CIE iGCSE Co-Ordinated Science P4.3.2 Series and parallel circuits Exam Style Questions Paper 2
▶️ Answer/Explanation
The circuit shows a parallel arrangement of a 4.0Ω resistor and a 6.0Ω resistor connected to a 6.0V supply. In a parallel circuit, the potential difference (p.d.) across each branch is equal to the supply voltage (6.0V). Using Ohm’s law \(I = \frac{V}{R}\), the current in the 4.0Ω resistor is \(\frac{6.0}{4.0} = 1.5 \text{ A}\). The current in the 6.0Ω resistor is \(\frac{6.0}{6.0} = 1.0 \text{ A}\). Therefore, option B is correct.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
From the circuit, \(I_1\) is the total current from the source, which is greater than branch current \(I_3\). Also, the combined resistance \(R_{\mathrm{T}}\) is less than each individual resistor, so \(R_1\) is smaller than \(R_2\). Thus row A is correct.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
For resistors in parallel: \(\frac{1}{R} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{1}{2.0}\), so \(R = 2.0\ \Omega\), which is less than 3.0 Ω.
Ammeter Q, positioned in the main circuit, measures the total combined current from both parallel branches.
Since the total current is always greater than the current in any single branch, ammeter Q gives the greater reading.
✅ Answer: (B)
Ammeter Q, positioned in the main circuit, measures the total combined current from both parallel branches.
Since the total current is always greater than the current in any single branch, ammeter Q gives the greater reading.
✅ Answer: (B)
▶️ Answer/Explanation
Lamp brightness increases with more current through it, which happens when the total circuit resistance is lowest.
Placing resistors in parallel (rather than in series) with the lamp reduces the overall resistance the battery has to drive current through, allowing more current to flow through the lamp.
Circuit D has the arrangement that gives the lowest effective resistance in the lamp’s branch, making the lamp brightest.
✅ Answer: (D)
Placing resistors in parallel (rather than in series) with the lamp reduces the overall resistance the battery has to drive current through, allowing more current to flow through the lamp.
Circuit D has the arrangement that gives the lowest effective resistance in the lamp’s branch, making the lamp brightest.
✅ Answer: (D)
▶️ Answer/Explanation
In parallel, each lamp gets the full battery voltage, stays lit independently if the other breaks, and can be switched individually — so A, B and C are genuine benefits.
However, in parallel the total current drawn from the battery is actually larger (it’s the sum of the branch currents), not smaller.
So “smaller current in the battery” is not a real benefit of parallel connection.
✅ Answer: (D)
However, in parallel the total current drawn from the battery is actually larger (it’s the sum of the branch currents), not smaller.
So “smaller current in the battery” is not a real benefit of parallel connection.
✅ Answer: (D)