Current Electricity

Analysis of the circuit shown in the diagram:

• The voltage of the source is $V = 2\,\text{V}$.
• The equivalent resistance of the resistor network connected across the battery is $R_{\text{eq}} = 2\,\Omega$.

By Ohm's law, the total current supplied by the source is:

$$I = \frac{V}{R_{\text{eq}}} = \frac{2\,\text{V}}{2\,\Omega} = 1\,\text{A}$$

The drift velocity of conduction electrons in a conductor of length $l$ under an applied potential difference $V$ is given by:

$$v_d = \frac{e E \tau}{m} = \frac{e V \tau}{m l}$$

where:

• $e$ is electron charge
• $m$ is electron mass
• $\tau$ is relaxation time
• $V$ is applied voltage
• $l$ is conductor length

For a given conductor at constant temperature, $v_d \propto V$. Hence, option 2 is the correct answer.

To maintain a constant speed $v$, the external force applied must balance the opposing magnetic force ($F_{\text{ext}} = F_m$).

The mechanical power supplied by the external agent is:

$$P = F_{\text{ext}} \times v = \frac{B^2 l^2 v^2}{R}$$

Alternatively, using electric power:

$$P = \frac{\varepsilon^2}{R} = \frac{(9.0 \times 10^{-3}\,\text{V})^2}{180 \times 10^{-3}\,\Omega} = \frac{81 \times 10^{-6}}{0.18} = 4.5 \times 10^{-4}\,\text{W} = 4.5 \times 10^{-3}\,\text{W}$$

When a cell of emf $E$ and internal resistance $r$ delivers a current $I$ to an external resistor:

• The potential drop across the internal resistance is:
  $$V_{\text{int}} = I r = (0.6\,\text{A})(3\,\Omega) = 1.8\,\text{V}$$
• The terminal potential difference $V$ across the external resistor is:
  $$V = E - I r = 12\,\text{V} - 1.8\,\text{V} = 10.2\,\text{V}$$