More Free Online MCQs (Click on the links and check yourself):
- 1. Units and Measurement
- 2. Motion In Straight Line
- 3. Motion In Plane
- 4. Laws of Motion
- 5. Work, Energy and Power
- 6. System of Particles and Rotational Motion
- 7. Gravitation
- 8. Mechanical Properties of Solids
- 9. Mechanical Properties of Fluids
- 10. Oscillations
- 11. Electric Charges, Fields & Capacitance
- 12. Current Electricity
- 13. Ray Optics
- 14. Wave Optics
- 15. Electromagnetic Induction
- 16. Dual Nature of of Radiation & Matter
- 17. Atoms
- 18. Semiconductor
Current Electricity!
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Total Marks: 100, Obtained Marks: 0
1. A wire of uniform cross-section has a resistance \( R \). If the length of the wire is doubled while keeping the material and temperature constant, what happens to the resistance?
A. \( \frac{R}{2} \)
B. \( 2R \)
C. \( 4R \)
D. \( R \)
The correct answer is \( 2R \). When the length of a wire is doubled and the cross-sectional area remains the same, the resistance increases. Resistance (\( R \)) is directly proportional to the length (\( L \)) of the wire, so if the length is doubled, the resistance also doubles.
2. Two resistors, \( R_1 = 4 \, \Omega \) and \( R_2 = 6 \, \Omega \), are connected in parallel. What is the equivalent resistance of the combination?
A. \( 2.4 \, \Omega \)
B. \( 10 \, \Omega \)
C. \( 24 \, \Omega \)
D. \( 5 \, \Omega \)
The correct answer is \( 2.4 \, \Omega \). For resistors in parallel, the equivalent resistance \( R \) is given by:
\[
\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}
\]
Substituting the values:
\[
\frac{1}{R} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}
\]
Therefore, \( R = \frac{12}{5} = 2.4 \, \Omega \).
3. A cell of EMF \( E \) and internal resistance \( r \) is connected to an external resistance \( R \). The maximum power is delivered to the external resistor when:
A. \( R = r \)
B. \( R > r \)
C. \( R < r \)
D. \( R = 0 \)
The correct answer is \( R = r \). According to the maximum power transfer theorem, the power delivered to the load is maximum when the external resistance (\( R \)) is equal to the internal resistance (\( r \)) of the source. Mathematically, the power delivered to the external resistor is:
\[
P = \frac{E^2 R}{(R + r)^2}
\]
This power is maximized when \( R = r \).
4. A potentiometer wire has a length of 2 meters and a resistance of 10 ohms. If the length of the wire is increased to 3 meters while keeping the same cross-sectional area and material, what will be the new resistance of the wire?
A. \( 15 \, \Omega \)
B. \( 5 \, \Omega \)
C. \( 30 \, \Omega \)
D. \( 20 \, \Omega \)
The correct answer is \( 15 \, \Omega \). Resistance of a wire is directly proportional to its length. If the length is increased from 2 meters to 3 meters, the new resistance can be calculated using the ratio of lengths:
\[
\frac{R_2}{R_1} = \frac{L_2}{L_1}
\]
Substituting the values:
\[
\frac{R_2}{10} = \frac{3}{2}
\]
Therefore, \( R_2 = 10 \times \frac{3}{2} = 15 \, \Omega \).
5. A current of \( 3 \, \text{A} \) flows through a resistor of \( 4 \, \Omega \) and an internal resistance of \( 2 \, \Omega \). What is the EMF of the cell supplying the current?
A. \( 12 \, \text{V} \)
B. \( 6 \, \text{V} \)
C. \( 8 \, \text{V} \)
D. \( 18 \, \text{V} \)
The correct answer is \( 18 \, \text{V} \). The EMF (\( E \)) of the cell is the sum of the potential drop across the external resistor and the internal resistance. The potential drop across the external resistor is:
\[
V_{\text{ext}} = I \cdot R_{\text{ext}} = 3 \, \text{A} \times 4 \, \Omega = 12 \, \text{V}
\]
The potential drop across the internal resistance is:
\[
V_{\text{int}} = I \cdot r = 3 \, \text{A} \times 2 \, \Omega = 6 \, \text{V}
\]
Thus, the EMF of the cell is:
\[
E = V_{\text{ext}} + V_{\text{int}} = 12 \, \text{V} + 6 \, \text{V} = 18 \, \text{V}
\]
6. If the current through a \( 10 \, \Omega \) resistor is \( 2 \, \text{A} \), what is the power dissipated in the resistor?
