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Current Electricity Past Papers
Solved past paper MCQs for Current Electricity from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Download Solved Current Electricity Past Papers (Print Edition)
Official printable A4 booklet containing all 92 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
Which one of the following materials has negative temperature coefficient of resistance?
A
Copper
B
Germanium
C
Sulphur
D
Zinc
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A negative temperature coefficient means the material's resistance drops as temperature goes up.
Solution:
Metals (like Copper and Zinc) have positive coefficients because heat increases atomic vibrations and scattering.
Semiconductors, like Germanium (and Silicon), have negative coefficients. Heating them frees massive amounts of valence electrons across the band gap, vastly improving conductivity and plummeting resistance.
Why other options are incorrect:
Opt A & D: Metals. Resistance increases with heat.
The magnitude of the current in metals is proportional to the potential difference cross it as long as temperature of conductor is kept constant is known as:
A
Joule's law
B
Gauss law
C
Ohm's law
D
Ampere's law
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The foundational law of basic electrical circuits connects Voltage, Current, and Resistance.
Solution:
The precise statement: "Current is directly proportional to voltage (\( I \propto V \)) provided physical state and temperature are constant" is the exact definition of Ohm's law.
Why other options are incorrect:
Opt A: Joule's Law deals with power and heat (\( P = I^2R \)).
Opt B & D: Deals with electrostatics and magnetism, not circuit resistance.
Which graph explains non-ohmic material whose resistance decreases?
A
Graph showing a straight line curving horizontally towards the V-axis (Slope decreases)
B
Graph showing a straight line curving vertically towards the I-axis (Slope increases)
C
Graph showing a perfect linear straight line from the origin
D
Graph showing a horizontal flat line
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
In an I-V graph (Current on the y-axis, Voltage on the x-axis), the slope of the curve at any point dictates the conductance. Resistance is the inverse of that slope.
The substance having negative temperature coefficient are:
A
Insulators
B
Conductors
C
Semi-conductor
D
Alloys
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Materials respond to heat differently based on their electron valence bands.
Solution:
Semi-conductors naturally have very few free charge carriers at room temperature.
When heated, the thermal energy breaks bonds, suddenly flooding the material with electrons and holes.
This surge in charge carriers vastly decreases the overall resistance.
Because resistance drops as temperature rises, this mathematical relationship is termed a negative temperature coefficient.
Why other options are incorrect:
Opt B & D: Conductors and most alloys have positive coefficients (resistance increases with heat).
Opt A: While insulators do technically decrease resistance slightly when heated, semiconductors are the defining textbook standard for this specific measurable coefficient property.
No current flows between two charged bodies if they have same ____.
A
charge
B
potential
C
capacity
D
density
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Current strictly requires a gradient (a difference) to flow, much like water needs a height difference.
Solution:
Electric current is driven exclusively by a Potential Difference (Voltage).
If two bodies are connected and sit at the exact same electric potential, the voltage difference between them is zero (\( \Delta V = 0 \)).
By Ohm's Law (\( I = \Delta V / R \)), if \( \Delta V = 0 \), then current \( I = 0 \).
Why other options are incorrect:
Opt A: Two bodies can have identical charge amounts but different potentials if their physical sizes (capacitance) differ. Current will still flow until potentials equalize.
Opt C & D: These do not inherently stop current flow if a potential difference exists.
Ohm's law state that electric current through a conductor is proportional to the applied voltage provided:
A
electric current is constant
B
electric field is constant
C
resistance is constant
D
electric charge is constant
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Ohm's Law holds true only under strict linear conditions.
Solution:
The law states \( V \to I \) linearly.
The mathematical constant of proportionality that makes this a firm equation (\( V = IR \)) is the Resistance (\(R\)).
For the relationship to remain perfectly proportional (a straight line graph), the resistance must remain absolutely constant. (This usually implies keeping physical states like temperature constant).
Why other options are incorrect:
Opt A, B, D: These are not the defined physical constraints required to maintain the proportionality of Ohm's Law.
