Question

(i) Answer the following giving reason:

(I) The electron drift speed is estimated to be only a few mm/s for currents in the range of a few amperes. How, then, is the current established almost the instant a circuit is closed?

(II) A low voltage supply from which one needs high currents must have very low internal resistance. Why?

(III) The assertion that V = IR is a statement of Ohm's law is not true. Why?

(ii) Two cells of emfs 12 V and 6 V are connected in parallel as shown in the figure. Their internal resistances are 1 \(\Omega\) and 0.5 \(\Omega\) respectively. Calculate the emf and internal resistance of the equivalent cell between points A and B.

Hint:

Answer each part with its reason. For (ii) use E_eq = (E1 r2 + E2 r1)/(r1 + r2) and r_eq = r1 r2/(r1 + r2) for cells in parallel.

Answer 1

Step 1: Part (i)(I): current is set up almost instantly:
When the switch is closed, an electric field is set up along the whole wire almost at the speed of light. The field travels as an electromagnetic wave along the circuit.
Free electrons are present everywhere in the conductor. All of them feel the force \(-e\vec{E}\) at nearly the same instant and start drifting together.
So a current flows in every part of the circuit almost at once, even though each electron drifts only a few mm/s.

Step 2: Part (i)(II): low internal resistance for high current:
The terminal voltage of a source is \(V = \varepsilon - Ir\), where \(r\) is the internal resistance.
If a large current \(I\) is drawn and \(r\) is large, the drop \(Ir\) inside the source becomes big, and the voltage available to the circuit falls a lot.
The largest current possible is \(\varepsilon/r\). To get a high current at a steady low voltage, \(r\) must be very small.

Step 3: Part (i)(III): V = IR is not the statement of Ohm's law:
Ohm's law says that, at constant temperature, the current through a conductor is proportional to the potential difference, that is, \(V \propto I\) with a constant \(R\).
The relation \(V = IR\) only defines the resistance \(R = V/I\) and holds for every conductor, even those where \(R\) changes with \(V\).
Many devices, such as a diode, a thermistor or a gas discharge tube, give a non-linear V-I graph. They do not obey Ohm's law, yet \(V = IR\) still defines their resistance at each point.

Step 4: Part (ii): read the circuit:
From the figure, the two cells are joined in parallel between A and B with their positive terminals on the same side. So \(\varepsilon_1 = 12\) V, \(r_1 = 1\ \Omega\) and \(\varepsilon_2 = 6\) V, \(r_2 = 0.5\ \Omega\).

Step 5: Find the current in the loop:
The two emfs oppose each other in the closed loop formed by the cells. The net emf is \(12 - 6 = 6\) V and the total resistance is \(1 + 0.5 = 1.5\ \Omega\).
\[ I = \frac{12 - 6}{1 + 0.5} = \frac{6}{1.5} = 4\ \text{A} \]

Step 6: Find the potential difference between A and B:
Go through the 12 V cell. It sends current \(I\) out, so its terminal voltage is
\[ V_{AB} = \varepsilon_1 - I r_1 = 12 - 4\times 1 = 8\ \text{V} \]
Check with the other cell, which is being charged: \(V_{AB} = \varepsilon_2 + I r_2 = 6 + 4\times 0.5 = 8\) V. Both agree.

Step 7: Equivalent cell:
The emf of the equivalent cell is the open-circuit voltage between A and B, so \(\varepsilon_{eq} = 8\) V.
\[ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{1\times 0.5}{1.5} = \frac{1}{3}\ \Omega \approx 0.33\ \Omega \]

Final Answer:
(I) The field is set up almost at light speed. (II) A large current needs small internal resistance to avoid the drop Ir. (III) V = IR only defines resistance. The equivalent cell has emf 8 V and internal resistance 1/3 ohm. \[ \boxed{\varepsilon_{eq} = 8\ \text{V},\quad r_{eq} = \frac{1}{3}\ \Omega \approx 0.33\ \Omega} \]