A. \( 40 \, \text{W} \)
B. \( 10 \, \text{W} \)
C. \( 4 \, \text{W} \)
D. \( 2 \, \text{W} \)
The correct answer is \( 40 \, \text{W} \). The power dissipated in a resistor is given by:
\[
P = I^2 \cdot R
\]
Substituting the given values:
\[
P = (2 \, \text{A})^2 \times 10 \, \Omega = 4 \times 10 = 40 \, \text{W}
\]
7. In a circuit with three resistors \( R_1 \), \( R_2 \), and \( R_3 \) in series, Kirchhoff's loop law states that the sum of the potential differences (voltages) around any closed loop in a circuit is equal to:
A. The resistance of the loop
B. The power supplied by the source
C. The total current through the loop
D. Zero
The correct answer is Zero. Kirchhoff's loop law states that the sum of the potential differences around any closed loop in a circuit is equal to zero. This is because the total energy supplied by the source is used up by the resistors and other components in the loop.
8. In a Wheatstone bridge, if the bridge is balanced, the ratio of the resistances in one arm is equal to the ratio in the other arm. If \( R_1 \) and \( R_2 \) are the resistances in one arm, and \( R_3 \) and \( R_4 \) are the resistances in the other arm, the condition for balance is:
A. \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \)
B. \( R_1 + R_2 = R_3 + R_4 \)
C. \( R_1 \times R_2 = R_3 \times R_4 \)
D. \( \frac{R_1}{R_3} = \frac{R_2}{R_4} \)
The correct answer is \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \). In a Wheatstone bridge, the condition for balance is that the ratio of the resistances in one arm is equal to the ratio in the other arm.
9. In a potentiometer experiment, if the length of the wire is increased while keeping the current constant, what happens to the potential gradient along the wire?
A. It increases
B. It decreases
C. It remains constant
D. It becomes zero
The correct answer is \( \text{It decreases} \). The potential gradient (\( k \)) of a potentiometer is given by:
\[
k = \frac{V}{L}
\]
where \( V \) is the potential difference across the wire and \( L \) is the length of the wire. If the length \( L \) increases while keeping the potential difference \( V \) constant, the potential gradient decreases.
10. A potentiometer wire has a length of \( 1 \, \text{m} \) and a potential difference of \( 10 \, \text{V} \) is applied across it. If the length of the wire where the potential difference is measured is \( 0.5 \, \text{m} \), what is the potential difference across this segment?
A. \( 5 \, \text{V} \)
B. \( 10 \, \text{V} \)
C. \( 15 \, \text{V} \)
D. \( 20 \, \text{V} \)
The correct answer is \( 5 \, \text{V} \). The potential gradient (\( k \)) is:
\[
k = \frac{V}{L}
\]
where \( V = 10 \, \text{V} \) and \( L = 1 \, \text{m} \). So,
\[
k = \frac{10 \, \text{V}}{1 \, \text{m}} = 10 \, \text{V/m}
\]
The potential difference across \( 0.5 \, \text{m} \) is:
\[
V_{\text{segment}} = k \times 0.5 \, \text{m} = 10 \, \text{V/m} \times 0.5 \, \text{m} = 5 \, \text{V}
\]
11. In a circuit, a cell of EMF \( E \) is connected to a resistor \( R \) and a galvanometer with a high resistance \( G \). If a galvanometer shows zero deflection, which of the following conditions must be true?
