Using the formula for parallel resistors, 1/R_eq = 1/R1 + 1/R2 = 1/4 + 1/6 = 5/12, R_eq = 12/5 = 2.4Ω.

According to KCL, I = total entering - total leaving = (2A + 3A) - 5A = 0A.

According to KCL, I = Total entering - Total leaving = (3A + 2A) - 5A = 0A.

Using voltage division, V_R1 = (R1 / (R1 + R2)) * V_total = (4Ω / (4Ω + 6Ω)) * 12V = 4.8V.

Using voltage division, V_R1 = V_total * (R1 / (R1 + R2)) = 12V * (4Ω / (4Ω + 6Ω)) = 12V * (4/10) = 4.8V.

According to Kirchhoff's voltage law, the sum of the voltage drops must equal the source voltage. Here, 4V + 5V = 9V, which is correct.

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points, which can be expressed as V = IR.

Increasing the emf of the cell increases the balancing length, as it requires a longer length to balance the higher voltage.

According to Ohm's Law, if resistance increases while voltage remains constant, the current decreases.

According to Ohm's Law, if voltage (V) is tripled and resistance (R) remains constant, the current (I) will also triple: I = V / R.

When the Wheatstone bridge is balanced, the current in the galvanometer becomes zero.

When the bridge is balanced, the current through the galvanometer becomes zero.

When the Wheatstone bridge is unbalanced, there is a non-zero potential difference across the galvanometer, causing current to flow through it.

When the Wheatstone bridge is balanced, the potential difference across the galvanometer is zero, resulting in zero current through it.

When the Wheatstone bridge is perfectly balanced, the current through the galvanometer is zero.

When the Wheatstone bridge is balanced, the potential difference across the galvanometer is zero, resulting in zero current.

Increasing the resistance of the wire decreases the current, which causes the null point to move away from the battery.

Increasing the resistance of the wire decreases the current, which in turn decreases the potential difference across any segment of the wire.

If the length of the segment is halved, the potential difference across that segment also halves, assuming the potential gradient remains constant.

The reading of a potentiometer is independent of the resistance of the wire as long as the potential difference remains constant.

As temperature increases, the resistance of the wire typically increases, which can affect the potential gradient and thus the reading.

If the wire is heated and its resistance increases, the reading of the potentiometer decreases because the potential gradient will be affected.

Increasing the resistance of the galvanometer reduces the current through it, which can lead to a decrease in the reading on the potentiometer.

For most conductors, resistance increases with an increase in temperature due to increased atomic vibrations.

Resistance R is inversely proportional to the area; halving the diameter quarters the area, thus quadrupling the resistance.

The resistivity of a superconductor approaches zero as it reaches absolute zero.

In the superconducting state, the resistivity of a superconductor becomes zero.

If the resistances are made equal, the sensitivity of the Wheatstone bridge becomes infinite, as even a small change will cause a large deflection.

If one branch is disconnected, the total current decreases because there is less path for the current to flow.

In a series circuit, the total resistance increases as more resistors are added.