A battery has an emf of 6.0V and an internal resistance of 0.4 \(\Omega\). It is connected to a 2.60 resistor through a switch. When switch is open, the potential difference across the switch is:
A
0V
B
6.0 V
C
2.6 V
D
5.2 V
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
An open switch effectively creates a circuit with infinite resistance, halting all current flow.
Solution:
When the switch is Open, Current \( I = 0 \text{ A} \).
Because there is no current, there is zero voltage drop across the internal resistance of the battery (\(Ir = 0 \times 0.4 = 0\text{V}\)) and zero voltage drop across the external 2.6\(\Omega\) resistor (\(IR = 0 \times 2.6 = 0\text{V}\)).
Therefore, the entirety of the battery's EMF is "waiting" across the break in the circuit (the open switch).
The potential difference across the switch is exactly the full EMF: 6.0 V.
Why other options are incorrect:
Opt A: A closed (on) switch has 0V across it, an open switch blocks the full voltage.
Opt D: This would be the terminal voltage if the switch was closed (\( I = 6/(2.6+0.4) = 2\text{A} \), \( V = 2 \times 2.6 = 5.2\text{V} \)).
The following formula can be used to determine the resistance of a length of conductor. \( R = \rho l/A \). In the formula, the symbol \( \rho \) stands for the:
A
Cross-sectional area of the conductor in \( \text{m}^2 \)
B
Product of the length of the conductor in metes
C
Resistivity of the material in units of ohm-meters
D
Resistance of the conductor in units of ohms per meter
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The formula connects bulk resistance to intrinsic material properties.
Solution:
In the equation \( R = \rho \frac{l}{A} \), \( R \) is resistance, \( l \) is length, and \( A \) is the cross-sectional area.
The symbol \( \rho \) (rho) represents the constant of proportionality known as Resistivity (or specific resistance).
Its SI unit is Ohm-meters (\(\Omega \cdot \text{m}\)).
Why other options are incorrect:
Opt A: Area is represented by \( A \).
Opt D: Resistance per meter is \( R/l \), not \( \rho \).
The magnitude of the current in metals is proportional to the applied voltage as long as temperature of conductor is kept constant. It is statement of:
A
Joule's Law
B
Gauss Law
C
Ohm's Law
D
Ampere's Law
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The linear relationship between current and voltage under constant physical conditions is a fundamental principle of basic circuits.
Solution:
The statement "\( V \propto I \) at constant temperature" is the exact textbook definition of Ohm's Law.
Why other options are incorrect:
Opt A: Joule's Law deals with heat dissipation (\(P = I^2R\)).
Opt B & D: Gauss's and Ampere's laws deal with electromagnetics and flux, not basic circuit resistance.
In circuit models, a real battery is modeled as an ideal voltage source (EMF) paired with a resistor.
Solution:
Because 100% of the current leaving the battery MUST first fight through the battery's own internal electrolytes, this resistance physically lies in the exact path of the load.
Therefore, it acts completely in series with the external load.
Why other options are incorrect:
Opt A: If it were parallel, current could completely bypass it.
Opt D: Only ideal theoretical batteries have zero internal resistance.
The electric potential sets across the terminals of a battery is called:
A
Potential difference
B
EMF
C
Internal potential drop
D
Terminal voltage
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
While EMF represents the battery's theoretical maximum capability, the actual measurable voltage across its output nodes under operation gets a distinct name.
Solution:
The specific voltage physically measurable across the outside connecting points (terminals) of a functioning cell is explicitly called the Terminal voltage (\(V_t\)).
Why other options are incorrect:
Opt A: This is a general physics term, not specific to battery nodes.
Opt B: EMF is the voltage when the circuit is entirely open (no current).
Opt C: This is the voltage lost inside the battery (\(Ir\)), not what is available at the terminals.
A soft cylindrical electrical conducting wire has resistance R. It is stretched so that its length is doubled, but its radius stays constant. What would be the new resistance?
A
R/2
B
R
C
4R
D
2R
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Resistance is proportionally linked to length and cross-sectional area. Usually stretching implies volume remains constant (decreasing area), but the prompt places a strict hypothetical constraint.
