If the current in the resistor R in diagram above is 0.05A, calculate the p.d. across the inductor.
2.5V
25.0V
49.0V
50.0V
250.0V
Correct answer is A
V= i × \(_l\)
X\(_l\) = inductive reactance → 50
I = current → 0.05A
V = 0.05 × 50
V = 2.5volt
Calculate the inductive reactance of the circuit shown above.
50.00
5.00
0.50
0.05
0.02
Correct answer is A
No explanation has been provided for this answer.
3.40x106V
6.75x105V
6.75x1012V
3.40x105V
8.30 x 102V
Correct answer is B
No explanation has been provided for this answer.
Calculate the inductance L of the coil in the circuit shown above.
14.4H
3.8H
0.6H
0.4H
0.2H
Correct answer is C
XL = \(\frac{V}{I} = \frac{240}{2} = 120\Omega\)
L = \(\frac{X_L}{2 \pi f}\)
= \(\frac{120 \times \pi}{2 \times \pi \times 100}\)
= 0.6H
526cm3
546cm3
556cm
566cm3
819cm3
Correct answer is D
No explanation has been provided for this answer.
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