Question: This question tests students’ knowledge of electrical circuits.
(a) In the first part of the question, students have to rank the magnitudes of the potential differences across five light bulbs (A, B, C, D, and E) from highest to lowest.
In circuit 1, there is only a light bulb that is connected to a battery. In circuit 2, there are two lightbulbs that are connected in series in which the same electric current flows through the circuit. In circuit 3, there are two lightbulbs that are connected in parallel in which the same potential difference is applied across both light bulbs.
In a sense, the flow of charge-carriers through light bulbs in the three circuits is similar to the flow of water through sponge filters in tubes as shown below:
We may assume the same water level is maintained in the three water tanks as shown above. As an example, there could be a total of 4 water drops per second that flows through the two tubes (a sponge filter each) as shown in Fig 1A. If there are one tube and a sponge filter, we expect only 2 water drops per second as shown in Fig 1B. However, if there are two sponge filters in a tube, we expect an increase in the resistance to the flow of water and there could be only 1 water drop per second as shown in Fig 1C.
(b) In the second part of the question, students have to identify the circuit that runs out of electrical energy first and the circuit that run out of electrical energy last. Students need to assume the batteries have the same amount of electrical energy and they are all connected to the light bulbs at the same time.
Possible answers:
1 (a)
In the first circuit, the light bulb A has the same potential difference (V) as the battery.
In the second circuit, the two light bulbs divide the potential difference (V) of the battery equally. Thus, the potential differences of light bulbs B and C are both equal to V/2.
In the third circuit, the two light bulbs D and E have the same potential difference (V) because they are connected in parallel to each other.
Thus, VA = VD = VE = V, whereas VB = VC = V/2. (We assume this is an ideal battery that has no internal resistance.)
(b)
For example, we may assume the potential difference of the battery is 1.0 V and the electrical resistance of the light bulb is 1 ohm.
In circuit 1, the electric current = V/R = (1.0)/1 = 1.0 A.
The power consumed in circuit 1 = IV = (1.0)(1.0) = 1.0 W.
In circuit 2, the electric current = V/R = (1.0)/(1 + 1) = 0.5 A.
The power consumed in circuit 2 = (0.5)(0.5) + (0.5)(0.5) = 0.5 W.
(The potential difference is halved because it is “shared” by the two lightbulbs.)
In circuit 3, the electric current = V/R = 1.0 A.
The power consumed in circuit 3 = (1.0)(1.0) + (1.0)(1.0) = 2.0 W.
In summary, P1 = 1.0 W, P2 = 0.5 W, P3 = 2.0 W.
The battery will run out of usable energy first in circuit 3 because it consumes relatively more power.
The battery will run out of usable energy last in circuit 2 because it consumes relatively lesser power.
Below is a comparison of a series circuit and parallel circuit.
Feynman’s insights or goofs?:
Feynman explains a series circuit as follows: “Suppose we have a more complicated circuit which has two pieces, which by themselves have certain impedances, Z1 and Z2 and we put them in series and apply a voltage. What happens? It is now a little more complicated, but if I is the current through Z1, the voltage difference across Z1, is V1 = IZ1; similarly, the voltage across Z2 is V2 = IZ2. The same current goes through both. Therefore the total voltage is the sum of the voltages across the two sections and is equal to V = V1 + V2 = (Z1 + Z2)I (Feynman et al., 1963, section 25–5 Series and parallel impedances).” Essentially, the electric current is the same through any components (resistors, capacitors, or inductors) that are in series. Thus, the total voltage through two components can be expressed as V = (Z1 + Z2) I.
In general, when we say that two resistors are in series, it means that they are connected end to end and the same electric current passes through them. In a sense, an electrical circuit that consists of only a battery and a resistor are also in series and it can be redrawn as shown below. Students should realize that the electric current that emerges from the battery (IA) is equal to the electric current after passing through the resistor (IB) and the electric current before passing through the resistor (IC).
In addition, Feynman also explains a parallel circuit as follows: “[w]e may also connect them in another way, called a parallel connection. Now we see that a given voltage across the terminals if the connecting wires are perfect conductors, is effectively applied to both of the impedances, and will cause currents in each independently. Therefore the current through Z1 is equal to I1= V/Z1. The current in Z2 is I2 = V/Z2. It is the same voltage. Now the total current which is supplied to the terminals is the sum of the currents in the two sections: I = V/Z1 + V/Z2 (Feynman et al., 1963, section 25–5 Series and parallel impedances).” In other words, it is the potential differences (or voltage differences) across the components in parallel that are the same instead of the electric current.
On the other hand, when we say that two resistors are in parallel, it means that the terminals of the two resistors are connected to the same two points such that their potential differences are the same. Interestingly, the electrical circuit that consists of a battery and a resistor are also in parallel and it can be redrawn as shown below. Students should realize that the potential differences across the battery and the resistor are the same. Furthermore, the potential difference between any point in blue and any point in red are the same.
References:
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
