A student set up the electrical circuit shown in Figure 9 - AQA - GCSE Physics - Question 7 - 2018 - Paper 1
Question 7
A student set up the electrical circuit shown in Figure 9.
Figure 9
The ammeter displays a reading of 0.10 A.
Calculate the potential difference across the 45 Ω r... show full transcript
Worked Solution & Example Answer:A student set up the electrical circuit shown in Figure 9 - AQA - GCSE Physics - Question 7 - 2018 - Paper 1
Step 1
The ammeter displays a reading of 0.10 A. Calculate the potential difference across the 45 Ω resistor.
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Answer
To find the potential difference (V) across the 45 Ω resistor, we can use Ohm's Law, which states that V = I × R. Here, I = 0.10 A and R = 45 Ω.
Therefore,
V=0.10imes45=4.5extV
Step 2
Calculate the resistance of the resistor labelled R.
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Answer
To calculate the resistance R, we first find the total resistance of the circuit. The total resistance (R_total) in series is:
Rtotal=R+60Ω
Knowing that total current (I) is 0.10 A and using V = 12 V, we can first compute R_total:
From Ohm's Law, we have:
Rtotal=IV=0.1012=120Ω
Now we substitute in the series equation:
120=R+60⟹R=120−60=60Ω
Step 3
State what happens to the total resistance of the circuit and the current through the circuit when switch S is closed.
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Answer
When switch S is closed, the total resistance of the circuit decreases because the circuit will short the resistor labelled R. As a result, the total current will increase due to the reduced resistance, according to Ohm's Law. This is summarized as:
