296.7\(\Omega\)
400.0\(\Omega\)
512.2\(\Omega\)
806.7\(\Omega\)
Correct answer is D
P = \(\frac{v^2}{P} = \frac{220^2}{100} = 806.7\Omega\)
7.5A
2.5A
1.2A
0.8A
Correct answer is C
P = IV
I = \(\frac{P}{V} = \frac{3 \times 100}{250} = 1.2A\)
What is the total energy stored by the capacitors?
2.0 x 10-4J
1.0 x 10-4J
9.0 x 10-2J
1.0 x 10-2J
Correct answer is D
Energy = \(\frac{1}{2}CV^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 100^2\)
= 1.0 x 10-2J
What is the effective capacitance in the circuit?
2\(\mu F\)
6\(\mu F\)
18\(\mu F\)
216\(\mu F\)
Correct answer is A
\(\frac{1}{c} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6}\)
= \(\frac{1 + 1 + 1}{6} = \frac{3}{6} = \frac{1}{2}\)
= 2\(\mu F\)
increased
decrease
remains the same
is doubled
Correct answer is B
No explanation has been provided for this answer.
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