100 Ω
200 Ω
300 Ω
500 Ω
Correct answer is B
Power of the lamp = 60W = (V2)/R and the voltage rate = 120v.
R = (V2)/60 = (120 X 120)/60 = 240 Ω, the fuse wire that should ensure current does not exceed this value should be rated 200 Ω
The instrument that measures both a.c and d.c is?
a current balance
a moving current ammeter
a moving iron ammeter
an inverter
Correct answer is C
No explanation has been provided for this answer.
4.0 x 102W
5.0 x 102W
4.0 x 103W
5.0 x 103W
Correct answer is D
In general, power = IV; 40kw = IV.
∴ 40,000 = 1 x 800
1 = (40.000)/800 = 50A. tune the current through resistor = 50A.
∴ power loss = 12R = 502 X 2 = 2500 X 2 = 5.0 X 103W
2.55 x 105 Ωm
2.55 x 102 Ωm
3.93 x 10-6 Ωm
3.93 x 10-8 Ωm
Correct answer is D
R = (1)/a, P = (R.a)/1 = (0.1 x 3.14)/2 (10-3/2)3
∴ P = (0.1 x 3.14 x 10-6)/2 x 4
= 0.3925 x 10-7 OR -3.930 X 10-8 Ωm
0.8 J
1.8 J
9.0 J
18.0 J
Correct answer is B
Work done = product of the charge Q, and the p.d V.
(i)/e work done = Q.V = 600 x 10-6 x 3.0 x 10-3 = 1.8 J.