JAMB Physics Past Questions & Answers - Page 153

761.

A 120V, 60W lamp is to be operated on 220V ac supply mains. calculate the value of non inductive resistance that would be required to ensure that the lamp is run on correct value

A.

100Ω

B.

200Ω

C.

300Ω

D.

500Ω

Correct answer is B

Power of the lamp = 60W;  

\(I = {\frac{P}{V}}\)

\(I = {\frac{60}{120}}\) = \(0.5A\)

\(R = {\frac{V}{I}}\)

\(R_mains\) = \(\frac{220}{0.5}\) = \(440\Omega\)

\(R_lamp\) = \(\frac{120}{0.5}\) = \(240\Omega\)

\(\therefore\) The non-inductive resistance to keep the lamp = \((440 - 240)\Omega\)

= \(200\Omega\)

762.

Lenz's law is a law of conservation of

A.

electric charge

B.

electric current

C.

Energy

D.

Momentum

Correct answer is C

Lenz's lew of electrical magnetic induction is usefully referred to as the law of conservation of energy

763.

Two long parallel wires X and Y carry currents 3A and 5A each. if the force experienced by unit length is 5 x 10-5N. the force per unit length experienced by wire Y is

A.

3 x 10-5

B.

3 x 10-6

C.

5 x 105

D.

5 x 10-5

Correct answer is D

since the wires are parallel to each other, the forces which they exert on each other are equal and opposite, since they must agree with newton's law of action an reaction. Hence the force exerted by the wire is equally
5 x 10-5

764.

The instrument that measures both ac and dc is called

A.

A current balance

B.

a moving coil ammeter

C.

a moving iron ammeter

D.

an inverter

Correct answer is C

No explanation has been provided for this answer.

765.

A 40KW electric cable was uses to transmit electricity through a resistor of resistance 2.00Ω at 800V. The power loss as internal energy is

A.

4.0 x 102W

B.

5.0 x 102W

C.

4.0 x 103W

D.

5.0 x 103W

Correct answer is D

In general, Power = IV; \( \implies 40KW = IV \\
\text{Therefore } 40000 = 1 \times 800\\

\implies I = \frac{40000}{800} = 50A,\text{tune the current through} \)

Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W