Formula:
$$ R = \rho \frac{L}{A} $$
Solution:
The prompt explicitly states "but its radius stays constant". This means Area (\( A = \pi r^2 \)) does not change.
If Area is constant, Resistance is directly and linearly proportional to Length (\( R \propto L \)).
Since Length is doubled (\( L' = 2L \)), the resistance simply doubles.
New resistance = 2R.
Why other options are incorrect:
Opt C: 4R would be correct ONLY IF the wire's volume was conserved (meaning the area naturally shrank by half as it stretched). The prompt's constraint explicitly overrides this.
A 3 V battery is connected in series with ammeter and 2 ohm resistance after short circuiting. What will be reading of ammeter?
A
1 A
B
1.5 A
C
5 A
D
6 A
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
"After short circuiting" implies any other parallel load was bypassed, leaving only the 2 \(\Omega\) resistor in series with the ideal ammeter across the battery.
Formula:
$$ I = \frac{V}{R} $$
Solution:
Voltage \( V = 3 \text{ V} \).
Total circuit resistance \( R = 2 \, \Omega \) (assuming the ammeter is ideal and has 0 resistance).
Current \( I = \frac{3}{2} = 1.5 \text{ A} \).
The ammeter reads the total current, which is 1.5 A.
Why other options are incorrect:
Opt A, C, D: Simple math errors or applying incorrect formulas (like \( I = V \times R \)).
Which of the following statement is NOT CORRECT Kirchhoff's rule?
A
Kirchhoff's current rule based upon the law of conservation of charge
B
Wheatstone bridge is an application of Kirchhoff's rule
C
Kirchhoff's rules are more suitable in AC circuits
D
Kirchhoff's voltage rule based upon the law of conservation of energy
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Kirchhoff's circuit laws are foundational to analyzing complex circuits, mostly modeled around DC circuits.
Solution:
Opt A: True. KCL states sum of currents entering a junction equals sum leaving (conservation of charge).
Opt B: True. The bridge formula is derived using Kirchhoff's rules.
Opt D: True. KVL states the sum of voltages around a closed loop is zero (conservation of energy).
Opt C:False. Kirchhoff's rules are primarily designed and most suitable for DC circuits. In high-frequency AC circuits, parasitic capacitance and inductance make standard KVL/KCL assumptions (like instantaneous propagation) inaccurate without complex modifications.
Why other options are incorrect:
The question asks for the statement that is NOT correct, making C the target answer.
The total resistance of wire is inversely proportional to:
A
Length
B
Area
C
Temperature
D
Time
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Resistance is dictated by the physical dimensions of the conductor.
Formula:
$$ R = \rho \frac{L}{A} $$
Solution:
From the formula, \( R \) is in the numerator opposite to Area \( A \) in the denominator.
This mathematically proves that Resistance is inversely proportional to the Cross-sectional Area. (A thicker wire provides more path for electrons, reducing resistance).
Why other options are incorrect:
Opt A & C: Resistance is directly proportional to length and (typically) temperature.
A conductor has resistance R. If its length is stretched to twice the actual value and its radius is reduced to one third of its original values, the new resistance will be:
A
3R
B
9R
C
18R
D
27R
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Resistance changes drastically when both length and radius are altered.
Formula:
$$ R = \rho \frac{L}{\pi r^2} $$
Solution:
Original Resistance: \( R = \rho \frac{L}{\pi r^2} \).
New Length \( L' = 2L \).
New radius \( r' = \frac{r}{3} \).
New Area \( A' = \pi (r')^2 = \pi \left(\frac{r}{3}\right)^2 = \frac{\pi r^2}{9} = \frac{A}{9} \).
Two electric bulbs "A" and "B" of powers 500W and 2000W respectively are connected to 240V supply. The ratio of current passing through bulb "A" to the current passing through bulb "B" is:
A
1:2
B
1:4
C
1:8
D
1:16
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
When components are connected across the same voltage supply (parallel by default for rated household appliances unless stated otherwise), current is directly proportional to power.
