ONE-EIGHTH STAGE 2026
Aburi Girl’s SHS: 38 points
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St. Augustine’s College
Anlo SHS
Kumasi High School
ROUND 1
PREAMBLE
Find the EMF ($E$) of a battery whose terminal voltage is $12\text{ V}$ when it supplies current $I_1$, and $15\text{ V}$ when it supplies current $I_2$.
FORMULA
By solving the simultaneous equations $V_1 = E - I_1r$ and $V_2 = E - I_2r$, we get a direct formula for EMF:
$E = \dfrac{(I_1 \times V_2) - (I_2 \times V_1)}{I_1 - I_2}$
Note: Since current units appear in both the numerator and denominator, they cancel out.
You can keep the units in mA or A directly without converting!
1. $I_1$ is $3.0\text{ mA}$ and $I_2$ is $2.0\text{ mA}$.
SOLUTION:
$E = \dfrac{(3.0 \times 15) - (2.0 \times 12)}{3.0 - 2.0}$
$E = \dfrac{45 - 24}{1}$
$E = 21\text{ V}$
ANSWER: $21\text{ V}$
2. $I_1$ is $5.0\text{ mA}$, and $I_2$ is $4.5\text{ mA}$.
SOLUTION:
$E = \dfrac{(5.0 \times 15) - (4.5 \times 12)}{5.0 - 4.5}$
$E = \dfrac{75 - 54}{0.5}$
$E = \dfrac{21}{0.5}$
$E = \dfrac{21}{5\times10^{-1}}$
$E = \dfrac{21}{5} \times10^{1}$
$E = 42\text{ V}$
ANSWER: $42\text{ V}$
3. $I_1$ is $6.4\text{ A}$ and $I_2$ is $5.8\text{ A}$.
SOLUTION:
$E = \dfrac{(6.4 \times 15) - (5.8 \times 12)}{6.4 - 5.8}$
$E = \dfrac{96 - 69.6}{0.6}$
$E = \dfrac{26.4}{0.6}$
$E = \dfrac{264}{6}$
$E = 44\text{ V}$
ANSWER: $44\text{ V}$
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PRACTICE QUESTIONS
1. QUESTION: A battery has a terminal voltage of 11.2 V when it supplies a current of 2.5 A, and 11.8 V when it supplies a current of 1.5 A. Find the EMF of the battery.
ANSWER: 13 V or $1.3\times10^{1}$ V
SOLUTION:
$E = V + Ir$
$E = 11.2 + 2.5r$
$E = 11.8 + 1.5r$
$11.2 + 2.5r = 11.8 + 1.5r$
$r = \dfrac{11.8 - 11.2}{2.5 - 1.5}$
$r = 0.60\ \Omega$
$E = 11.8 + 1.5 \times 0.60$
$E \approx 13\text{ V}$
2. QUESTION: A battery has a terminal voltage of 8.4 V when it supplies a current of 3.2 A, and 9.0 V when it supplies a current of 1.2 A. Find the EMF of the battery.
ANSWER: 9.4 V
SOLUTION:
$E = V + Ir$
$E = 8.4 + 3.2r$
$E = 9.0 + 1.2r$
$8.4 + 3.2r = 9.0 + 1.2r$
$r = \dfrac{9.0 - 8.4}{3.2 - 1.2}$
$r = 0.30\ \Omega$
$E = 9.0 + 1.2 \times 0.30$
$E \approx 9.4\text{ V}$
3. QUESTION: A battery has a terminal voltage of 5.6 V when it supplies a current of 1.8 A, and 6.2 V when it supplies a current of 0.60 A. Find the EMF of the battery.
ANSWER: 6.5 V
SOLUTION:
$E = V + Ir$
$E = 5.6 + 1.8r$
$E = 6.2 + 0.60r$
$5.6 + 1.8r = 6.2 + 0.60r$
$r = \dfrac{6.2 - 5.6}{1.8 - 0.60}$
$r = 0.50\ \Omega$
$E = 6.2 + 0.60 \times 0.50$
$E = 6.5\text{ V}$
4. QUESTION: A battery has a terminal voltage of 11.2 V when it supplies a current of 3.2 A, and 11.8 V when it supplies a current of 1.2 A. Find the EMF of the battery.
ANSWER: 12 V or $1.2\times10^{1}$ V
SOLUTION:
$E = V + Ir$
$E = 11.2 + 3.2r$
$E = 11.8 + 1.2r$
$11.2 + 3.2r = 11.8 + 1.2r$
$r = \dfrac{V_2 - V_1}{I_1 - I_2}$
$r = \dfrac{11.8 - 11.2}{3.2 - 1.2}$
$r = \dfrac{0.60}{2.0}$
$r = \dfrac{6\times10^{-1}}{2}$
$r = 0.30\ \Omega$
$E = 11.8 + 1.2 \times 0.30$
$E = 11.8 + 12\times10^{-1} \times 3\times10^{-1}$
$E = 11.8 + 12 \times 3 \times10^{-2}$
$E = 11.8 + 0.36$
$E \approx 12\text{ V}$
5. QUESTION: A battery has a terminal voltage of 2.7 V when it supplies a current of 0.85 A, and 2.9 V when it supplies a current of 0.35 A. Find the EMF of the battery.
ANSWER: 3.0 V
SOLUTION:
$E = V + Ir$
$E = 2.7 + 0.85r$
$E = 2.9 + 0.35r$
$2.7 + 0.85r = 2.9 + 0.35r$
$r = \dfrac{2.9 - 2.7}{0.85 - 0.35}$
$r = 0.40\ \Omega$
$E = 2.9 + 0.35 \times 0.40$
$E \approx 3.0\text{ V}$
6. QUESTION: A battery has a terminal voltage of 2.7 V when it supplies a current of 6.4 A, and 2.9 V when it supplies a current of 2.4 A. Find the EMF of the battery.