A. The EMF of the cell is equal to the potential difference across the resistor.
B. The resistance \( R \) is equal to the resistance \( G \).
C. The EMF of the cell is equal to the potential difference across the galvanometer.
D. The cell must have an internal resistance equal to the resistance \( R \).
The correct answer is: The EMF of the cell is equal to the potential difference across the resistor. When a galvanometer shows zero deflection, it indicates that the potential difference across the galvanometer is zero, implying that the EMF of the cell is equal to the potential difference across the resistor, and the current through the galvanometer is zero.
12. In a circuit with a potentiometer, if the wire is made longer and the current through the wire is kept constant, what happens to the potential difference across the wire?
A. It decreases
B. It increases
C. It remains the same
D. It becomes zero
The correct answer is: It increases. The potential difference (\( V \)) across a potentiometer wire is given by:
\[
V = I \cdot R
\]
where \( I \) is the current and \( R \) is the resistance of the wire. Since the resistance \( R \) increases with the length of the wire, the potential difference across the wire will also increase if the current remains constant.
13. A galvanometer with a full-scale deflection of \( 1 \, \text{mA} \) is connected in series with a resistor of \( 99 \, \Omega \) to form a voltmeter. If the maximum potential difference the voltmeter can measure is \( 10 \, \text{V} \), what is the value of the resistor used in the voltmeter?
A. \( 10 \, \Omega \)
B. \( 9 \, \text{K}\Omega \)
C. \( 9.9 \, \text{K}\Omega \)
D. \( 100 \, \Omega \)
The correct answer is: \( 9.9 \, \text{K}\Omega \). To measure a potential difference of \( 10 \, \text{V} \), the total resistance \( R \) required for the voltmeter can be found using Ohm's Law:
\[
V = I \cdot R
\]
where \( I \) is the full-scale deflection current (\( 1 \, \text{mA} = 0.001 \, \text{A} \)) and \( V \) is the maximum potential difference (\( 10 \, \text{V} \)):
\[
R = \frac{V}{I} = \frac{10 \, \text{V}}{0.001 \, \text{A}} = 10 \, \text{K}\Omega
\]
The resistor in series with the galvanometer is \( 99 \, \Omega \), so the total resistance needed is:
\[
R_{\text{total}} = 10 \, \text{K}\Omega - 99 \, \Omega \approx 9.9 \, \text{K}\Omega
\]
14. In a circuit where a cell of EMF \( E \) is connected in series with a resistor \( R \) and a galvanometer, the deflection in the galvanometer is observed to be zero. If the galvanometer shows zero deflection with a cell of EMF \( E \) and a resistance \( R \), what is the relationship between \( E \), \( R \), and the internal resistance \( r \) of the cell?
A. \( E = I \cdot (R + r) \)
B. \( E = I \cdot R \)
C. \( E = I \cdot r \)
D. \( E = I \cdot (R - r) \)
The correct answer is: \( E = I \cdot (R + r) \). When the galvanometer shows zero deflection, the potential difference across it is zero. This means that the EMF of the cell \( E \) is equal to the potential drop across the series combination of the resistor \( R \) and the internal resistance \( r \) of the cell.
15. In a Wheatstone bridge, if the bridge is balanced, which of the following conditions must be satisfied?
A. The ratio of the resistances in one branch is equal to the ratio of the resistances in the other branch.
B. The current through the galvanometer is maximum.
C. The potential difference across the bridge is maximum.
D. The resistance of the bridge is equal to zero.
The correct answer is: The ratio of the resistances in one branch is equal to the ratio of the resistances in the other branch. For a Wheatstone bridge to be balanced, the following condition must be satisfied:
\[
\frac{R_1}{R_2} = \frac{R_3}{R_4}
\]
where \( R_1 \) and \( R_2 \) are the resistances in one branch, and \( R_3 \) and \( R_4 \) are the resistances in the other branch. In this balanced condition, the current through the galvanometer is zero, not maximum.
16. In a circuit with Kirchhoff’s laws, if the total current entering a junction is \( 5 \, \text{A} \) and the currents leaving the junction are \( 2 \, \text{A} \) and \( 1 \, \text{A} \), what is the current through the remaining branch?
