A student investigates the resistance of a resistor and a lamp.
The student records VSV_{\text{S}}VS, the potential difference across the resistor R\text{R}R, and the current ISI_{\text{S}}IS, in the circuit:
VS=2.6 V V_{\text{S}} = 2.6\text{ V} VS=2.6 V IS=0.40 A I_{\text{S}} = 0.40\text{ A} IS=0.40 A(i) Calculate RSR_{\text{S}}RS, the resistance of resistor R\text{R}R, using the equation RS=VSISR_{\text{S}} = \frac{V_{\text{S}}}{I_{\text{S}}}RS=ISVS. Include the unit.
The student replaces the resistor with a lamp. She records VLV_{\text{L}}VL, the potential difference across the lamp, and the current ILI_{\text{L}}IL in the circuit:
VL=1.9 V V_{\text{L}} = 1.9\text{ V} VL=1.9 V IL=0.35 A I_{\text{L}} = 0.35\text{ A} IL=0.35 ACalculate RLR_{\text{L}}RL, the resistance of the lamp, using the equation RL=VLILR_{\text{L}} = \frac{V_{\text{L}}}{I_{\text{L}}}RL=ILVL.
Describe how the voltmeter must be connected to measure the combined potential difference VCV_{\text{C}}VC across the series combination of both the resistor and the lamp.
The student connects the resistor R\text{R}R and the lamp in series and records the combined potential difference VCV_{\text{C}}VC and the current ICI_{\text{C}}IC:
VC=4.5 V V_{\text{C}} = 4.5\text{ V} VC=4.5 V IC=0.32 A I_{\text{C}} = 0.32\text{ A} IC=0.32 ACalculate RCR_{\text{C}}RC, the combined resistance of the resistor and the lamp connected in series, using the equation RC=VCICR_{\text{C}} = \frac{V_{\text{C}}}{I_{\text{C}}}RC=ICVC.
State and explain briefly whether the results show that RS+RL=RCR_{\text{S}} + R_{\text{L}} = R_{\text{C}}RS+RL=RC within the limits of experimental accuracy (taken as 10%10\%10%).
132 exam-style questions on CIE IGCSE Physics 4.2.4a Resistance. Each one has a worked solution and a mark scheme showing where the marks go.