Pick out wrong statement about the Kirchhoff’s laws of electric circuit.
1. Outgoing currents adds up and are equal to incoming currents at a junction
2. Starting with any point if we come back to same point, total potential change must be zero
3. Junction rule is based on conservation of energy law
4. Bending or reorienting the wire does not change the validity of Kirchhoff’s junction rule.
View Answer
Kirchhoff's first law (the junction rule) is based on the law of conservation of charge, not energy. The second law (loop rule) is based on the conservation of energy.
Consider the following statements and choose the correct option
Statement (A): Electromotive force (emf) is not a force, it is the voltage difference between two terminal of a battery in open circuit.
Statement (B): The potential difference across terminals of a battery having some internal resistance can never be greater than the emf of the battery.
1. Statement (A) is correct while statement (B) is incorrect
2. Statement (A) is incorrect while statement (B) is correct
3. Both the statements are correct
4. Both the statements are incorrect
View Answer
Statement (A) is correct as EMF is indeed a potential difference under open circuit. Statement (B) is incorrect because during charging, the terminal potential difference is \(V = E + Ir\), which is greater than the EMF \(E\).
The energy used by a 100 W bulb which is on for 10 hours is:
1. 1 kWh
2. \( 3.6 \times 10^6 \, \text{J} \)
3. \( 4.8 \times 10^5 \, \text{J} \)
4. Both (1) and (2)
View Answer
Energy is calculated as \( E = P \times t = 100 \, \text{W} \times 10 \, \text{h} = 1000 \, \text{Wh} = 1 \, \text{kWh} \). In standard SI units, this is \( 1000 \, \text{W} \times 3600 \, \text{s} = 3.6 \times 10^6 \, \text{J} \). Thus, both Options (1) and (2) are correct.
Assertion (A): When constant current is passing through a conductor of variable area of cross section, electric field inside conductor is inversely proportional to cross sectional area.
Reason (R): Microscopic form of Ohm’s law is \( \vec{E} = \rho \vec{J} \), where \( \vec{E} \) stands for electric field, \( \rho \) stands for resistivity and \( \vec{J} \) stands for current density.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is true because \( I = JA \) and \( E = \rho J \) imply \( E = \frac{\rho I}{A} \). Reason (R) is true as it's the microscopic form of Ohm's law. Reason correctly explains assertion.
Assertion (A): The current density \( \vec{J} \) at any point in ohmic resistor is in direction of electric field \( \vec{E} \) at that point.
Reason (R): A point charge when released from rest in a region having only electrostatic field always moves along electric lines of force.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is true from \( \vec{J} = \sigma \vec{E} \) where \( \sigma \) is conductivity. Reason (R) is false because a charge released from rest moves along an electric field line only if the field line is straight, which is not always true.
Assertion (A): The rate at which energy is being delivered to a light bulb is lower after it has been on for a few seconds than just after it is turned on.
Reason (R): As the filaments warms up, its resistance rises and the current falls.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Concept: Resistance of metals increases with temperature. Power \(P = V^2/R\).
Formula: \(P = V^2/R\), \(R \propto T\).
Solution: As filament warms, its resistance \(R\) increases. For constant voltage \(V\), current \(I = V/R\) decreases, so power \(P = V^2/R\) delivered to the bulb also decreases. Thus, A and R are true and R explains A.