Current, Potential Difference and Resistance

Solution

1. 1

   Draw the main loop (rectangle of wire).

   $\text{Cell — switch — resistor — bulb — back to cell, all in series.}$

2. 2

   Insert ammeter.

   $\text{Ammeter goes IN series with the bulb (in the main loop).}$

3. 3

   Insert voltmeter.

   $\text{Voltmeter goes in PARALLEL across the resistor only.}$

Solution

1. 1

   (i)

   $\text{Voltmeter — measures potential difference; placed in parallel.}$

2. 2

   (ii)

   $\text{Variable resistor — resistance can be changed; controls current.}$

3. 3

   (iii)

   $\text{Diode — allows current to flow in one direction only.}$

Solution

1. 1

   Convert time.

   $t = 90 \times 60 = 5400\,\text{s}$

2. 2

   Apply Q = It.

   $Q = 0.4 \times 5400 = 2160\,\text{C}$

Solution

1. 1

   Convert time.

   $t = 5 \times 60 = 300\,\text{s}$

2. 2

   Rearrange Q = It.

   $I = Q/t = 90 / 300 = 0.3\,\text{A}$

Solution

1. 1

   Apply the series rule.

   $\text{Current is the same at every point in a series circuit.}$

2. 2

   Therefore both ammeters read the same value.

   $I_2 = I_1 = 0.25\,\text{A}$

Solution

1. 1

   Rearrange.

   $R = V/I$

2. 2

   Substitute.

   $R = 6 / 0.25 = 24\,\text{Ω}$

Solution

1. 1

   Combine series resistors.

   $R = 12 + 4 = 16\,\text{Ω}$

2. 2

   Apply V = IR rearranged.

   $I = V/R = 8/16 = 0.5\,\text{A}$

Solution

1. 1

   Use the gradient.

   $R = \text{gradient} \times L = 2.5 \times 60$

2. 2

   Calculate.

   $R = 150\,\text{Ω}$

Solution

1. 1

   Interpret a straight line through origin.

   $I \propto V \Rightarrow R \text{ is constant} \Rightarrow \text{ohmic behaviour}$

2. 2

   Conclude.

   $\text{The component is a fixed resistor at constant temperature.}$

Solution

1. 1

   Apply R = V/I at each point.

   $R_1 = 2.0 / 0.20 = 10\,\text{Ω};\quad R_2 = 12 / 0.50 = 24\,\text{Ω}$

2. 2

   Comment.

   $\text{R rises from 10 to 24 Ω as V increases — the lamp is non-ohmic.}$