QUARTER FINAL STAGE 2025
Mfantsipim School: 64 points (Winner)
Prempeh College: 40 points
Adisadel College: 40 points
ROUND 2 - SPEED RACE
QUESTION
Find the current drawn by a resistor network composed of a 5.5 Ω resistor in series with a parallel connection of a 12 Ω and a 13 Ω resistor, from a 12 V battery with internal resistance 0.26 Ω.
Keep all values and final answers to 2 significant figures.
ANSWER: 1.0 A
SOLUTION 1:
Calculate the parallel combination using the 25-multiplier trick:
$R_p = \dfrac{12 \times 13}{12 + 13}$
$R_p = \dfrac{156}{25}$
Multiply numerator and denominator by 4.
$ = \dfrac{156 \times 4}{25 \times 4}$
$ = \dfrac{39 \times 4}{25}$
$ = 6.24$
Sum all series components (including internal resistance) to find the total resistance:
$R_{\text{total}} = 5.5 + 6.24 + 0.26$
$R_{\text{total}} = 12.0\ \Omega$
Apply Ohm's law to evaluate the total circuit current:
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{12}{12.0}$
$I = 1.0\text{ A}$
SOLUTION 2:
Determine the equivalent resistance of the parallel combination branch:
$\dfrac{1}{R_p} = \dfrac{1}{12} + \dfrac{1}{13}$
$\dfrac{1}{R_p} = \dfrac{13 + 12}{156}$
$\dfrac{1}{R_p} = \dfrac{25}{156}$
$R_p = 6.24\ \Omega$
Add the remaining in-line series resistance parameter and the battery internal resistance $r$:
$R_{\text{total}} = R_{\text{series}} + R_p + r$
$R_{\text{total}} = 5.5 + 6.24 + 0.26$
$R_{\text{total}} = 12.0\ \Omega$
Use the total electromotive force equation to calculate the circuit loop current:
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{12\text{ V}}{12.0\ \Omega}$
$I = 1.0\text{ A}$
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PRACTICE QUESTIONS
1. QUESTION: Find the current drawn by a resistor network made of a 5.5 Ω resistor in series with a parallel connection of a 6.0 Ω and a 12 Ω resistor, from a 12 V battery with internal resistance 0.50 Ω.
ANSWER: 1.2 A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 18.0\ \Omega$
$R_p = \dfrac{6.0 \times 12}{18.0}$
$R_p = \dfrac{12}{3}$
$R_p = 4.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 5.5 + 4.0 + 0.50$
$R_{\text{total}} = 10.0\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{12}{10.0}$
$I = \dfrac{6}{5}$
$I = 1.2\text{ A}$
2. QUESTION: Find the current drawn by a resistor network made of a 3.8 Ω resistor in series with a parallel connection of a 4.0 Ω and a 12 Ω resistor, from a 14 V battery with internal resistance 0.20 Ω.
ANSWER: 2.0 A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 16.0\ \Omega$
$R_p = \dfrac{4.0 \times 12}{16.0}$
$R_p = \dfrac{12}{4}$
$R_p = 3.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 3.8 + 3.0 + 0.20$
$R_{\text{total}} = 7.00\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{14}{7.00}$
$I = 2.0\text{ A}$
3. QUESTION: Find the current drawn by a resistor network made of a 2.7 Ω resistor in series with a parallel connection of a 3.0 Ω and a 6.0 Ω resistor, from a 6.0 V battery with internal resistance 0.30 Ω.
ANSWER: 1.2 A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 9.0\ \Omega$
$R_p = \dfrac{3.0 \times 6.0}{9.0}$
$R_p = \dfrac{6}{3}$
$R_p = 2.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 2.7 + 2.0 + 0.30$
$R_{\text{total}} = 5.00\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{6.0}{5.00}$
$I = 1.2\text{ A}$
4. QUESTION: Find the current drawn by a resistor network made of a 1.5 Ω resistor in series with a parallel connection of a 12 Ω and a 24 Ω resistor, from a 9.0 V battery with internal resistance 0.50 Ω.
