101. Which set correctly matches the material type with its usual resistance-temperature behaviour?
| Material type | Usual behaviour when temperature increases |
|---|
| P. Metal | 1. Resistance generally decreases |
| Q. Semiconductor | 2. Resistance generally increases |
| R. Alloy such as manganin | 3. Resistance changes only slightly |
Choose the correct matching.
ⓐ. P-2, Q-1, R-3
ⓑ. P-1, Q-2, R-3
ⓒ. P-2, Q-3, R-1
ⓓ. P-3, Q-1, R-2
Correct Answer: P-2, Q-1, R-3
Explanation: For metals, resistance generally increases with temperature because lattice vibrations become stronger and collisions increase. For semiconductors, resistance generally decreases with temperature because more charge carriers become available. Alloys such as manganin usually have small temperature coefficients, so their resistance changes only slightly over moderate temperature ranges. This difference is important in selecting materials for resistors and measuring devices. The most common mistake is to assume all materials behave like metals when heated.
102. The resistance of a conductor at temperature \(T\) is given by \(R_T=R_0[1+\alpha(T-T_0)]\). What does \(\alpha\) represent?
ⓐ. The resistivity of the material at all temperatures
ⓑ. The temperature coefficient of resistance
ⓒ. The conductivity of the conductor
ⓓ. The current flowing through the conductor
Correct Answer: The temperature coefficient of resistance
Explanation: In \(R_T=R_0[1+\alpha(T-T_0)]\), the symbol \(\alpha\) tells how sensitively resistance changes with temperature. It is called the temperature coefficient of resistance. A positive \(\alpha\) means resistance increases as temperature rises, which is typical for metals. A negative \(\alpha\) means resistance decreases as temperature rises, which is typical for semiconductors. The coefficient is not the resistance itself; it controls the fractional change of resistance per degree change in temperature. The sign of \(\alpha\) is as important as its magnitude.
103. A metallic resistor has resistance \(20\,\Omega\) at \(20^\circ\text{C}\). Its temperature coefficient of resistance is \(4.0\times10^{-3}\,^\circ\text{C}^{-1}\). What is its resistance at \(70^\circ\text{C}\)?
ⓐ. \(16\,\Omega\)
ⓑ. \(20.4\,\Omega\)
ⓒ. \(24\,\Omega\)
ⓓ. \(40\,\Omega\)
Correct Answer: \(24\,\Omega\)
Explanation: \( \textbf{Given:} \) \(R_0=20\,\Omega\), \(T_0=20^\circ\text{C}\), \(T=70^\circ\text{C}\), and \(\alpha=4.0\times10^{-3}\,^\circ\text{C}^{-1}\).
\( \textbf{Temperature change:} \) \(\Delta T=T-T_0=70^\circ\text{C}-20^\circ\text{C}=50^\circ\text{C}\).
\( \textbf{Formula:} \) \(R_T=R_0[1+\alpha\Delta T]\).
\( \textbf{Substitution:} \) \(R_T=20[1+(4.0\times10^{-3})(50)]\).
\( \textbf{Coefficient product:} \) \((4.0\times10^{-3})(50)=0.20\).
\( \textbf{Bracket value:} \) \(1+0.20=1.20\).
\( \textbf{Resistance:} \) \(R_T=20(1.20)=24\,\Omega\).
\( \textbf{Final answer:} \) \(R_T=24\,\Omega\); for a metal with positive \(\alpha\), the final resistance should be greater than the initial resistance.
104. A semiconductor has resistance \(500\,\Omega\) at \(30^\circ\text{C}\) and an average temperature coefficient \(\alpha=-2.0\times10^{-2}\,^\circ\text{C}^{-1}\) over a small range. Estimate its resistance at \(40^\circ\text{C}\) using the linear relation.
ⓐ. \(400\,\Omega\)
ⓑ. \(500\,\Omega\)
ⓒ. \(600\,\Omega\)
ⓓ. \(1000\,\Omega\)
Correct Answer: \(400\,\Omega\)
Explanation: \( \textbf{Given:} \) \(R_0=500\,\Omega\), \(T_0=30^\circ\text{C}\), \(T=40^\circ\text{C}\), and \(\alpha=-2.0\times10^{-2}\,^\circ\text{C}^{-1}\).
\( \textbf{Temperature change:} \) \(\Delta T=40^\circ\text{C}-30^\circ\text{C}=10^\circ\text{C}\).
\( \textbf{Linear relation:} \) \(R_T=R_0[1+\alpha\Delta T]\).
\( \textbf{Substitution:} \) \(R_T=500[1+(-2.0\times10^{-2})(10)]\).