A. \( 1 \, \text{A} \)
B. \( 2 \, \text{A} \)
C. \( 3 \, \text{A} \)
D. \( 4 \, \text{A} \)
The correct answer is: \( 2 \, \text{A} \). According to Kirchhoff’s Current Law (KCL), the total current entering a junction must equal the total current leaving the junction. Given:
\[
I_{\text{in}} = 5 \, \text{A}
\]
\[
I_{\text{out}} = 2 \, \text{A} + 1 \, \text{A} + I_{\text{remaining}}
\]
\[
5 \, \text{A} = 2 \, \text{A} + 1 \, \text{A} + I_{\text{remaining}}
\]
\[
I_{\text{remaining}} = 5 \, \text{A} - 3 \, \text{A} = 2 \, \text{A}
\]
17. In a circuit, if the current through a resistor \( R \) is \( 3 \, \text{A} \) and the potential difference across the resistor is \( 6 \, \text{V} \), what is the power dissipated by the resistor?
A. \( 6 \, \text{W} \)
B. \( 9 \, \text{W} \)
C. \( 18 \, \text{W} \)
D. \( 24 \, \text{W} \)
The correct answer is: \( 18 \, \text{W} \). The power dissipated by a resistor is given by:
\[
P = I^2 \cdot R
\]
where \( I \) is the current through the resistor, and \( R \) is the resistance. First, find the resistance \( R \) using Ohm’s Law:
\[
R = \frac{V}{I} = \frac{6 \, \text{V}}{3 \, \text{A}} = 2 \, \Omega
\]
Now, calculate the power dissipated:
\[
P = I^2 \cdot R = (3 \, \text{A})^2 \cdot 2 \, \Omega = 9 \cdot 2 = 18 \, \text{W}
\]
18. In a potentiometer setup, if the total length of the potentiometer wire is \( 2 \, \text{m} \) and the length of the wire used for balancing a cell of EMF \( 1.5 \, \text{V} \) is \( 0.5 \, \text{m} \), what is the EMF of another cell if the length of the wire used to balance it is \( 1.0 \, \text{m} \)?
A. \( 1.0 \, \text{V} \)
B. \( 1.5 \, \text{V} \)
C. \( 3.0 \, \text{V} \)
D. \( 6.0 \, \text{V} \)
The correct answer is: \( 3.0 \, \text{V} \). In a potentiometer, the EMF of a cell is proportional to the length of the wire segment used to balance it. Given that the length of the wire used to balance the cell with \( 1.5 \, \text{V} \) is \( 0.5 \, \text{m} \) and the length of the wire used to balance the other cell is \( 1.0 \, \text{m} \), the relationship is:
\[
\frac{E}{1.5 \, \text{V}} = \frac{1.0 \, \text{m}}{0.5 \, \text{m}}
\]
\[
E = 1.5 \, \text{V} \times \frac{1.0 \, \text{m}}{0.5 \, \text{m}} = 1.5 \, \text{V} \times 2 = 3.0 \, \text{V}
\]
19. A galvanometer of resistance \( 100 \, \Omega \) gives full-scale deflection for a current of \( 1 \, \text{mA} \). To convert it into an ammeter capable of measuring up to \( 5 \, \text{A} \), what is the value of the series resistance required?
A. \( 0.02 \, \Omega \)
B. \( 0.02 \, \text{k}\Omega \)
C. \( 0.20 \, \Omega \)
D. \( 2.0 \, \Omega \)
The correct answer is: \( 0.02 \, \Omega \). To convert a galvanometer into an ammeter, a series resistance \( R_s \) is connected. The galvanometer shows full-scale deflection for \( 1 \, \text{mA} \) and has a resistance of \( 100 \, \Omega \). The voltage across the galvanometer when it is fully deflected is:
\[
V_g = I_g \cdot R_g = 1 \, \text{mA} \times 100 \, \Omega = 0.1 \, \text{V}
\]
When measuring \( 5 \, \text{A} \), the series resistance \( R_s \) can be found using:
\[
R_s = \frac{0.1 \, \text{V}}{5 \, \text{A}} = 0.02 \, \Omega
\]
20. A galvanometer of resistance \( 200 \, \Omega \) is used to measure a current of \( 2 \, \text{A} \). The galvanometer shows full-scale deflection for \( 5 \, \text{mA} \). What is the value of the shunt resistance required to convert the galvanometer into an ammeter?
