The positive terminals of the batteries are connected together.
One battery has electromotive force (e.m.f.) 4.5 V and internal resistance 0.80 Ω.
The other battery has e.m.f. 2.4 V and internal resistance 0.50 Ω.
R is a coil of insulated wire of resistance 1.2 Ω at room temperature.
The switch S is closed.
[1]
I = ...................................................... A [3]
The number density of charge carriers in R is 4.2 × 1028 m–3.
Calculate the mean drift velocity v of the charge carriers in R.
v = ................................................ m s–1 [2]
Without any changes to the circuit itself, state and explain what practically can be done to make the measured current the same as the calculated current.
[2]
The circuit diagram of the arrangement is shown below.
The student changes the resistance of the variable resistor. The potential difference V across the variable resistor and the current I in the circuit are measured.
The V against I graph plotted by the student is shown below.
There is an incomplete table next to the graph.
R is the resistance of the variable resistor and P is the power dissipated by the variable resistor.
• Use the graph to determine E and r. Explain your reasoning.
• Calculate R and P to complete the table. Describe how P depends on R.
[6]
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