\( \textbf{Coefficient product:} \) \((-2.0\times10^{-2})(10)=-0.20\).
\( \textbf{Bracket value:} \) \(1-0.20=0.80\).
\( \textbf{Resistance:} \) \(R_T=500(0.80)=400\,\Omega\).
\( \textbf{Final answer:} \) \(R_T=400\,\Omega\); the negative sign of \(\alpha\) is the key reason resistance decreases when temperature rises.
105. Use the graph description below. The vertical axis represents resistance \(R\), and the horizontal axis represents temperature \(T\). For a metal over a moderate temperature range, the graph is approximately a straight line with positive slope. What does the positive slope indicate?
ⓐ. The metal has negative temperature coefficient of resistance
ⓑ. The resistance decreases when temperature increases
ⓒ. The resistance increases when temperature increases
ⓓ. The resistance is independent of temperature
Correct Answer: The resistance increases when temperature increases
Explanation: In an \(R\)-versus-\(T\) graph, the slope shows how resistance changes with temperature. A positive slope means \(R\) increases as \(T\) increases. This is typical for metallic conductors over ordinary temperature ranges. The microscopic reason is that increased lattice vibrations cause more frequent collisions of drifting electrons. A negative slope would be expected for many semiconductors, not for ordinary metals. The graph-interpretation mistake is to read “positive slope” as merely “positive resistance,” but it actually describes the trend with temperature.
106. If the resistance-temperature graph of a material slopes downward as temperature increases, which material type is most likely represented?
ⓐ. A pure metal
ⓑ. A copper connecting wire at constant temperature
ⓒ. A perfect conductor with fixed resistance
ⓓ. A semiconductor
Correct Answer: A semiconductor
Explanation: A downward \(R\)-versus-\(T\) graph means resistance decreases as temperature increases. This corresponds to a negative temperature coefficient of resistance. Semiconductors commonly show this behaviour because heating can produce more charge carriers, increasing conductivity and reducing resistance. Metals generally show the opposite trend because higher temperature increases lattice scattering. A connecting wire held at constant temperature would not show a downward temperature trend. The sign of the slope is a quick way to distinguish semiconductor-like behaviour from metallic behaviour.
107. A resistance thermometer uses a platinum wire. Which property makes platinum suitable for this purpose?
ⓐ. Its resistance is exactly zero at room temperature
ⓑ. Its resistance decreases sharply with temperature like a semiconductor
ⓒ. Its resistance is independent of temperature
ⓓ. Its resistance changes predictably with temperature
Correct Answer: Its resistance changes predictably with temperature
Explanation: A resistance thermometer works by measuring temperature through the change in resistance of a material. Platinum is useful because its resistance changes in a regular and reproducible way with temperature over a wide range. For a metal like platinum, resistance generally increases with temperature. The material should not have zero resistance at ordinary temperature, because then small temperature changes would not be measured through ordinary resistance changes. The key idea is controlled and predictable variation, not absence of resistance.
108. A wire has resistance \(10\,\Omega\) at \(0^\circ\text{C}\) and \(12\,\Omega\) at \(100^\circ\text{C}\). Assuming a linear temperature dependence, what is the temperature coefficient of resistance referred to \(0^\circ\text{C}\)?
ⓐ. \(2.0\times10^{-4}\,^\circ\text{C}^{-1}\)
ⓑ. \(2.0\times10^{-3}\,^\circ\text{C}^{-1}\)
ⓒ. \(1.2\times10^{-2}\,^\circ\text{C}^{-1}\)
ⓓ. \(5.0\times10^{-3}\,^\circ\text{C}^{-1}\)
Correct Answer: \(2.0\times10^{-3}\,^\circ\text{C}^{-1}\)
Explanation: \( \textbf{Given:} \) \(R_0=10\,\Omega\) at \(0^\circ\text{C}\), \(R_T=12\,\Omega\) at \(100^\circ\text{C}\), and \(\Delta T=100^\circ\text{C}\).
\( \textbf{Formula:} \) \(R_T=R_0[1+\alpha\Delta T]\).
\( \textbf{Rearrangement:} \) \(\frac{R_T}{R_0}=1+\alpha\Delta T\).
\( \textbf{Substitution:} \) \(\frac{12}{10}=1+\alpha(100)\).
\( \textbf{Simplification:} \) \(1.2=1+100\alpha\).
\( \textbf{Solving:} \) \(100\alpha=0.2\), so \(\alpha=\frac{0.2}{100}=0.002\).
\( \textbf{Result:} \) \(\alpha=2.0\times10^{-3}\,^\circ\text{C}^{-1}\).
