101. A device has \(V=2\,\text{V}\) when \(I=0.20\,\text{A}\), but \(V=8\,\text{V}\) when \(I=0.40\,\text{A}\). What conclusion follows from these two readings?
ⓐ. The ratio \(\frac{V}{I}\) changes, so it is non-ohmic in this range
ⓑ. The resistance remains constant at \(10\,\Omega\) for both readings
ⓒ. The device follows direct proportionality \(V\propto I\) in this range
ⓓ. The current unit changes between the two readings
Correct Answer: The ratio \(\frac{V}{I}\) changes, so it is non-ohmic in this range
Explanation: For an ohmic conductor under fixed physical conditions, the ratio \(\frac{V}{I}\) should remain constant. In the first reading, \(\frac{V}{I}=\frac{2}{0.20}=10\,\Omega\). In the second reading, \(\frac{V}{I}=\frac{8}{0.40}=20\,\Omega\). The resistance value inferred from the two readings is not the same. This shows that \(V\) is not proportional to \(I\) over the given range, even though both readings are valid measurements.
102. A graph of \(V\) versus \(I\) for an ohmic resistor is a straight line through the origin. If the slope of the line is \(8\,\text{V A}^{-1}\), the resistance is
ⓐ. \(0.125\,\Omega\)
ⓑ. \(4\,\Omega\)
ⓒ. \(8\,\Omega\)
ⓓ. \(16\,\Omega\)
Correct Answer: \(8\,\Omega\)
Explanation: \( \textbf{Graph type:} \) The vertical axis is \(V\), and the horizontal axis is \(I\).
\( \textbf{Slope of graph:} \)
\[
\text{slope}=\frac{\Delta V}{\Delta I}
\]
For a \(V\)-versus-\(I\) graph of an ohmic resistor, this slope equals \(R\).
\( \textbf{Given slope:} \)
\[
\frac{\Delta V}{\Delta I}=8\,\text{V A}^{-1}
\]
\( \textbf{Unit conversion:} \)
\[
1\,\text{V A}^{-1}=1\,\Omega
\]
\( \textbf{Resistance:} \)
\[
R=8\,\Omega
\]
The reciprocal slope would apply only if the graph were \(I\) versus \(V\).
\( \textbf{Final answer:} \) The resistance is \(8\,\Omega\).
103. Compare the two resistor graphs described below.
Two ohmic resistors \(P\) and \(Q\) have straight \(V\)-versus-\(I\) graphs through the origin. The line for \(P\) is steeper than the line for \(Q\).
The comparison of their resistances is
ⓐ. \(R_P\gt R_Q\)
ⓑ. \(R_P\lt R_Q\)
ⓒ. \(R_P=R_Q=0\)
ⓓ. \(R_P=\frac{1}{R_Q}\)
Correct Answer: \(R_P\gt R_Q\)
Explanation: In a \(V\)-versus-\(I\) graph, the slope is \(\frac{\Delta V}{\Delta I}\). For an ohmic resistor, this slope equals resistance \(R\). A steeper line on this graph means a larger value of \(\frac{V}{I}\). Therefore, resistor \(P\) has greater resistance than resistor \(Q\). The comparison would reverse only for an \(I\)-versus-\(V\) graph, where the slope represents conductance.
104. Compare the two current-voltage graphs described below.
Two ohmic resistors \(P\) and \(Q\) have straight \(I\)-versus-\(V\) graphs through the origin. The line for \(P\) is steeper than the line for \(Q\).
The comparison of their resistances is
ⓐ. \(R_P\gt R_Q\)
ⓑ. \(R_P=R_Q\)
ⓒ. \(R_P+R_Q=0\)
ⓓ. \(R_P\lt R_Q\)
Correct Answer: \(R_P\lt R_Q\)
Explanation: In an \(I\)-versus-\(V\) graph, the slope is \(\frac{\Delta I}{\Delta V}\). From Ohm's law, \(I=\frac{V}{R}\), so the slope of this graph is \(\frac{1}{R}\). A steeper \(I\)-versus-\(V\) line means a larger value of \(\frac{1}{R}\). A larger reciprocal resistance means smaller resistance. Thus the steeper line in an \(I\)-versus-\(V\) plot belongs to the smaller resistor.