A. \( 0.2 \, \Omega \)
B. \( 0.25 \, \Omega \)
C. \( 0.50 \, \Omega \)
D. \( 1.0 \, \Omega \)
The correct answer is: \( 0.50 \, \Omega \). To convert the galvanometer into an ammeter, a shunt resistance \( R_s \) is connected in parallel. The galvanometer shows full-scale deflection for \( 5 \, \text{mA} \) and has a resistance of \( 200 \, \Omega \). The voltage across the galvanometer at full-scale deflection is:
\[
V_g = I_g \cdot R_g = 5 \, \text{mA} \times 200 \, \Omega = 1 \, \text{V}
\]
The current through the shunt is:
\[
I_s = I - I_g = 2 \, \text{A} - 5 \, \text{mA} = 1.995 \, \text{A}
\]
The value of the shunt resistance \( R_s \) is:
\[
R_s = \frac{V_s}{I_s} = \frac{1 \, \text{V}}{1.995 \, \text{A}} \approx 0.50 \, \Omega
\]
21. A galvanometer of resistance \( 50 \, \Omega \) is used to measure current. It shows full-scale deflection for a current of \( 10 \, \text{mA} \). To convert this galvanometer into an ammeter capable of measuring up to \( 100 \, \text{mA} \), what should be the value of the shunt resistance?
A. \( 0.45 \, \Omega \)
B. \( 0.50 \, \Omega \)
C. \( 0.55 \, \Omega \)
D. \( 0.60 \, \Omega \)
The correct answer is: \( 0.50 \, \Omega \). To convert a galvanometer into an ammeter, a shunt resistance \( R_s \) is connected in parallel. The galvanometer shows full-scale deflection for \( 10 \, \text{mA} \) and has a resistance of \( 50 \, \Omega \). The voltage across the galvanometer at full-scale deflection is:
\[
V_g = I_g \cdot R_g = 10 \, \text{mA} \times 50 \, \Omega = 0.5 \, \text{V}
\]
The current through the shunt is:
\[
I_s = I - I_g = 100 \, \text{mA} - 10 \, \text{mA} = 90 \, \text{mA} = 0.09 \, \text{A}
\]
The value of the shunt resistance \( R_s \) is:
\[
R_s = \frac{V_s}{I_s} = \frac{0.5 \, \text{V}}{0.09 \, \text{A}} \approx 5.56 \, \Omega
\]
22. A galvanometer of resistance \( 10 \, \Omega \) shows full-scale deflection for a current of \( 2 \, \text{mA} \). To convert this galvanometer into a voltmeter that can measure up to \( 20 \, \text{V} \), what should be the value of the series resistance required?
A. \( 9.99 \, \text{k}\Omega \)
B. \( 9.90 \, \text{k}\Omega \)
C. \( 10.0 \, \text{k}\Omega \)
D. \( 10.1 \, \text{k}\Omega \)
The correct answer is: \( 9.99 \, \text{k}\Omega \). To convert a galvanometer into a voltmeter, a series resistance \( R_s \) is connected. The galvanometer shows full-scale deflection for \( 2 \, \text{mA} \) and has a resistance of \( 10 \, \Omega \). The voltage across the galvanometer at full-scale deflection is:
\[
V_g = I_g \cdot R_g = 2 \, \text{mA} \times 10 \, \Omega = 0.02 \, \text{V}
\]
To measure \( 20 \, \text{V} \), the total resistance \( R_{\text{total}} \) is:
\[
R_{\text{total}} = \frac{V_{\text{max}}}{I_g} = \frac{20 \, \text{V}}{2 \, \text{mA}} = 10 \, \text{k}\Omega
\]
The series resistance \( R_s \) required is:
\[
R_s = R_{\text{total}} - R_g = 10 \, \text{k}\Omega - 10 \, \Omega = 9.99 \, \text{k}\Omega
\]
23. A galvanometer of resistance \( 20 \, \Omega \) is capable of measuring a maximum current of \( 15 \, \text{mA} \) for full-scale deflection. To convert this galvanometer into a voltmeter capable of measuring up to \( 25 \, \text{V} \), what should be the value of the series resistance required?
