Physics

Problem

12\,\text{V}

battery is connected in series with resistors of

4\,\Omega

and

6\,\Omega

. Points

P

Q

and

R

are located such that

P

is the positive terminal of the battery,

Q

is the junction between the two resistors and

R

is the negative terminal of the battery. a. (a) Calculate the current flowing in the circuit. b. (b) Determine the potential difference between points

Q

and

R

. c. (c) The student wishes to measure only the potential difference across the

6\,\Omega

resistor. Identify the two points between which the voltmeter should be connected. Justify your answer. d. (d) If the voltmeter is mistakenly connected between points

P

and

Q

, what numerical value will it show? Explain why this value differs from your answer in part (b).

Intuitive explanation

This is about voltage distribution in a series circuit—understanding that the battery's voltage doesn't stay constant, but instead "drops" across each resistor in proportion to its resistance. The key insight is that the voltage at any point depends on how much resistance you've already passed through: use Ohm's law ( $V = IR$ ) to find the current first, then track how much voltage remains as you move from the positive terminal toward the negative one.

Intuitive explanation

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