Question

The circuit depicted in the figure is employed to study Ohm's Law. Rather than employing a conventional resistor, a glass tube of length 'l' and cross-sectional area 'a', which is half-filled with mercury of resistivity 'd', is connected to the circuit via two electrodes, E₁ and E₂. These electrodes are linked to a battery with an emf of 'E' and negligible internal resistance. Provide the answer in terms of 'a', 'l', 'd', and 'E'.

1. Resistance of mercury in the tube.
2. The ammeter reading.
3. Voltmeter reading.
4. Which of the measurements are altered when the tube is entirely filled with mercury?

Answer 1

The mercury column fills half the length of the tube, so its length is `l/2` and its cross-sectional area is a.

(a) Resistance of mercury: R = `d(l/2)/a = dl/(2a)`

(b) Ammeter reading: I = `E/R = E ÷ (dl/(2a)) = 2Ea/(dl)`

(c) Voltmeter reading: Since the battery has negligible internal resistance and the voltmeter is connected across the mercury column, it reads the full emf of the battery. Voltmeter reading = E.

(d) When the tube is completely filled with mercury, the length of the mercury column becomes l. So, the new resistance is `dl/a`, and the new current is `Ea/(dl)`. Thus, the resistance and the ammeter reading change, while the voltmeter reading remains E.