2024 A/L Physics Past Paper · Question 18 (English Medium)

Answer: (2) $0{\cdot}8\ \text{V}$

Given information

This question is about a circuit of five resistors joined to one battery.

The battery pushes a current round the circuit. The current splits at the junctions and joins again. Each resistor takes part of the battery voltage.

What we must find: the voltage across the resistor $R_4$.

How we find it: join the resistors step by step until one resistor is left. Find the current from the battery. Then go back into the circuit, step by step, to $R_4$.

$E=6.0\ \text{V}$, and the battery has negligible internal resistance. $R_1=50\ \Omega$, $R_2=50\ \Omega$, $R_3=60\ \Omega$, $R_4=30\ \Omega$, $R_5=30\ \Omega$.

What the question asks

$E$ ($\text{V}$): e.m.f. of the battery.

$R_1$ to $R_5$ ($\Omega$): the five resistances.

$R$ ($\Omega$): equivalent resistance.

$I$ ($\text{A}$): current.

$V$ ($\text{V}$): potential difference (voltage).

Relevant theory

If a number of resistors are connected so that the same current flows through them, then it is a series connection. The resistance of the single resistor which carries the same current but is maintained across it, the total potential difference of the system is called the equivalent resistance of the system.

If a number of resistors are connected so that they maintain the same potential difference, then it is a parallel connection. The resistance of the single resistance which carries the total current passing through the system but maintaining the same potential difference across it is called the equivalent resistance of the system.

V =IR, where R is a constant.

What is happening: $R_3$ and $R_4$ sit side by side, so they have the same voltage. Together they are in series with $R_5$, so the same current flows through that pair and through $R_5$.

This whole $R_3$, $R_4$, $R_5$ part sits side by side with $R_2$. Then all of it is in series with $R_1$ and the battery.

So we shrink the circuit to one resistor, find the battery current, and then open it up again to reach $R_4$.

For example, think of students leaving a school hall at the end of the day.

They all walk down one corridor. Then the corridor meets two doors that are just as easy to pass, and the crowd splits in half.

Behind one door, the path splits again into a wide stair and a narrow stair. More students take the wide one.

In words: a crowd splits at each junction, and the easier way takes the bigger share. Current in a circuit splits the same way.

Required equations

$R=R_1+R_2$

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$V=IR$

Applying it to this question

$20$

$R_{34}=\dfrac{60\times30}{60+30}$

$50$

$R_{345}=20+30$

$25$

$R_{2345}=\dfrac{50\times50}{50+50}$

$75$

$R=50+25$

$0.08$

$I=\dfrac{6.0}{75}$

$2.0$

$V_2=0.08\times25$

$0.04$

$I_{345}=\dfrac{2.0}{50}$

$0.8$

$V_4=0.04\times20$

Reasoning and calculation

$R_3$ and $R_4$ are in parallel, so the voltage across $R_4$ is the voltage across the pair, $0.8\ \text{V}$. This is option (2).

No step of this circuit gives $0.7\ \text{V}$. To get it, the current through $R_3$ and $R_4$ would have to be $0.035\ \text{A}$, but it is $0.04\ \text{A}$.

$1.2\ \text{V}$ is the voltage across $R_5$: $0.04\times30$. The question asks about $R_4$, which shares the current with $R_3$.

$2.0\ \text{V}$ is the voltage across $R_2$, and across the whole $R_3$, $R_4$, $R_5$ part. That part splits it again: $1.2\ \text{V}$ on $R_5$ and only $0.8\ \text{V}$ on $R_4$.

$2.4\ \text{V}$ comes from sharing the whole $6.0\ \text{V}$ between $20\ \Omega$ and $30\ \Omega$, as if $R_1$ and $R_2$ were not there. $R_1$ takes $4.0\ \text{V}$ first.

Checks

$E/(50+25)*25/50*20$

$V_1+V_5+V_4=4.0+1.2+0.8$

$4.0+1.2+0.8=6.0\ \text{V}$

Going round $R_1$, $R_5$ and $R_4$ back to the battery, the voltages add to the e.m.f., $6.0\ \text{V}$, as they must.

$V=IR$ is $\text{A}\times\Omega$, which is $\text{V}$, a voltage.

Conclusion

$V_4=0.8\ \text{V}$

The voltage across $R_4$ is $0.8\ \text{V}$. The answer is (2).