105. The same resistor is represented on two different graphs. One graph plots \(V\) against \(I\), and the other plots \(I\) against \(V\). If the resistor has resistance \(5\,\Omega\), the slopes of the two graphs respectively are
ⓐ. \(5\,\Omega\) and \(0.20\,\Omega^{-1}\)
ⓑ. \(5\,\Omega\) and \(5\,\Omega\)
ⓒ. \(0.20\,\Omega^{-1}\) and \(5\,\Omega\)
ⓓ. \(0.20\,\Omega^{-1}\) and \(0.20\,\Omega^{-1}\)
Correct Answer: \(5\,\Omega\) and \(0.20\,\Omega^{-1}\)
Explanation: \( \textbf{Given resistance:} \) \(R=5\,\Omega\).
\( \textbf{\(V\)-against-\(I\) graph:} \)
\[
\text{slope}=\frac{\Delta V}{\Delta I}=R
\]
So the first slope is \(5\,\Omega\).
\( \textbf{\(I\)-against-\(V\) graph:} \)
\[
\text{slope}=\frac{\Delta I}{\Delta V}=\frac{1}{R}
\]
\( \textbf{Second slope:} \)
\[
\frac{1}{R}=\frac{1}{5}=0.20\,\Omega^{-1}
\]
The two graphs use interchanged axes, so their slopes are reciprocals.
\( \textbf{Final answer:} \) The slopes are \(5\,\Omega\) and \(0.20\,\Omega^{-1}\), respectively.
106. A graph of \(I\) on the vertical axis and \(V\) on the horizontal axis passes through the origin and the point \((V,I)=(12\,\text{V},3\,\text{A})\). The resistance of the conductor is
ⓐ. \(0.25\,\Omega\)
ⓑ. \(4\,\Omega\)
ⓒ. \(3\,\Omega\)
ⓓ. \(36\,\Omega\)
Correct Answer: \(4\,\Omega\)
Explanation: \( \textbf{Graph axes:} \) Vertical axis is \(I\), and horizontal axis is \(V\).
\( \textbf{Point on graph:} \) \(V=12\,\text{V}\), \(I=3\,\text{A}\).
\( \textbf{Slope of \(I\)-versus-\(V\) graph:} \)
\[
\text{slope}=\frac{I}{V}=\frac{3}{12}=0.25\,\text{A V}^{-1}
\]
This slope is conductance, equal to \(\frac{1}{R}\).
\( \textbf{Resistance:} \)
\[
R=\frac{1}{0.25}=4\,\Omega
\]
The value \(0.25\,\Omega\) would confuse conductance with resistance.
\( \textbf{Final answer:} \) The resistance is \(4\,\Omega\).
107. Consider the following statements about ohmic graph slopes.
I. Slope of a \(V\)-versus-\(I\) graph gives \(R\).
II. Slope of an \(I\)-versus-\(V\) graph gives \(\frac{1}{R}\).
III. A steeper \(I\)-versus-\(V\) line represents larger \(R\).
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I and II only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is true because \(V=IR\), so \(\frac{V}{I}=R\) for an ohmic resistor. Statement II is also true because \(I=\frac{V}{R}\), so \(\frac{I}{V}=\frac{1}{R}\). Statement III is false because a steeper \(I\)-versus-\(V\) line means a larger value of \(\frac{1}{R}\), not larger \(R\). Therefore, in an \(I\)-\(V\) plot, the steeper line belongs to the smaller resistance. The axis order must always be checked before interpreting slope.
108. A resistor is tested at constant temperature. Its \(V\)-\(I\) graph passes through \((0.5\,\text{A},2\,\text{V})\) and \((2.0\,\text{A},8\,\text{V})\). The slope and resistance are
ⓐ. \(2\,\text{V A}^{-1}\) and \(2\,\Omega\)
ⓑ. \(6\,\text{V A}^{-1}\) and \(6\,\Omega\)
ⓒ. \(4\,\text{V A}^{-1}\) and \(4\,\Omega\)
ⓓ. \(8\,\text{V A}^{-1}\) and \(8\,\Omega\)
Correct Answer: \(4\,\text{V A}^{-1}\) and \(4\,\Omega\)
Explanation: \( \textbf{First point:} \) \(I_1=0.5\,\text{A}\), \(V_1=2\,\text{V}\).