Formula:
$$ P = VI \implies I = \frac{P}{V} $$
Solution:
Because both are connected to the same 240V supply, \( V \) is constant.
Current \( I \propto P \).
Ratio \( \frac{I_A}{I_B} = \frac{P_A}{P_B} = \frac{500}{2000} = \frac{1}{4} \).
The ratio is 1:4.
Why other options are incorrect:
Opt D: Would be the ratio if the relationship depended on the square of the power.
In a conducting electric wire, the electric current flows due to:
A
Protons
B
Ions
C
Holes
D
Electrons
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Current flow depends on the nature of the medium. In solid metallic conductors, the atoms are locked in a lattice, leaving only specific particles free to move.
Solution:
In metallic electric wires (like copper), only the outer valence electrons are loosely bound and free to drift.
Therefore, electric current in typical metallic conductors is exclusively due to the flow of free electrons.
Why other options are incorrect:
Opt A: Protons are tightly bound in the atomic nucleus and cannot move.
Opt B: Ions are the primary charge carriers in liquid electrolytes, not solid wires.
Opt C: Holes act as charge carriers in semiconductors, not typical metallic wires.
An Ohmic device strictly obeys Ohm's Law (\(V \propto I\)), meaning its resistance remains constant over a wide range of voltages and currents, provided physical conditions (like temperature) are stable.
Solution:
Copper wire is a classic metallic conductor that behaves linearly (ohmic) under normal operating conditions.
Why other options are incorrect:
Opt A: A filament bulb heats up drastically as current flows, causing its resistance to increase (non-ohmic).
Opt B & C: Semiconductors, diodes, and transistors inherently possess non-linear I-V characteristics.
When different resistors are connected across the terminal of a battery:
A
Its emf remains same but terminal potential difference changes
B
Both emf and terminal potential difference changes
C
Its emf changes but terminal potential difference remains the same
D
Both emf and terminal potential difference remains the same
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Electromotive force (EMF) is an intrinsic property of the battery chemistry, whereas terminal potential difference (\(V_t\)) depends on the current flowing through the circuit.
Formula:
$$ V_t = E - Ir $$
Solution:
Because EMF (E) is fixed by the chemical reaction of the cell, it remains the same regardless of the external circuit.
When you connect different external resistors, the total equivalent resistance changes, causing the total current (\(I\)) to change.
Since \(I\) changes, the internal voltage drop (\(Ir\)) changes, meaning the terminal potential difference changes. This makes Option A correct.
Why other options are incorrect:
Opt B, C, D: All incorrectly assert that EMF changes or that terminal voltage remains immune to current variations.
Resistance of 5 \(\Omega\) and 10 \(\Omega\) are connected in parallel. If the P.D across 5 \(\Omega\) resistor is 20V then current through 10 \(\Omega\) resistor will be:
A
0.5 A
B
2 A
C
4 A
D
10 A
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
In a parallel circuit, the potential difference (voltage) across all branches is identical.
Formula:
$$ I = \frac{V}{R} $$
Solution:
Because the resistors are in parallel, the voltage across the 10 \(\Omega\) resistor is the exact same as the 5 \(\Omega\) resistor, which is 20V.
Apply Ohm's law to the 10 \(\Omega\) resistor: \( I = \frac{20 \text{ V}}{10 \, \Omega} \).
\( I = 2 \text{ A} \) (Option B).
Why other options are incorrect:
Opt A: Inverting the calculation \( 10/20 \).
Opt C: This is the current flowing through the 5 \(\Omega\) resistor (\(20/5 = 4\text{A}\)).
A conducting wire of resistivity '\(\rho\)' is cut into two equal parts. The resistivity of each part will be:
A
The same
B
One fourth
C
\( \sqrt{2} \) times
D
Halved
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Resistivity (\(\rho\)) is an intrinsic property of the material and its temperature. It is strictly independent of physical dimensions like length or area.
Solution:
Cutting the wire halves its Resistance (\(R\)), because resistance depends on length.
