iGCSE Physics (0625) 4.3.3 Action and use of circuit components -Exam Style Questions Paper 2 - New Syllabus
▶️ Answer/Explanation
Detailed solution:
To vary the potential difference (p.d.) from $0\text{ V}$ up to the full source voltage of $6.0\text{ V}$, a potentiometer (potential divider) arrangement is required.
In Circuit C, the lamp is connected in parallel with a variable length of the resistance wire using a sliding contact.
When the slider is at the far left, the lamp is short-circuited, resulting in a p.d. of $0\text{ V}$.
Moving the slider to the far right places the lamp across the entire resistance, giving it the full $6.0\text{ V}$ of the battery.
Circuits A and B use variable resistors in series, which cannot achieve $0\text{ V}$ as the fixed resistor or lamp always retains some p.d.
Circuit D is incorrectly wired as the lamp is in the main branch, preventing the full range of control.
▶️ Answer/Explanation
Detailed solution:
In a series circuit acting as a potential divider, the current $I$ flowing through both resistors $R_1$ and $R_2$ is identical.
According to Ohm’s Law ($V = IR$), the potential difference across each resistor is proportional to its resistance.
This relationship is expressed by the ratio $\frac{V_1}{R_1} = \frac{V_2}{R_2}$, as both sides represent the constant current $I$.
To solve for $V_1$, we rearrange the equation by multiplying both sides by $R_1$.
This yields the final expression: $V_1 = \frac{V_2 \times R_1}{R_2}$.
Thus, option B is the correct mathematical representation for calculating $V_1$ based on the given circuit parameters.
▶️ Answer/Explanation
Detailed solution:
In a series circuit, the current \(I\) is the same through both resistors. According to Ohm’s law, \(V = IR\).
Therefore, \(V_{1} = IR_{1}\) and \(V_{2} = IR_{2}\). Dividing these two equations eliminates the common current \(I\).
This gives the potential divider equation: \(\frac{V_{1}}{V_{2}} = \frac{IR_{1}}{IR_{2}} = \frac{R_{1}}{R_{2}}\).
Thus, the ratio of the voltages is equal to the ratio of the resistances, which matches Option D.