\( \textbf{Second point:} \) \(I_2=2.0\,\text{A}\), \(V_2=8\,\text{V}\).
\( \textbf{Slope of \(V\)-\(I\) graph:} \)
\[
\text{slope}=\frac{\Delta V}{\Delta I}
\]
\( \textbf{Substitution:} \)
\[
\text{slope}=\frac{8-2}{2.0-0.5}
\]
\( \textbf{Calculation:} \)
\[
\text{slope}=\frac{6}{1.5}=4\,\text{V A}^{-1}
\]
For a \(V\)-versus-\(I\) graph, this slope equals \(R\).
\( \textbf{Resistance:} \)
\[
R=4\,\Omega
\]
Using only \(\frac{8}{2}\) also works here because the straight line passes through the origin, but the slope method is safer for graph data.
\( \textbf{Final answer:} \) The slope is \(4\,\text{V A}^{-1}\), and the resistance is \(4\,\Omega\).
109. A straight \(V\)-\(I\) graph for a conductor does not pass through the origin. The most careful conclusion is that
ⓐ. it always represents an ideal ohmic resistor
ⓑ. it is not the standard ohmic-resistor graph through the origin
ⓒ. the ratio \(\frac{V}{I}\) is automatically constant for all readings
ⓓ. its slope must be zero
Correct Answer: it is not the standard ohmic-resistor graph through the origin
Explanation: For an ideal ohmic resistor under constant physical conditions, the relation is \(V=IR\). This gives a straight line passing through the origin on a \(V\)-\(I\) graph. If a straight graph does not pass through the origin, \(V\) is not simply proportional to \(I\) in the standard resistor form. The slope may still have a mathematical value, but the graph is not the usual origin-passing ohmic relation. A nonzero intercept needs a separate physical explanation before treating it as a simple resistor.
110. The unit of slope of a \(V\)-versus-\(I\) graph is
ⓐ. \(\text{A V}^{-1}\)
ⓑ. \(\text{V A}^{-1}\)
ⓒ. \(\text{C s}^{-1}\)
ⓓ. \(\text{J C}^{-1}\)
Correct Answer: \(\text{V A}^{-1}\)
Explanation: A \(V\)-versus-\(I\) graph has potential difference on the vertical axis and current on the horizontal axis. Its slope is \(\frac{\Delta V}{\Delta I}\). The unit is therefore \(\frac{\text{V}}{\text{A}}=\text{V A}^{-1}\). Since \(1\,\text{V A}^{-1}=1\,\Omega\), this slope is measured in ohm. The unit \(\text{A V}^{-1}\) belongs to the slope of an \(I\)-versus-\(V\) graph.
111. Study the table and choose the row that matches the graph type with the slope meaning.
| Row | Graph | Slope meaning for ohmic resistor |
|---|
| P | \(V\) versus \(I\) | \(R\) |
| Q | \(I\) versus \(V\) | \(R\) |
| R | \(V\) versus \(I\) | \(\frac{1}{R}\) |
| S | \(I\) versus \(V\) | \(\frac{1}{R}\) |
ⓐ. Q and R only
ⓑ. P and Q only
ⓒ. P and S only
ⓓ. R and S only
Correct Answer: P and S only
Explanation: For a \(V\)-versus-\(I\) graph, the slope is \(\frac{\Delta V}{\Delta I}\), which equals \(R\). Therefore, row P is suitable. For an \(I\)-versus-\(V\) graph, the slope is \(\frac{\Delta I}{\Delta V}\), which equals \(\frac{1}{R}\). Therefore, row S is suitable. Rows Q and R interchange the meanings of the two graph slopes, which is a common axis-reading error.
112. A resistor \(P\) has an \(I\)-versus-\(V\) graph slope of \(0.50\,\text{A V}^{-1}\), while resistor \(Q\) has an \(I\)-versus-\(V\) graph slope of \(0.20\,\text{A V}^{-1}\). The resistance ratio \(\frac{R_P}{R_Q}\) is
ⓐ. \(0.50\)
ⓑ. \(2.5\)
ⓒ. \(5.0\)
ⓓ. \(0.40\)
Correct Answer: \(0.40\)
Explanation: \( \textbf{Slope for \(P\):} \) \(m_P=0.50\,\text{A V}^{-1}\).