ANSWER: 3.0 V
SOLUTION:
$E = V + Ir$
$E = 2.7 + 6.4r$
$E = 2.9 + 2.4r$
$2.7 + 6.4r = 2.9 + 2.4r$
$r = \dfrac{V_2 - V_1}{I_1 - I_2}$
$r = \dfrac{2.9 - 2.7}{6.4 - 2.4}$
$r = \dfrac{0.20}{4.0}$
$r = \dfrac{2\times10^{-1}}{4}$
$r = \dfrac{1}{2} \times10^{-1}$
$r = 0.050\ \Omega$
$E = 2.9 + 2.4 \times 0.050$
$E = 2.9 + 24\times10^{-1} \times 5\times10^{-2}$
$E = 2.9 + 24 \times 5 \times10^{-3}$
$E = 2.9 + 0.12$
$E \approx 3.0\text{ V}$
7. QUESTION: A battery has a terminal voltage of 4.2 V when it supplies a current of 2.2 A, and 4.6 V when it supplies a current of 1.2 A. Find the EMF of the battery.
ANSWER: 5.1 V
SOLUTION:
$E = V + Ir$
$E = 4.2 + 2.2r$
$E = 4.6 + 1.2r$
$4.2 + 2.2r = 4.6 + 1.2r$
$r = \dfrac{4.6 - 4.2}{2.2 - 1.2}$
$r = 0.40\ \Omega$
$E = 4.6 + 1.2 \times 0.40$
$E \approx 5.1\text{ V}$
8. QUESTION: A battery has a terminal voltage of 1.35 V when it supplies a current of 0.45 A, and 1.45 V when it supplies a current of 0.25 A. Find the EMF of the battery.
ANSWER: 1.6 V
SOLUTION:
$E = V + Ir$
$E = 1.35 + 0.45r$
$E = 1.45 + 0.25r$
$1.35 + 0.45r = 1.45 + 0.25r$
$r = \dfrac{V_2 - V_1}{I_1 - I_2}$
$r = \dfrac{1.45 - 1.35}{0.45 - 0.25}$
$r = \dfrac{0.10}{0.20}$
$r = \dfrac{1}{2}$
$r = 0.50\ \Omega$
$E = 1.45 + 0.25 \times 0.50$
$E = 1.45 + 25\times10^{-2} \times 5\times10^{-1}$
$E = 1.45 + 25 \times 5 \times10^{-3}$
$E = 1.45 + 0.125$
$E \approx 1.6\text{ V}$
9. QUESTION: A battery has a terminal voltage of 7.5 V when it supplies a current of 3.6 A, and 8.1 V when it supplies a current of 1.6 A. Find the EMF of the battery.
ANSWER: 8.6 V
SOLUTION:
$E = V + Ir$
$E = 7.5 + 3.6r$
$E = 8.1 + 1.6r$
$7.5 + 3.6r = 8.1 + 1.6r$
$r = \dfrac{8.1 - 7.5}{3.6 - 1.6}$
$r = 0.30\ \Omega$
$E = 8.1 + 1.6 \times 0.30$
$E \approx 8.6\text{ V}$
10. QUESTION: A battery has a terminal voltage of 23.4 V when it supplies a current of 2.4 A, and 24.0 V when it supplies a current of 1.2 A. Find the EMF of the battery.
ANSWER: 25 V or $2.5\times10^{1}$ V
SOLUTION:
$E = V + Ir$
$E = 23.4 + 2.4r$
$E = 24.0 + 1.2r$
$23.4 + 2.4r = 24.0 + 1.2r$
$r = \dfrac{24.0 - 23.4}{2.4 - 1.2}$
$r = 0.50\ \Omega$
$E = 24.0 + 1.2 \times 0.50$
$E \approx 25\text{ V}$
11. QUESTION: A battery has a terminal voltage of 7.5 V when it supplies a current of 4.5 A, and 8.1 V when it supplies a current of 1.5 A. Find the EMF of the battery.
ANSWER: 8.4 V
SOLUTION:
$E = V + Ir$
$E = 7.5 + 4.5r$
$E = 8.1 + 1.5r$
$7.5 + 4.5r = 8.1 + 1.5r$
$r = \dfrac{V_2 - V_1}{I_1 - I_2}$
$r = \dfrac{8.1 - 7.5}{4.5 - 1.5}$
$r = \dfrac{0.60}{3.0}$
$r = \dfrac{6\times10^{-1}}{3}$
$r = 0.20\ \Omega$
$E = 8.1 + 1.5 \times 0.20$
$E = 8.1 + 15\times10^{-1} \times 2\times10^{-1}$
$E = 8.1 + 15 \times 2 \times10^{-2}$
$E = 8.1 + 0.30$
$E = 8.4\text{ V}$
12. QUESTION: A battery has a terminal voltage of 9.6 V when it supplies a current of 1.5 A, and 10.4 V when it supplies a current of 0.50 A. Find the EMF of the battery.
ANSWER: 11 V or $1.1\times10^{1}$ V
SOLUTION:
$E = V + Ir$
$E = 9.6 + 1.5r$
$E = 10.4 + 0.50r$
$9.6 + 1.5r = 10.4 + 0.50r$
$r = \dfrac{10.4 - 9.6}{1.5 - 0.50}$
$r = 0.80\ \Omega$
$E = 10.4 + 0.50 \times 0.80$
$E \approx 11\text{ V}$