\( \textbf{Final answer:} \) \(\alpha=2.0\times10^{-3}\,^\circ\text{C}^{-1}\); divide the fractional change in resistance by the temperature change.
109. Which statement about alloys used in standard resistors is most appropriate?
ⓐ. They are preferred because their resistance changes enormously with small temperature changes
ⓑ. They are preferred because their resistance is always zero
ⓒ. They are preferred because they behave exactly like semiconductors
ⓓ. They are preferred because their temperature coefficient of resistance is usually small
Correct Answer: They are preferred because their temperature coefficient of resistance is usually small
Explanation: Standard resistors should keep their resistance nearly constant during ordinary temperature changes. Many alloys have a small temperature coefficient of resistance, so their resistance changes only slightly with temperature. This makes them useful in resistance boxes and measuring instruments. If the resistance changed greatly with temperature, the resistor value would become unreliable during use. Alloys are not chosen because they have zero resistance or semiconductor-like negative temperature behaviour. The practical requirement is stability of resistance.
110. A metal wire has resistance \(R_0\) at temperature \(T_0\). If its temperature is increased by \(\Delta T\) and its temperature coefficient is positive, which expression correctly gives the fractional increase in resistance?
ⓐ. \(\frac{\Delta R}{R_0}=\alpha\Delta T\)
ⓑ. \(\frac{\Delta R}{R_0}=\frac{\Delta T}{\alpha}\)
ⓒ. \(\frac{\Delta R}{R_0}=\frac{\alpha}{\Delta T}\)
ⓓ. \(\frac{\Delta R}{R_0}=1+\alpha\Delta T\)
Correct Answer: \(\frac{\Delta R}{R_0}=\alpha\Delta T\)
Explanation: \( \textbf{Temperature relation:} \) \(R_T=R_0[1+\alpha\Delta T]\).
\( \textbf{Expanding:} \) \(R_T=R_0+R_0\alpha\Delta T\).
\( \textbf{Change in resistance:} \) \(\Delta R=R_T-R_0\).
\( \textbf{Substitution:} \) \(\Delta R=(R_0+R_0\alpha\Delta T)-R_0\).
\( \textbf{Simplification:} \) \(\Delta R=R_0\alpha\Delta T\).
\( \textbf{Fractional change:} \) \(\frac{\Delta R}{R_0}=\alpha\Delta T\).
\( \textbf{Final answer:} \) \(\frac{\Delta R}{R_0}=\alpha\Delta T\); the expression \(1+\alpha\Delta T\) gives the ratio \(\frac{R_T}{R_0}\), not the fractional increase.
111. Consider the following statements about temperature coefficient of resistance.
Statement I: A positive \(\alpha\) means resistance increases with temperature.
Statement II: A negative \(\alpha\) means resistance decreases with temperature.
Statement III: If \(\alpha=0\), resistance is approximately independent of temperature over the range considered.
Which statements are correct?
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I, II and III
Explanation: Statement I is correct because \(R_T=R_0[1+\alpha(T-T_0)]\) gives a larger resistance for a temperature rise when \(\alpha\) is positive. Statement II is correct because a negative \(\alpha\) makes the bracket smaller for a temperature rise, so resistance decreases. Statement III is also correct within the linear approximation because \(\alpha=0\) makes \(R_T\approx R_0\). Metals usually have positive \(\alpha\), semiconductors often have negative \(\alpha\), and some alloys have very small \(\alpha\). The sign of \(\alpha\) gives the direction of change, while its magnitude gives how strong the change is.
112. A resistor of \(40\,\Omega\) at \(25^\circ\text{C}\) is heated to \(75^\circ\text{C}\). Its temperature coefficient is \(5.0\times10^{-3}\,^\circ\text{C}^{-1}\). If this resistor is connected to a constant \(10\,\text{V}\) supply at \(75^\circ\text{C}\), what current flows?
ⓐ. \(0.20\,\text{A}\)
ⓑ. \(0.25\,\text{A}\)
ⓒ. \(0.16\,\text{A}\)
ⓓ. \(0.50\,\text{A}\)
Correct Answer: \(0.20\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(R_0=40\,\Omega\), \(T_0=25^\circ\text{C}\), \(T=75^\circ\text{C}\), \(\alpha=5.0\times10^{-3}\,^\circ\text{C}^{-1}\), and \(V=10\,\text{V}\).
\( \textbf{Temperature change:} \) \(\Delta T=75^\circ\text{C}-25^\circ\text{C}=50^\circ\text{C}\).
\( \textbf{Resistance relation:} \) \(R_T=R_0[1+\alpha\Delta T]\).