\( \textbf{Slope for \(Q\):} \) \(m_Q=0.20\,\text{A V}^{-1}\).
For an \(I\)-versus-\(V\) graph, slope is conductance:
\[
m=\frac{1}{R}
\]
So resistance is the reciprocal of the slope.
\( \textbf{Resistance of \(P\):} \)
\[
R_P=\frac{1}{0.50}=2\,\Omega
\]
\( \textbf{Resistance of \(Q\):} \)
\[
R_Q=\frac{1}{0.20}=5\,\Omega
\]
\( \textbf{Ratio:} \)
\[
\frac{R_P}{R_Q}=\frac{2}{5}=0.40
\]
The larger \(I\)-\(V\) slope belongs to the smaller resistance.
\( \textbf{Final answer:} \) The ratio \(\frac{R_P}{R_Q}\) is \(0.40\).
113. In a graph-based comparison, line \(X\) is steeper than line \(Y\). Without knowing the axes, this information alone is insufficient to decide which resistor is larger because
ⓐ. current has no definite SI unit in graph analysis
ⓑ. ohmic resistors never produce origin-passing straight lines
ⓒ. slope depends on whether \(V\) or \(I\) is on each axis
ⓓ. resistance cannot be compared from voltage-current data
Correct Answer: slope depends on whether \(V\) or \(I\) is on each axis
Explanation: A steeper line has a larger slope, but the physical meaning of slope depends on the axes. In a \(V\)-versus-\(I\) graph, slope equals \(R\), so a steeper line means larger resistance. In an \(I\)-versus-\(V\) graph, slope equals \(\frac{1}{R}\), so a steeper line means smaller resistance. Therefore, the axes must be known before comparing resistors from graph steepness. The same visual word steeper can lead to opposite resistance conclusions in the two graph types.
114. The resistance of a uniform conductor of length \(l\), cross-sectional area \(A\), and resistivity \(\rho\) is
ⓐ. \(R=\rho\frac{l}{A}\)
ⓑ. \(R=\rho\frac{A}{l}\)
ⓒ. \(R=\frac{lA}{\rho}\)
ⓓ. \(R=\frac{\rho}{lA}\)
Correct Answer: \(R=\rho\frac{l}{A}\)
Explanation: The resistance of a uniform conductor depends on its material and geometry. The material is represented by resistivity \(\rho\), while the geometry is represented by length \(l\) and cross-sectional area \(A\). The relation is \(R=\rho\frac{l}{A}\). This shows that resistance increases with length and decreases with area. A long thin wire has more resistance than a short thick wire made of the same material.
115. A wire is made longer while its material and cross-sectional area remain unchanged. Its resistance increases because
ⓐ. electrons must drift through a longer conducting path
ⓑ. charge carriers get more conducting paths in parallel
ⓒ. resistivity becomes zero when length increases
ⓓ. current density stops depending on area
Correct Answer: electrons must drift through a longer conducting path
Explanation: For a uniform wire, \(R=\rho\frac{l}{A}\). If the material is unchanged, \(\rho\) remains the same. If the area is also unchanged, resistance is directly proportional to length \(l\). A longer wire offers more opposition to the motion of charge carriers because the conducting path is longer. The increase comes from geometry, not from resistivity becoming zero or current density losing meaning.
116. Two wires are made of the same material and have the same length. Wire \(P\) has twice the cross-sectional area of wire \(Q\). The resistance comparison is
ⓐ. \(R_P=2.0R_Q\)
ⓑ. \(R_P=1.0R_Q\)
ⓒ. \(R_P=4.0R_Q\)
ⓓ. \(R_P=0.5R_Q\)
Correct Answer: \(R_P=0.5R_Q\)
Explanation: For wires of the same material and length, \(\rho\) and \(l\) are the same. The resistance relation is \(R=\rho\frac{l}{A}\), so resistance is inversely proportional to area. If wire \(P\) has twice the area of wire \(Q\), its resistance is half as large. A larger cross-section provides more conducting area for charge flow. The comparison is inverse, not direct, because area appears in the denominator of the resistance formula.
