2020 A/L Physics Past Paper · Question 47 (English Medium)

Answer: (5) $2\ \Omega < R < 3\ \Omega$

What the question asks

We must find: the range in which the equivalent resistance $R$ between $A$ and $B$ of the endless ladder lies.

Given information

Given: every resistor is $1\ \Omega$; each section has one on the top rail, one on the bottom rail, then a rung across; this pattern continues without end.

Relevant theory

V =IR, where R is a constant.

"The algebraic sum of the currents flowing across a junction of a circuit is zero".

Reasoning and calculation

Structure: from $A$, a top $1\ \Omega$ to junction $C$; from $B$, a bottom $1\ \Omega$ to junction $D$; between $C$ and $D$, the first rung $1\ \Omega$; the rest of the ladder continues from the same $C$, $D$.

Symmetry (self-similarity): the ladder is endless; removing the first section leaves, from $C$, $D$, the exact same endless ladder. So the resistance of the remaining ladder between $C$ and $D$ is also $R$.

Nodes: $A \to C$ on the top rail and $D \to B$ on the bottom rail carry the whole current (series). Between $C$ and $D$ there are two paths: the rung $1\ \Omega$ and the rest of the ladder $R$ — the same potential difference across both.

Reducing: replacing the rest of the ladder by one resistance $R$, the network becomes $1\ \Omega$ — ($1\ \Omega \parallel R$) — $1\ \Omega$ in series. By Kirchhoff's first law at junction $C$, $I = \dfrac{V_{CD}}{1} + \dfrac{V_{CD}}{R}$; so the equivalent resistance between $C$ and $D$ is $\dfrac{V_{CD}}{I} = \dfrac{R}{1 + R}$.

Required equations

$R_{\text{series}} = R_1 + R_2 + \cdots$

$I = \frac{V}{R_1} + \frac{V}{R_2} \;\Rightarrow\; R_{\parallel} = \frac{R_1R_2}{R_1 + R_2}$

$R = 1 + \frac{1 \times R}{1 + R} + 1$

$R(1 + R) = 2(1 + R) + R \;\Rightarrow\; R^2 - 2R - 2 = 0$

$2.73$

$R = \frac{2 + \sqrt{4 + 8}}{2} = 1 + \sqrt{3}$

Physical root: the other root of $R^2 - 2R - 2 = 0$ is $1 - \sqrt3 < 0$ — a resistance cannot be negative; so $R = 1 + \sqrt3 \approx 2.73\ \Omega$, i.e. $2\ \Omega < R < 3\ \Omega$.

Reading the options

Comparing with the options: $R \approx 2.73\ \Omega$ — option (5) $2\ \Omega < R < 3\ \Omega$.

Checks

$1 + sqrt(3)$

Boundary check: a single section alone is $1 + 1 + 1 = 3\ \Omega$ (the rest an open circuit — an upper bound); the rest as a short circuit gives $2\ \Omega$ (a lower bound). The true $R$ lies between these. Second way (finite ladders): $n = 1$: $3$; $n = 2$: $2 + \dfrac{3}{4} = 2.75$; $n = 3$: $2 + \dfrac{2.75}{3.75} = 2.733$ — converging toward $2.732$ ✓.

Conclusion

$R = 1 + \sqrt{3} \approx 2.73\ \Omega$

The rest of the ladder is also $R$: $R = 2 + \dfrac{R}{1 + R}$, giving $R = 1 + \sqrt3 \approx 2.73\ \Omega$ — the answer is (5) $2\ \Omega < R < 3\ \Omega$.

Assuming the resistance keeps increasing as more sections are added to the endless ladder.

Every new section joins in parallel with the previous rung: the resistance decreases and settles at $1 + \sqrt3$.