A. \( 1.67 \, \text{k}\Omega \)
B. \( 1.80 \, \text{k}\Omega \)
C. \( 1.90 \, \text{k}\Omega \)
D. \( 2.00 \, \text{k}\Omega \)
The correct answer is: \( 1.67 \, \text{k}\Omega \). To convert a galvanometer into a voltmeter, a series resistance \( R_s \) is connected. The galvanometer shows full-scale deflection for \( 15 \, \text{mA} \) and has a resistance of \( 20 \, \Omega \). The voltage across the galvanometer at full-scale deflection is:
\[
V_g = I_g \cdot R_g = 15 \, \text{mA} \times 20 \, \Omega = 0.3 \, \text{V}
\]
To measure \( 25 \, \text{V} \), the total resistance \( R_{\text{total}} \) is:
\[
R_{\text{total}} = \frac{V_{\text{max}}}{I_g} = \frac{25 \, \text{V}}{15 \, \text{mA}} \approx 1.67 \, \text{k}\Omega
\]
The series resistance \( R_s \) required is:
\[
R_s = R_{\text{total}} - R_g = 1.67 \, \text{k}\Omega - 20 \, \Omega = 1.67 \, \text{k}\Omega
\]
24. In a circuit with three resistors \( R_1 \), \( R_2 \), and \( R_3 \) connected in series, and a battery of \( 12 \, \text{V} \) connected across the series combination, the following conditions are observed:
- The potential drop across \( R_1 \) is \( 4 \, \text{V} \).
- The potential drop across \( R_2 \) is \( 3 \, \text{V} \).
A. \( 5 \, \text{V} \), \( 2 \, \Omega \)
B. \( 5 \, \text{V} \), \( 3 \, \Omega \)
C. \( 6 \, \text{V} \), \( 2 \, \Omega \)
D. \( 6 \, \text{V} \), \( 3 \, \Omega \)
The correct answer is: \( 5 \, \text{V} \) and \( 3 \, \Omega \). Using Kirchhoff's Voltage Law (KVL):
\[
V_{\text{total}} = V_{R_1} + V_{R_2} + V_{R_3}
\]
Substituting the values:
\[
12 \, \text{V} = 4 \, \text{V} + 3 \, \text{V} + V_{R_3}
\]
\[
V_{R_3} = 12 \, \text{V} - 4 \, \text{V} - 3 \, \text{V} = 5 \, \text{V}
\]
The current \( I \) in the circuit is:
\[
I = \frac{V_{R_1}}{R_1} = \frac{4 \, \text{V}}{2 \, \Omega} = 2 \, \text{A}
\]
The resistance \( R_3 \) is:
\[
R_3 = \frac{V_{R_3}}{I} = \frac{5 \, \text{V}}{2 \, \text{A}} = 2.5 \, \Omega
\]
Rounding to the nearest provided options, \( R_3 \) is closest to \( 3 \, \Omega \).
25. In a circuit with two loops, resistors \( R_1 \), \( R_2 \), \( R_3 \), and \( R_4 \) are connected as follows:
- Loop 1 includes \( R_1 \) and \( R_2 \).
- Loop 2 includes \( R_3 \) and \( R_4 \).
A. \( 0.5 \, \text{A} \)
B. \( 1 \, \text{A} \)
C. \( 1.5 \, \text{A} \)
D. \( 2 \, \text{A} \)
The correct answer is: \( 1 \, \text{A} \). To find \( I_2 \):
- Calculate the voltage drop across \( R_4 \) using: \[ V_{R_4} = V_{\text{battery}} - V_{R_3} = 10 \, \text{V} - 4 \, \text{V} = 6 \, \text{V} \]
- Then: \[ I_2 = \frac{V_{R_4}}{R_4} = \frac{6 \, \text{V}}{6 \, \Omega} = 1 \, \text{A} \]
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