IB myp 4-5 physics – Practice Questions- All Topics
Topic :Electromagnetism-Electric Circuits
Topic :Electromagnetism– Weightage : 21 %
All Questions for Topic : Electric Fields,Static Electricity ,Magnetism,Magnetic Fields,Current,Power,Voltage,Electric Circuits,Electric and Magnetic Fields,Electromagnetic forces and induction
Question
A student investigates an electrical circuit containing a variable power supply, an ammeter and a filament lamp (bulb) in which the current through the bulb can be changed.
One of the circuit diagrams below is incorrect, the other circuit diagram is correct.
Explain why the current through the bulb could not be measured using the incorrect circuit.
▶️Answer/Explanation
Ans:
The current through the bulb could not be measured using the incorrect circuit where the ammeter is connected in parallel. Here’s why:
1. Parallel Connection: When the ammeter is connected in parallel with the bulb, it forms a separate branch or path for current to flow. In this arrangement, the current will divide between the bulb and the ammeter. As a result, the ammeter will measure only a fraction of the total current passing through the circuit, which does not accurately represent the current flowing through the bulb.
2. Incorrect Reading: Since the ammeter measures only a fraction of the total current in the parallel configuration, the reading obtained would not reflect the actual current passing through the bulb. This incorrect measurement hinders the student from obtaining accurate data about the current flowing through the bulb in the circuit.
Therefore, connecting the ammeter in parallel with the bulb in the circuit would lead to inaccurate readings and prevent the proper measurement of the current through the bulb.
Question:
A current of 0.25A flows through lamp A in the circuit below.
a) How much charge flows through the circuit in 10minutes?
▶️Answer/Explanation
Ans: 0.25 × 10 × 60 = 150 C
b) If the current through lamp B is 0.15A, what current must flow through lamp C?
▶️Answer/Explanation
Ans: 0.25 – 0.15 = 0.10 A