117. A wire of resistance \(R\) is stretched uniformly so that its length becomes \(2l\), while its volume remains unchanged. Its new resistance is
ⓐ. \(\frac{R}{2}\)
ⓑ. \(4R\)
ⓒ. \(R\)
ⓓ. \(2R\)
Correct Answer: \(4R\)
Explanation: \( \textbf{Original resistance:} \)
\[
R=\rho\frac{l}{A}
\]
\( \textbf{New length:} \) \(l'=2l\).
\( \textbf{Volume condition:} \) The wire is stretched uniformly, so volume remains constant.
\[
Al=A'l'
\]
\( \textbf{New area:} \)
\[
A'=\frac{Al}{2l}=\frac{A}{2}
\]
\( \textbf{New resistance:} \)
\[
R'=\rho\frac{l'}{A'}=\rho\frac{2l}{A/2}
\]
\( \textbf{Simplification:} \)
\[
R'=4\rho\frac{l}{A}=4R
\]
The area change is as important as the length change in a stretching problem.
\( \textbf{Final answer:} \) The new resistance is \(4R\).
118. A uniform wire is compressed without changing its volume so that its length becomes \(\frac{l}{2}\). If its original resistance was \(R\), the new resistance is
ⓐ. \(\frac{R}{2}\)
ⓑ. \(2R\)
ⓒ. \(4R\)
ⓓ. \(\frac{R}{4}\)
Correct Answer: \(\frac{R}{4}\)
Explanation: \( \textbf{Original relation:} \)
\[
R=\rho\frac{l}{A}
\]
\( \textbf{New length:} \)
\[
l'=\frac{l}{2}
\]
\( \textbf{Volume remains constant:} \)
\[
Al=A'l'
\]
\( \textbf{New area:} \)
\[
A'=\frac{Al}{l/2}=2A
\]
\( \textbf{New resistance:} \)
\[
R'=\rho\frac{l'}{A'}=\rho\frac{l/2}{2A}
\]
\( \textbf{Simplification:} \)
\[
R'=\frac{1}{4}\rho\frac{l}{A}=\frac{R}{4}
\]
Shortening alone would suggest \(\frac{R}{2}\), but the doubled area gives another factor of \(\frac{1}{2}\).
\( \textbf{Final answer:} \) The new resistance is \(\frac{R}{4}\).
119. Two wires of the same material have the same volume. Wire \(P\) is twice as long as wire \(Q\). The ratio \(\frac{R_P}{R_Q}\) is
ⓐ. \(1\)
ⓑ. \(4\)
ⓒ. \(2\)
ⓓ. \(\frac{1}{4}\)
Correct Answer: \(4\)
Explanation: \( \textbf{Same material:} \) Both wires have the same resistivity \(\rho\).
\( \textbf{Same volume condition:} \)
\[
V_{\text{wire}}=Al
\]
So \(A=\frac{V_{\text{wire}}}{l}\).
\( \textbf{Resistance formula:} \)
\[
R=\rho\frac{l}{A}
\]
\( \textbf{Substitute area from volume:} \)
\[
R=\rho\frac{l}{V_{\text{wire}}/l}
\]
\( \textbf{Simplify:} \)
\[
R=\rho\frac{l^2}{V_{\text{wire}}}
\]
For the same material and same volume, \(R\propto l^2\).
\( \textbf{Length comparison:} \) \(l_P=2l_Q\).
\[
\frac{R_P}{R_Q}=\left(\frac{l_P}{l_Q}\right)^2=2^2=4
\]
The equal-volume condition changes the comparison from \(R\propto l\) to \(R\propto l^2\).
\( \textbf{Final answer:} \) The resistance ratio is \(4\).
120. A cylindrical wire is replaced by another wire of the same material and same length, but with twice the radius. The resistance of the new wire is
ⓐ. \(\frac{R}{2}\)
ⓑ. \(2R\)
ⓒ. \(4R\)
ⓓ. \(\frac{R}{4}\)
Correct Answer: \(\frac{R}{4}\)
Explanation: For a wire, \(R=\rho\frac{l}{A}\). The material and length are unchanged, so \(\rho\) and \(l\) remain constant. The cross-sectional area of a cylindrical wire is \(A=\pi r^2\). If the radius becomes \(2r\), the new area becomes \(A'=\pi(2r)^2=4\pi r^2=4A\). Since resistance is inversely proportional to area, the new resistance becomes \(\frac{R}{4}\). The square on radius is the key reason the resistance does not merely become \(\frac{R}{2}\).
