ELECTRIC CIRCUITS — GRADE 11 & 12
Circuit Graphs
Six graphs that appear in NSC Physics exams — each one is just the EMF equation rearranged into a straight line. Every step of the algebra is shown, and each graph's slope and y-intercept are explained.
6 graphs
Grade 11 & 12
EMF & internal resistance
\(y = mx + c\) derivations
The master equation
\(\varepsilon = I(R + r)\)
\(\varepsilon\)EMF of the cell (V)
\(I\)current in circuit (A)
\(R\)external resistance (Ω)
\(r\)internal resistance (Ω)
01
V vs I — Ohm's Law (resistor only)
straight line through origin\(I\text{ (A)}\)
\(V\text{ (V)}\)
\(O\)
\(m = R\)
\(\Delta I\)
\(\Delta V\)
1
Start with Ohm's Law
\[V = IR\]
2
This is already in y = mx + c form (c = 0)
\[V = R \cdot I + 0\]
3
Identify each part
\[y \to V,\quad x \to I,\quad m \to R,\quad c \to 0\]
Compare to y = mx + c
| y | = | m | · | x | + | c |
| \(V\) | = | \(R\) | · | \(I\) | + | \(0\) |
Reading the graph
- The slope \(= R\) — steeper line means larger resistance
- Graph passes through the origin — no current, no voltage drop
02
I vs V — Ohm's Law (axes swapped)
straight line through origin\(V\text{ (V)}\)
\(I\text{ (A)}\)
\(O\)
\(m = \tfrac{1}{R}\)
\(\Delta V\)
\(\Delta I\)
1
Start with Ohm's Law
\[V = IR\]
2
Divide both sides by R to make I the subject
\[I = \frac{V}{R}\]
3
Rewrite to show the slope clearly
\[I = \frac{1}{R} \cdot V + 0\]
4
Identify each part
\[y \to I,\quad x \to V,\quad m \to \frac{1}{R},\quad c \to 0\]
Compare to y = mx + c
| y | = | m | · | x | + | c |
| \(I\) | = | \(\tfrac{1}{R}\) | · | \(V\) | + | \(0\) |
Reading the graph
- Slope \(= \tfrac{1}{R}\) — steeper line means smaller resistance (opposite to graph 01)
- Still through the origin
- Used when experimenters plot \(I\) on the y-axis against \(V\) on the x-axis
03
V vs time — switch closes at \(t_1\)
step function\(t\text{ (s)}\)
\(V\text{ (V)}\)
\(\varepsilon\)
\(V_t\)
\(t_1\)
switch open
\(I=0,\; V=\varepsilon\)
switch closed
\(I \neq 0,\; V < \varepsilon\)
\(Ir\)
1
Start with the EMF equation, expand
\[\varepsilon = I(R+r) = IR + Ir\]
2
Terminal voltage V = IR — substitute
\[\varepsilon = V + Ir \implies V = \varepsilon - Ir\]
3
Before t₁: switch open, I = 0
\[V = \varepsilon - (0)r = \varepsilon\]
4
After t₁: switch closes, current flows
\[V = \varepsilon - Ir < \varepsilon\]
What the graph shows
| Region | Voltage reading | |||||
| \(t < t_1\) | → | \(V = \varepsilon\) (EMF) | ||||
| \(t > t_1\) | → | \(V = \varepsilon - Ir\) (terminal) | ||||
Reading the graph
- The upper flat section gives the EMF \(\varepsilon\) — read the y-value
- The lower flat section is the terminal voltage \(V_t\)
- The drop at \(t_1\) equals \(Ir\) — voltage lost to internal resistance
- Larger \(r\) → bigger drop between the two levels
04
V vs I — Terminal voltage vs current
decreasing line · y-intercept = \(\varepsilon\)\(I\text{ (A)}\)
\(V\text{ (V)}\)
\(O\)
\(\varepsilon\)
