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) \((17 \times 10) \pm 10 \%\)

Short Answer

Expert verified
The correct value of resistance for the carbon resistor with color bands brown, violet, brown, and silver is (D) \((17 \times 10) \pm 10 \% \Rightarrow 170 \Omega \pm 10 \%\).

Step by step solution

01

Color Code Table

As a start, we need to know the color code table which assigns a numerical value to each color. Color Code Table: Black: 0 Brown: 1 Red: 2 Orange: 3 Yellow: 4 Green: 5 Blue: 6 Violet: 7 Gray: 8 White: 9
02

First Two Bands

The first two bands represent the significant digits' value. Brown corresponds to 1 and violet corresponds to 7. So, the first two significant digits are 17.
03

Third Band

The third band represents the multiplier value. A brown band represents a multiplier of 10. So, the resistance value will be multiplied by 10.
04

Resistance Value Calculation

Now, we can determine the resistance value by combining the significant digits and the multiplier. Resistance Value = Significant Digits × Multiplier = \(17 \times 10 = 170 \Omega\)
05

Fourth Band

Finally, the fourth band represents the tolerance or the percentage of error that the resistor can have. In our exercise, the fourth ring is silver, which corresponds to a 10% tolerance.
06

Final Solution

With the resistance value and tolerance, the final value of resistance is: \((17 \times 10) \pm 10 \% \Rightarrow 170 \Omega \pm 10 \%\) Comparing this result with the given options, the correct choice is (D) \((17 \times 10) \pm 10 \%\).

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Most popular questions from this chapter

A heater boils \(1 \mathrm{~kg}\) of water in time \(\mathrm{t}_{1}\) and another heater boils the same water in time \(\mathrm{t}_{2}\) If both are connected in series, the combination will boil the same water in time. (A) $\left\\{\left(\mathrm{t}_{1} \mathrm{t}_{2}\right) /\left(\mathrm{t}_{2}+\mathrm{t}_{1}\right)\right\\}$ (B) $\left\\{\left(\mathrm{t}_{1} \mathrm{t}_{2}\right) /\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)\right\\}$ (C) \(\mathrm{t}_{1}+\mathrm{t}_{2}\) (D) \(2\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)\)

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The resistance of the wire made of silver at \(27^{\circ} \mathrm{C}\) temperature is equal to \(2.1 \Omega\) while at \(100^{\circ} \mathrm{C}\) it is \(2.7 \Omega\) calculate the temperature coefficient of the resistivity of silver. Take the reference temperature equal to \(20^{\circ} \mathrm{C}\) (A) \(4.02 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (B) \(0.402 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (C) \(40.2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\) (D) \(4.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)

The effective resistance of a n number of resistors connected in parallel in \(\mathrm{x} \mathrm{ohm}\). When one of the resistors is removed, the effective resistance becomes y ohm. The resistance of the resistor that is removed is.... (A) \(\\{(\mathrm{xy}) /(\mathrm{x}+\mathrm{y})\\}\) (B) \(\\{(\mathrm{xy}) /(\mathrm{y}-\mathrm{x})\\}\) (C) \((\mathrm{y}-\mathrm{x})\) (D) \(\sqrt{x y}\)

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