You are provided with a voltmeter V, a chemical cell/ battery E; two standard resistors R, and R a potentiometer a key K a jockey, and other necessary materials.
(b)i. State the two devices in which ohm's law does not apply.
ii. A current of 1 A is supplied to two resistors of resistance 2\(\Omega\) and 3\(\Omega\) connected in parallel. Calculate the current in each resistor.
Practical: voltmeter reading against tapping length on a potentiometer
For balancing lengths \(x = 20,30,40,50,60,80\ \text{cm}\) the voltmeter reading \(V\) is recorded, and \(x^{-1}\) and \(V^{-1}\) evaluated and tabulated.
Expected results. A graph of \(V^{-1}\) (vertical) against \(x^{-1}\) (horizontal) is a straight line; determine its slope \(s\) and the intercept \(c\) on the vertical axis.
Two precautions:
- The jockey was tapped gently (not dragged) on the wire to avoid altering its cross-section.
- Connections were kept clean and tight, and the key opened between readings to prevent heating of the wire.
(b)(i) Two devices in which Ohm's law does not apply: a semiconductor diode (rectifier) and a thermionic (vacuum) diode/valve. (An electrolyte and a filament lamp are also acceptable non-ohmic conductors.)
(b)(ii) Current in each parallel resistor. With \(2\ \Omega\) and \(3\ \Omega\) in parallel carrying a total of \(1\ \text{A}\), the current divides in inverse ratio of the resistances:
\[ I_{2\Omega} = 1 \times \frac{3}{2+3} = 0.6\ \text{A}, \qquad I_{3\Omega} = 1 \times \frac{2}{2+3} = 0.4\ \text{A} \]
So \(0.6\ \text{A}\) flows through the \(2\ \Omega\) resistor and \(0.4\ \text{A}\) through the \(3\ \Omega\) resistor.