CIE iGCSE Co-Ordinated Science P4.2.4 Resistance Exam Style Questions Paper 4

(a)(i) Charge = 75 C.

\( Q = It = 0.25 \times 5.0 \times 60 = 75 \text{ C} \)

The equation linking charge (Q), current (I), and time (t) is:

\( Q = I \times t \)

Where Q is measured in coulombs (C), I in amperes (A), and t in seconds (s). The time given is 5.0 minutes, which must be converted to seconds: 5.0 × 60 = 300 seconds.
Therefore, \( Q = 0.25 \times 300 = 75 \text{ C} \).

(a)(ii) p.d. across the 20 Ω resistor = 3.0 V.

In a parallel circuit, the potential difference (voltage) across each branch is the same as the voltage across the power source. The cell provides 3.0 V across its terminals, so the p.d. across each resistor is also 3.0 V.

(a)(iii) Current in resistor R = 0.10 A.

\( I_R = 0.25 – 0.15 = 0.10 \text{ A} \)

In a parallel circuit, the total current supplied by the cell is the sum of the currents in each parallel branch:
\( I_{\text{total}} = I_{20\Omega} + I_R \)
\( 0.25 = 0.15 + I_R \)
\( I_R = 0.25 – 0.15 = 0.10 \text{ A} \)

(a)(iv) Total resistance = 12 Ω.

\( R_{\text{total}} = \frac{V}{I_{\text{total}}} = \frac{3.0}{0.25} = 12 \text{ Ω} \)

The total resistance of the circuit can be calculated using Ohm’s law:

\( R_{\text{total}} = \frac{V}{I_{\text{total}}} = \frac{3.0}{0.25} = 12 \text{ Ω} \)

(b)(i) The current in the cell decreases.

Explanation: Removing the branch containing resistor R increases the total resistance of the circuit. Since the voltage of the cell remains constant, the total current decreases (from \( I = V/R \)).

When the branch containing resistor R is removed, the circuit now consists only of the 20 Ω resistor. The total resistance of the circuit increases (from the combined parallel resistance to just the 20 Ω resistor). Since the voltage is constant, the total current from the cell decreases (Ohm’s law: \( I = V/R \)).

(b)(ii) The current in the 20 Ω resistor remains unchanged.

Explanation: The potential difference (p.d.) across the 20 Ω resistor remains the same (3.0 V, as it is still directly connected across the cell terminals). Since the resistance of the 20 Ω resistor is unchanged, the current through it remains \( I = V/R = 3.0/20 = 0.15 \text{ A} \).

The 20 Ω resistor is still directly connected across the cell terminals, so the potential difference across it remains 3.0 V. Since its resistance is unchanged, the current through it remains \( I = 3.0/20 = 0.15 \text{ A} \). The change in total current only affects the branch that was removed.