Fig. 1 shows two resistors connected in series to a 12 V battery in a science demonstration box. Resistor R1 has resistance 2.0 ohms and resistor R2 has res...
Question 1 Report
Fig. 1 shows two resistors connected in series to a 12 V battery in a science demonstration box. Resistor R1 has resistance 2.0 ohms and resistor R2 has resistance 4.0 ohms. Three digital voltmeters are available. The teacher asks a student to predict the readings before closing the switch, then to compare the readings with the rule for voltages in a series circuit.
(a) Complete the predicted readings: current in the circuit, voltage across R1, and voltage across R2. [3]
(b) Calculate the total resistance of the two resistors. [2]
(c) Explain why the two resistor voltages add to 12 V. [2]
(d) Describe how a voltmeter is connected to measure the voltage across R2. [2]
Answer Details
(a) The series resistance is \(2.0+4.0=6.0\ \Omega\), so:
\[I=\frac{V}{R}=\frac{12}{6.0}=2.0\ \mathrm{A}\]
The voltage across \(R_1\) is \(V=IR=2.0\times2.0=4.0\ \mathrm{V}\). The voltage across \(R_2\) is \(V=IR=2.0\times4.0=8.0\ \mathrm{V}\). Thus: current \(=2.0\ \mathrm{A}\); voltage across \(R_1=4.0\ \mathrm{V}\); voltage across \(R_2=8.0\ \mathrm{V}\) [3].
(b) In series, resistances add:
\[R_\text{total}=2.0\ \Omega+4.0\ \Omega=6.0\ \Omega\]
Total resistance = \(6.0\ \Omega\) [2].
(c) The battery transfers \(12\ \mathrm{J}\) of energy per coulomb of charge [1]. All of this energy is transferred in the two resistors, so the potential differences in series add to the supply voltage: \(4.0+8.0=12\ \mathrm{V}\) [1].
(d) Connect the voltmeter in parallel with \(R_2\) [1], across both ends of \(R_2\), with correct polarity [1].
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