Diagram 1 shows a circuit built by a student.
(a)(i) State the formula linking voltage, current and resistance.
(ii) Calculate the voltage across the \(17\,\Omega\) resistor.
(iii) State the voltage across the \(6.2\,\Omega\) resistor.
(iv) Calculate the current in the battery.
(b) Diagram 2 shows a second circuit built by the student using the same battery and resistors.
Explain how the current in the battery will change now the resistors are connected in series. You do not need to do any calculations in your answer.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
(a)(i) Ohm’s law [1 mark]
The formula linking voltage, current and resistance is:
\(\boxed{V=IR}\)
(a)(ii) Voltage across the \(17\,\Omega\) resistor [3 marks]
The current through the \(17\,\Omega\) resistor is \(83\,\mathrm{mA}\).
Convert the current into amperes:
\(I=\dfrac{83}{1000}=0.083\,\mathrm{A}\)
Using \(V=IR\):
\(V=0.083\times17\)
\(V=1.411\,\mathrm{V}\)
Therefore:
\(\boxed{V\approx1.4\,\mathrm{V}}\)
(a)(iii) Voltage across the \(6.2\,\Omega\) resistor [1 mark]
The two resistors are connected in parallel, so the potential difference across each resistor is the same.
Therefore:
\(\boxed{1.4\,\mathrm{V}}\)
(a)(iv) Current in the battery [2 marks]
In a parallel circuit, the current from the battery is the sum of the currents in the separate branches.
The current through the \(17\,\Omega\) resistor is \(83\,\mathrm{mA}\), and the current through the \(6.2\,\Omega\) resistor is \(228\,\mathrm{mA}\).
Therefore:
\(I_{\mathrm{battery}}=83+228\)
\(I_{\mathrm{battery}}=311\,\mathrm{mA}\)
Therefore:
\(\boxed{I_{\mathrm{battery}}=311\,\mathrm{mA}}\)
(b) Current when the resistors are connected in series [3 marks]
Therefore, the current supplied by the battery is smaller when the two resistors are connected in series.