However, since both pieces are made of the exact same material and are at the same temperature, their Resistivity remains exactly the same.
Why other options are incorrect:
Opt B, C, D, E: All incorrectly assume resistivity scales with physical dimensions.
A load of resistance 0.04 \(\Omega\) is attached to the cell having E.M.F 1.5 V and 15 A is drawn in the circuit. The internal resistance of cell will be
A
0.6 \(\Omega\)
B
0.06 \(\Omega\)
C
6 \(\Omega\)
D
0.3 \(\Omega\)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The total resistance of a circuit includes both external load resistance and the internal resistance of the battery.
Formula:
$$ I = \frac{E}{R + r} $$
Solution:
Given \( I = 15 \text{ A} \), \( E = 1.5 \text{ V} \), \( R = 0.04 \, \Omega \).
Rearrange for total resistance: \( R + r = \frac{E}{I} \).
When a wire is compressed and its radius becomes 2R then its resistance will be:
A
16R
B
4R
C
1/16 R
D
1/4 R
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
When a wire is mechanically compressed or stretched, its volume remains constant. Therefore, a change in radius forces a corresponding change in length.
Formula:
$$ \text{Volume} = A \cdot L = \text{constant} $$
$$ R = \rho \frac{L}{A} $$
Solution:
Original Area \( A = \pi r^2 \).
New radius \( r' = 2r \). New Area \( A' = \pi (2r)^2 = 4\pi r^2 = 4A \).
Since Volume is constant: \( A \cdot L = A' \cdot L' \implies L' = \frac{A \cdot L}{4A} = \frac{L}{4} \). The wire is 4 times thicker and 4 times shorter.
New Resistance \( R' = \rho \frac{L'}{A'} = \rho \frac{L/4}{4A} = \frac{1}{16} \rho \frac{L}{A} = \frac{1}{16} R \).
Why other options are incorrect:
Opt A: Occurs if you stretch the wire (making it thinner) rather than compress it.
Opt D: Occurs if you forget that length also shrinks by a factor of 4.
An Ohmic device strictly obeys Ohm's Law (\(V \propto I\)), meaning its resistance stays constant over a wide range of voltages and currents, provided temperature is stable.
Solution:
Copper wire is a classic metallic conductor that behaves linearly (ohmic) under normal conditions.
Why other options are incorrect:
Opt A: A filament bulb heats up drastically as current flows, causing its resistance to increase (non-ohmic).
Opt B & C: Semiconductors, diodes, and transistors inherently possess non-linear I-V characteristics.
The change in a resistance of metallic conductor below 0\(^\circ\text{C}\)?
A
Nonlinear
B
Curve
C
Linear
D
Curvilinear
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The resistance of typical metallic conductors (like copper or platinum) follows a specific empirical relationship with temperature.
Solution:
For typical metallic conductors, the change in resistance with respect to temperature is approximately linear over a wide range of temperatures, including somewhat below 0\(^\circ\text{C}\) (until very near absolute zero where it curves or drops to zero in superconductors).
The standard formula \( R_t = R_0(1 + \alpha t) \) is the equation of a straight (linear) line.
Why other options are incorrect:
Opt A, B, D: While strictly non-linear at cryogenic temperatures, standard syllabus models this behavior as linear for standard metallic conductors.
A copper wire has length L and cross-sectional area A. Its resistance is R. If we halved the length and halved the diameter of wire, then what will be the resistance of this wire?
A
R
B
2R
C
3R
D
4R
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Resistance depends on length and the square of the diameter (since Area \( \propto d^2 \)).
Formula:
$$ R = \rho \frac{L}{A} \propto \frac{L}{d^2} $$
Solution:
Original Resistance \( R \propto \frac{L}{d^2} \).
New length \( L' = \frac{L}{2} \).
New diameter \( d' = \frac{d}{2} \), so new Area \( A' \propto \left(\frac{d}{2}\right)^2 = \frac{d^2}{4} \). This means area is 1/4th of the original.
A car battery has EMF of 12 Volts and internal resistance \(5 \times 10^{-2} \text{ ohm}\). If it draws 60 ampere current, then terminal voltage of the battery will be
A
5 volts
B
3 volts
C
15 volts
D
9 volts
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
When a battery outputs heavy current (like starting a car), there is a significant voltage drop across its internal resistance.
Formula:
$$ V_t = E - Ir $$
Solution:
EMF \( E = 12 \text{ V} \).
Internal resistance \( r = 0.05 \, \Omega \).
Current \( I = 60 \text{ A} \).
Voltage drop \( Ir = 60 \times 0.05 = 3 \text{ V} \).
When potential difference is applied across the ends of uniform wire of length \(l\) and radius \(r\), a current \(I\) flow in the wire. If same potential difference is applied to the ends of another wire of the same material but of length \(2l\) and radius \(2r\), the current in the wire is
A
\( I/4 \)
B
\( I \)
C
\( 2I \)
D
\( I/2 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Changing the physical dimensions of a wire alters its resistance, which inversely affects current for a constant voltage.
Formula:
$$ R = \rho \frac{l}{\pi r^2} \quad \text{and} \quad I = \frac{V}{R} $$
Solution:
Original Resistance: \( R = \rho \frac{l}{\pi r^2} \).
New length \( l' = 2l \). New radius \( r' = 2r \), meaning new area \( A' = \pi (2r)^2 = 4\pi r^2 \).
Since voltage is constant, \( I' = \frac{V}{R'} = \frac{V}{R/2} = 2 \left( \frac{V}{R} \right) = 2I \).
Why other options are incorrect:
Opt A, B, D: Arise from failing to square the radius when calculating the new area, or directly applying the resistance ratio to current without taking the inverse.
A typical mobile phone of 5.0 V and an internal resistance of 200 m\(\Omega\). What is the terminal P.D of the battery when it supports a current of 500 mA?
A
4.8V
B
4.9V
C
5.0 V
D
5.1V
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
When a battery supplies current, the terminal potential difference drops below the EMF due to internal resistance.
A metal cube with sides of length "a" has electrical resistance R between opposite faces. What is the resistance between the opposite faces of a cube of the same metal with sides of length 3a?
A
9R
B
3R
C
R/3
D
R/9
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The resistance of a regular solid changes dynamically with its length and cross-sectional area.
Formula:
$$ R = \rho \frac{L}{A} $$
Solution:
For the first cube, distance \( L = a \) and area \( A = a \times a = a^2 \). Resistance \( R_1 = \rho \frac{a}{a^2} = \frac{\rho}{a} = R \).
For the new larger cube, distance \( L = 3a \) and area \( A = (3a) \times (3a) = 9a^2 \).
A cell of internal resistance 2.0\(\Omega\) and electromotive force (e.m.f) 1.5V is connected to a resistor of resistance 3.0\(\Omega\) what is the potential difference across 3\(\Omega\) resistor.
A
5V
B
1.2V
C
0.9 V
D
0.6V
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The potential difference across the external resistor is the same as the terminal voltage of the cell.
Formula:
$$ I = \frac{E}{R+r} \quad \text{and} \quad V = IR $$
A carbon resistor is connected to a battery of 50 volt and 2 ampere current is passing through it. If voltage is increased to 75 volt, current will become:
A
3Amp
B
4.5Amp
C
1.5Amp
D
6Amp
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A carbon resistor obeys Ohm's Law, maintaining a constant resistance.
Formula:
$$ R = \frac{V_1}{I_1} = \frac{V_2}{I_2} $$
Solution:
First, find the resistance: \( R = \frac{50 \text{ V}}{2 \text{ A}} = 25 \, \Omega \).
Now, apply the new voltage to find the new current: \( I_2 = \frac{V_2}{R} \).
Resistance between two opposite faces of square thin film of area 1mm\(^2\) having thickness of 1\(\mu\)m if resistivity of material is \(10^{-6}\Omega\text{m}\) will be:
A
1000 \(\Omega\)
B
1 \(\Omega\)
C
100 \(\Omega\)
D
10 \(\Omega\)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Resistance is determined by the length of the path the current travels through and the cross-sectional area perpendicular to that flow.