\(\tfrac{\varepsilon}{r}\)
\(m = {-r}\)
1
Start with the EMF equation
\[\varepsilon = I(R + r)\]
2
Expand the brackets
\[\varepsilon = IR + Ir\]
3
Terminal voltage V = IR — substitute
\[\varepsilon = V + Ir\]
4
Make V the subject
\[V = \varepsilon - Ir\]
5
Rewrite to match y = mx + c exactly
\[V = -r \cdot I + \varepsilon\]
Compare to y = mx + c
| y | = | m | · | x | + | c |
| \(V\) | = | \(-r\) | · | \(I\) | + | \(\varepsilon\) |
Reading the graph
- y-intercept \(= \varepsilon\) — the EMF of the cell
- Slope \(= -r\) — gradient is negative; \(|\text{slope}|\) gives the internal resistance
- x-intercept \(= \tfrac{\varepsilon}{r}\) — short-circuit current (when \(V = 0\))
- Flatter line → smaller internal resistance
05
R vs \(\tfrac{1}{I}\) — External resistance vs reciprocal current
positive slope · negative y-intercept\(\tfrac{1}{I}\)
\(R\text{ (}\Omega\text{)}\)
\(O\)
\(-r\)
\(m = \varepsilon\)
1
Start with the EMF equation
\[\varepsilon = I(R + r)\]
2
Divide both sides by I
\[\frac{\varepsilon}{I} = R + r\]
3
Subtract r from both sides to make R the subject
\[R = \frac{\varepsilon}{I} - r\]
4
Rewrite ε/I as ε × (1/I) to reveal the straight-line form
\[R = \varepsilon \cdot \frac{1}{I} - r\]
5
Identify each part
\[y \to R,\quad x \to \tfrac{1}{I},\quad m \to \varepsilon,\quad c \to {-r}\]
Compare to y = mx + c
| y | = | m | · | x | + | c |
| \(R\) | = | \(\varepsilon\) | · | \(\tfrac{1}{I}\) | + | \(-r\) |
Reading the graph
- Slope \(= \varepsilon\) — gradient gives the EMF directly
- y-intercept \(= -r\) — negative; the line extends below the x-axis, and the magnitude gives \(r\)
- The dashed section shows the line extended to reach the y-axis
06
\(\tfrac{1}{V}\) vs \(\tfrac{1}{R}\) — Reciprocal voltage vs reciprocal resistance
positive slope · positive y-intercept\(\tfrac{1}{R}\)
\(\tfrac{1}{V}\)
\(O\)
\(\tfrac{1}{\varepsilon}\)
\(m = \dfrac{r}{\varepsilon}\)
1
Start with the EMF equation
\[\varepsilon = I(R + r)\]
2
Terminal voltage V = IR, so I = V/R — substitute
\[\varepsilon = \frac{V}{R}(R + r)\]
3
Expand the brackets
\[\varepsilon = V + \frac{Vr}{R}\]
4
Factor V out on the right
\[\varepsilon = V\!\left(1 + \frac{r}{R}\right)\]
5
Divide both sides by Vε to isolate 1/V
\[\frac{1}{V} = \frac{1 + \dfrac{r}{R}}{\varepsilon}\]
6
Split the fraction to reveal y = mx + c
\[\frac{1}{V} = \frac{r}{\varepsilon} \cdot \frac{1}{R} + \frac{1}{\varepsilon}\]
Compare to y = mx + c
| y | = | m | · | x | + | c |
| \(\tfrac{1}{V}\) | = | \(\tfrac{r}{\varepsilon}\) | · | \(\tfrac{1}{R}\) | + | \(\tfrac{1}{\varepsilon}\) |
Reading the graph
- y-intercept \(= \tfrac{1}{\varepsilon}\) — take the reciprocal to find the EMF
- Slope \(= \tfrac{r}{\varepsilon}\) — divide slope by y-intercept to get \(r\)
- Both slope and intercept are positive — the line stays in the first quadrant
- This is the most algebraically complex graph — memorise the derivation step by step