ANSWER: 0.90 A or $9.0\times10^{-1}$ A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 36\ \Omega$
$R_p = \dfrac{12 \times 24}{36}$
$R_p = \dfrac{24}{3}$
$R_p = 8.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 1.5 + 8.0 + 0.50$
$R_{\text{total}} = 10.0\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{9.0}{10.0}$
$I = 0.90\text{ A}$
5. QUESTION: Find the current drawn by a resistor network made of a 4.6 Ω resistor in series with a parallel connection of a 8.0 Ω and a 24 Ω resistor, from a 22 V battery with internal resistance 0.40 Ω.
ANSWER: 2.0 A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 32.0\ \Omega$
$R_p = \dfrac{8.0 \times 24}{32.0}$
$R_p = \dfrac{24}{4}$
$R_p = 6.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 4.6 + 6.0 + 0.40$
$R_{\text{total}} = 11.0\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{22}{11.0}$
$I = 2.0\text{ A}$
6. QUESTION: Find the current drawn by a resistor network made of a 2.5 Ω resistor in series with a parallel connection of a 9.0 Ω and a 18 Ω resistor, from a 4.5 V battery with internal resistance 0.50 Ω.
ANSWER: 0.50 A or $5.0\times10^{-1}$ A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 27.0\ \Omega$
$R_p = \dfrac{9.0 \times 18}{27.0}$
$R_p = \dfrac{18}{3}$
$R_p = 6.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 2.5 + 6.0 + 0.50$
$R_{\text{total}} = 9.00\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{4.5}{9.00}$
$I = \dfrac{45\times10^{-1}}{9}$
$I = 0.50\text{ A}$
7. QUESTION: Find the current drawn by a resistor network made of a 0.80 Ω resistor in series with a parallel connection of a 36 Ω and a 12 Ω resistor, from a 15 V battery with internal resistance 0.20 Ω.
ANSWER: 1.5 A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 48\ \Omega$
$R_p = \dfrac{36 \times 12}{48}$
$R_p = \dfrac{3 \times 12}{4}$
$R_p = 3 \times 3$
$R_p = 9.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 0.80 + 9.0 + 0.20$
$R_{\text{total}} = 10.0\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{15}{10.0}$
$I = \dfrac{3}{2}$
$I = 1.5\text{ A}$
8. QUESTION: Find the current drawn by a resistor network made of a 1.7 Ω resistor in series with a parallel connection of a 4.0 Ω and a 4.0 Ω resistor, from a 6.0 V battery with internal resistance 0.30 Ω.
ANSWER: 1.5 A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 8.0\ \Omega$
$R_p = \dfrac{4.0 \times 4.0}{8.0}$
$R_p = \dfrac{4}{2}$
$R_p = 2.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 1.7 + 2.0 + 0.30$
$R_{\text{total}} = 4.00\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{6.0}{4.00}$
$I = \dfrac{3}{2}$
$I = 1.5\text{ A}$
9. QUESTION: Find the current drawn by a resistor network made of a 3.5 Ω resistor in series with a parallel connection of a 12 Ω and a 12 Ω resistor, from a 9.6 V battery with internal resistance 0.50 Ω.
ANSWER: 0.96 A or $9.6\times10^{-1}$ A
SOLUTION:
$R_p = \dfrac{R_aR_b}{R_a + R_b}$
$R_a + R_b = 24\ \Omega$
$R_p = \dfrac{12 \times 12}{24}$
$R_p = \dfrac{12}{2}$
$R_p = 6.0\ \Omega$
$R_{\text{total}} = R_1 + R_p + r$
$R_{\text{total}} = 3.5 + 6.0 + 0.50$
$R_{\text{total}} = 10.0\ \Omega$
$I = \dfrac{E}{R_{\text{total}}}$
$I = \dfrac{9.6}{10.0}$
$I = \dfrac{96\times10^{-1}}{10}$
$I = \dfrac{48}{5} \times10^{-1}$
$I = 0.96\text{ A}$