\( \textbf{Substitution:} \) \(R_T=40[1+(5.0\times10^{-3})(50)]\).
\( \textbf{Product:} \) \((5.0\times10^{-3})(50)=0.25\).
\( \textbf{New resistance:} \) \(R_T=40(1.25)=50\,\Omega\).
\( \textbf{Current relation:} \) At the final temperature, \(I=\frac{V}{R_T}\).
\( \textbf{Current calculation:} \) \(I=\frac{10}{50}=0.20\,\text{A}\).
\( \textbf{Final answer:} \) \(I=0.20\,\text{A}\); using the original \(40\,\Omega\) would ignore the heating effect.
113. In a four-band carbon resistor, what do the first, second, third, and fourth bands usually represent respectively?
ⓐ. Multiplier, first digit, second digit, tolerance
ⓑ. First digit, second digit, multiplier, tolerance
ⓒ. Tolerance, multiplier, first digit, second digit
ⓓ. First digit, multiplier, tolerance, second digit
Correct Answer: First digit, second digit, multiplier, tolerance
Explanation: In a standard four-band carbon resistor, the first band gives the first significant digit. The second band gives the second significant digit. The third band gives the power-of-ten multiplier. The fourth band gives tolerance, such as gold for \(\pm5\%\) and silver for \(\pm10\%\). A common mistake is to read the multiplier band as a third significant digit, which can change the resistance by a large factor.
114. A carbon resistor has colour bands brown, black, red, and gold. What is its resistance value with tolerance?
ⓐ. \(10\,\Omega \pm 5\%\)
ⓑ. \(100\,\Omega \pm 10\%\)
ⓒ. \(1.0\,\text{k}\Omega \pm 5\%\)
ⓓ. \(2.0\,\text{k}\Omega \pm 5\%\)
Correct Answer: \(1.0\,\text{k}\Omega \pm 5\%\)
Explanation: \( \textbf{Band meanings:} \) Brown is \(1\), black is \(0\), red is multiplier \(10^2\), and gold is tolerance \(\pm5\%\).
\( \textbf{Significant digits:} \) The first two bands form \(10\).
\( \textbf{Multiplier step:} \) Red means multiply by \(10^2\).
\( \textbf{Resistance calculation:} \) \(10\times10^2\,\Omega=1000\,\Omega\).
\( \textbf{Unit conversion:} \) \(1000\,\Omega=1.0\,\text{k}\Omega\).
\( \textbf{Tolerance:} \) Gold gives \(\pm5\%\).
\( \textbf{Final answer:} \) The resistor value is \(1.0\,\text{k}\Omega \pm 5\%\); treating red as the digit \(2\) would incorrectly give \(102\,\Omega\).
115. A resistor has colour bands yellow, violet, orange, and silver. What is the marked resistance?
ⓐ. \(4.7\,\text{k}\Omega \pm 10\%\)
ⓑ. \(47\,\Omega \pm 10\%\)
ⓒ. \(470\,\text{k}\Omega \pm 5\%\)
ⓓ. \(47\,\text{k}\Omega \pm 10\%\)
Correct Answer: \(47\,\text{k}\Omega \pm 10\%\)
Explanation: \( \textbf{First digit:} \) Yellow corresponds to \(4\).
\( \textbf{Second digit:} \) Violet corresponds to \(7\).
\( \textbf{Significant number:} \) The first two bands form \(47\).
\( \textbf{Multiplier:} \) Orange corresponds to \(10^3\).
\( \textbf{Resistance:} \) \(47\times10^3\,\Omega=47000\,\Omega\).
\( \textbf{In kilo-ohms:} \) \(47000\,\Omega=47\,\text{k}\Omega\).
\( \textbf{Tolerance:} \) Silver corresponds to \(\pm10\%\).
\( \textbf{Final answer:} \) The marked value is \(47\,\text{k}\Omega \pm 10\%\); orange is the multiplier band, not another digit after \(47\).
116. A carbon resistor is marked red, red, brown, and gold. What is the possible resistance range allowed by the tolerance?
ⓐ. \(198\,\Omega\) to \(242\,\Omega\)
ⓑ. \(209\,\Omega\) to \(231\,\Omega\)
ⓒ. \(220\,\Omega\) to \(270\,\Omega\)
ⓓ. \(22\,\Omega\) to \(220\,\Omega\)
Correct Answer: \(209\,\Omega\) to \(231\,\Omega\)
Explanation: \( \textbf{Colour values:} \) Red is \(2\), red is \(2\), brown is multiplier \(10^1\), and gold is tolerance \(\pm5\%\).
\( \textbf{Marked resistance:} \) The first two digits give \(22\).