Formula:
$$ R = \rho \frac{L}{A_{cross}} $$
Solution:
The film is a square of area \( 1 \text{ mm}^2 \), so each side length \( L = 1 \text{ mm} = 10^{-3} \text{ m} \).
Current flows between opposite faces (edges) of this square, meaning it travels a distance \( L = 10^{-3} \text{ m} \).
The cross-sectional area \( A_{cross} \) the current sees is the width \( \times \) thickness.
If the potential difference across a resistor is doubled:
A
Only the current is doubled
B
Only the current is halved
C
Only the resistance is doubled
D
Only the resistance is halved
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
For an ohmic resistor, Ohm's Law states that current is directly proportional to voltage, provided physical conditions (like temperature) remain constant.
Formula:
$$ I = \frac{V}{R} $$
Solution:
The resistance \( R \) is an intrinsic property of the resistor's dimensions and material; it does not change when voltage changes.
If \( V \) becomes \( 2V \), then the new current \( I' = \frac{2V}{R} = 2I \).
Thus, only the current doubles.
Why other options are incorrect:
Opt B: Current is directly, not inversely, proportional to voltage.
Opt C & D: Resistance is independent of applied voltage for ohmic devices.
12-volt battery is applied across 6\(\Omega\) resistance to have a steady flow of current. What must be the required potential difference across the same resistance to have a steady current of one ampere?
A
12V
B
1V
C
3V
D
6V
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Ohm's Law relates voltage, current, and resistance.
Formula:
$$ V = IR $$
Solution:
We want a steady current \( I_{new} = 1 \text{ A} \).
The resistance remains unchanged, so \( R = 6 \, \Omega \).
If 2 A current passes through a resistor when connected to a certain battery. If the resistance is replaced by the double resistance, then the current will become
A
2A
B
6A
C
4A
D
1A
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
According to Ohm's Law, current is inversely proportional to resistance when voltage is constant.
Formula:
$$ I = \frac{V}{R} $$
Solution:
Initial current \( I_1 = 2\text{A} \).
The battery remains the same, so \( V \) is constant. \( I \propto \frac{1}{R} \).
If new resistance \( R' = 2R \), then new current \( I' = \frac{V}{2R} = \frac{I_1}{2} \).
\( I' = \frac{2}{2} = 1\text{A} \).
Why other options are incorrect:
Opt A: Current would only stay 2A if resistance didn't change.
Opt B & C: Current would increase only if resistance was decreased (e.g., to half). Doubling resistance restricts the flow more.
The equivalent current which passes from a point at higher potential to a point at a lower potential as if it represented a movement of positive charges is
A
Electronic current
B
Magnetic lines
C
Electric current
D
Conventional current
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Historically, electric current was defined before the discovery of the electron.
Solution:
The flow of current from a higher potential (positive terminal) to a lower potential (negative terminal) mimics the theoretical flow of positive charges.
This standardized model is universally known as Conventional current.
Why other options are incorrect:
Opt A: Electronic current flows from lower potential (negative) to higher potential (positive).
Opt B & C: Magnetic lines aren't currents; "Electric current" is a general term that doesn't specify directionality based on positive charge flow.
The fractional change in resistance per Kelvin is known as:
A
Temperature coefficient of resistance
B
Linear coefficient of expansion
C
Thermal coefficient
D
Volumetric coefficient of expansion
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The fractional change in resistance \( \left(\frac{\Delta R}{R_0}\right) \) per unit change in temperature \( (\Delta T) \) is the formal definition of a specific electrical property.
Formula:
$$ \alpha = \frac{\Delta R}{R_0 \Delta T} $$
Solution:
By definition, this quantity \( \alpha \) is called the Temperature coefficient of resistance.
Why other options are incorrect:
Opt B & D: These refer to physical dimensional changes (length and volume) due to heat, not electrical resistance.
Opt C: A vague term that does not specifically denote resistance change.
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