Question

A 12-volt battery is connected to two light bulbs, as drawn in figure 1 light bulb 1 has resistance 3 ohms, while light bulb 2 has resistance 6 ohms. The battery has essentially no internal resistance, and all the wires are essentially resistanceless, too. When a light bulb is unscrewed, no current flows through that branch of the circuit. For instance, if light bulb 2 is unscrewed, current flows only around the lower loop of the circuit, which consist of the battery and light bulb 1. The more current flows through a light bulb, their equivalent resistance is $R_{eq}=R_1+R_2$. By contrast, when two reisistors are wired in parallel, their net resisance is given by
$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$ .
Which of the following statement is false?

• A. Some of the energy produced by the light bulb takes the form of heat
• B. The battery is the source of all the electrons flowing around the circuit
• C. The current entering the light bulb equals the current leaving the light bulb
• D. The potential in the wire to the left of the light bulb differs from the potential in the wire to the right of that bulb

Answer 1

Answer: (B)

The battery is the source of all the electrons flowing around the circuit

Most of the charges flowing around the circuit are valence electrons stripped off the metal atoms in the wires and light bulbs. A battery doesn't "supply" all of the charges. It merely pushes around charges already present in the circuit. So the statement (C) and (D) are both true but (B) is wrong. All charges flowing into the light bulb also flow back out; no current gets "used up" But inside the bulb, those charges lose energy. This lost electrical energy converts into light and heat. So, the current has lower "potential"after flowing through the bulb. Thus statement (A) is also true.

Hence according to question the statement which is wrong is B.