\( \textbf{Applying multiplier:} \) \(22\times10^1\,\Omega=220\,\Omega\).
\( \textbf{Tolerance fraction:} \) \(\pm5\%\) means \(\pm\frac{5}{100}\) of the marked value.
\( \textbf{Tolerance amount:} \) \(5\%\) of \(220\,\Omega\) is \(0.05\times220=11\,\Omega\).
\( \textbf{Lower limit:} \) \(220\,\Omega-11\,\Omega=209\,\Omega\).
\( \textbf{Upper limit:} \) \(220\,\Omega+11\,\Omega=231\,\Omega\).
\( \textbf{Final answer:} \) The possible range is \(209\,\Omega\) to \(231\,\Omega\); tolerance changes the allowed range, not the central marked value.
117. Match the resistor colour with its digit or tolerance meaning.
| Column I | Column II |
|---|
| P. Green | 1. \(2\) |
| Q. Blue | 2. \(5\) |
| R. Red | 3. \(6\) |
| S. Gold | 4. \(\pm5\%\) |
Choose the correct matching.
ⓐ. P-2, Q-3, R-1, S-4
ⓑ. P-3, Q-2, R-1, S-4
ⓒ. P-2, Q-1, R-3, S-4
ⓓ. P-4, Q-3, R-1, S-2
Correct Answer: P-2, Q-3, R-1, S-4
Explanation: In the standard colour code, green represents the digit \(5\). Blue represents the digit \(6\). Red represents the digit \(2\), and as a multiplier band it would mean \(10^2\). Gold is not used as an ordinary significant digit in the standard resistor colour code; it commonly represents tolerance \(\pm5\%\). The key is to identify whether a band is being used as a digit, multiplier, or tolerance band.
118. A resistor is labelled \(3.3\,\text{k}\Omega \pm 10\%\). Which four-band colour code is correct?
ⓐ. Orange, orange, red, silver
ⓑ. Orange, orange, brown, gold
ⓒ. Red, orange, orange, silver
ⓓ. Orange, red, orange, silver
Correct Answer: Orange, orange, red, silver
Explanation: \( \textbf{Marked value:} \) \(3.3\,\text{k}\Omega=3300\,\Omega\).
\( \textbf{Scientific form for colour code:} \) \(3300\,\Omega=33\times10^2\,\Omega\).
\( \textbf{First two digits:} \) The digits are \(3\) and \(3\), both represented by orange.
\( \textbf{Multiplier:} \) The multiplier is \(10^2\), represented by red.
\( \textbf{Tolerance:} \) \(\pm10\%\) is represented by silver.
\( \textbf{Final answer:} \) The correct bands are orange, orange, red, silver; the multiplier band must represent \(10^2\), not the digit after the decimal point.
119. A carbon resistor has bands brown, green, green, and no tolerance band. What is its marked value and tolerance?
ⓐ. \(15\,\text{k}\Omega \pm 20\%\)
ⓑ. \(15\,\text{M}\Omega \pm 5\%\)
ⓒ. \(150\,\text{k}\Omega \pm 10\%\)
ⓓ. \(1.5\,\text{M}\Omega \pm 20\%\)
Correct Answer: \(1.5\,\text{M}\Omega \pm 20\%\)
Explanation: \( \textbf{Colour values:} \) Brown is \(1\), green is \(5\), and green as the multiplier is \(10^5\).
\( \textbf{Significant digits:} \) The first two bands give \(15\).
\( \textbf{Multiplier step:} \) The third band green gives \(10^5\).
\( \textbf{Resistance:} \) \(15\times10^5\,\Omega=1.5\times10^6\,\Omega\).
\( \textbf{Unit conversion:} \) \(1.5\times10^6\,\Omega=1.5\,\text{M}\Omega\).
\( \textbf{Tolerance:} \) No tolerance band usually means \(\pm20\%\).
\( \textbf{Final answer:} \) The resistor is \(1.5\,\text{M}\Omega \pm 20\%\); no colour does not mean zero tolerance.
120. Which statement about the third band in a four-band carbon resistor is correct?
ⓐ. It gives the third significant digit of the resistance
ⓑ. It gives the tolerance of the resistor
ⓒ. It gives the power-of-ten multiplier
ⓓ. It gives the maximum current through the resistor
Correct Answer: It gives the power-of-ten multiplier
Explanation: The third band in a standard four-band resistor gives the multiplier. The first two bands give the two significant digits of the resistance value. The multiplier tells how many powers of ten multiply those two digits. The fourth band gives tolerance, not the third band. This distinction is important because reading the multiplier as an extra digit gives a completely different resistance value.
