Fig. 22.1 shows a series circuit with a 12.0 V battery, a switch and three resistors: R1 = 6 Ω, R2 = 12 Ω and R3 = 18 Ω. A voltmeter is connected across R2 ...

(a) In a series circuit, the same current flows through every component. First find the total resistance:

\[ R_{\text{total}} = R_1 + R_2 + R_3 = 6 + 12 + 18 = 36\;\Omega \]

Then calculate the current using Ohm's law:

\[ I = \frac{V}{R_{\text{total}}} = \frac{12.0}{36} = 0.333\;\text{A} \]

The voltmeter reads the p.d. across R₂ alone:

\[ V_{R_2} = I \times R_2 = 0.333 \times 12 = 4.0\;\text{V} \] [2]

(b) Using the same current:

\[ V_{R_1} = I \times R_1 = 0.333 \times 6 = 2.0\;\text{V} \] [1]

(c) Rather than recalculating from scratch, use the fact that the three p.d.s must sum to the battery voltage:

\[ V_{R_3} = 12.0 - 4.0 - 2.0 = 6.0\;\text{V} \] [1]

(d) The sum of the potential differences across all components in a series circuit equals the e.m.f. of the supply:

\[ V_{R_1} + V_{R_2} + V_{R_3} = V_{\text{battery}} \]

Energy given to charges by the battery is shared among the resistors, so none is lost or gained around the loop (Kirchhoff's second law). [1]

(e) Current is found from \( V = IR \) rearranged:

\[ I = \frac{V}{R_{\text{total}}} = \frac{12.0}{36} = 0.33\;\text{A} \] [2]

(f) Possible reasons include: the battery has internal resistance (reducing the terminal p.d. below 12.0 V), the connecting wires have small but non-zero resistance, or the resistors may not be exactly their labelled values. Any of these would cause measured p.d.s to differ slightly from calculated values. [1]

(g) The equation linking potential difference, current and resistance is:

\[ V = I \times R \]

where \( V \) is the potential difference in volts, \( I \) is the current in amperes, and \( R \) is the resistance in ohms. [1]