 June 2019 Q.1. Solutions - Grad Plus

# June 2019 Q.1. Solutions

Q.1.a) Define resistance and state its unit. What is the effect of temperature on resistance of the i] metallic conductor ii] alloys and iii] insulator. [06 M]

Ans.

Resistance is defined as the opposition to the flow of current.

The property of a substance due to which it opposes (or restricts) the flow of electricity i.e. electrons) through it.

The unit is ohm (Ω).

Effect of temperature on resistance of the

i) metallic conductor – To increase the resistance of pure metals like metallic conductor. The increase is large and fairly regular for normal ranges of temperature. The temperature /resistance graph is a straight line . Metals have a positive temperature coefficient of resistance.

ii) alloys – To increase the resistance of alloys, the increase is relatively smalland irregular. For some high resistance alloys like Eureka(60% Cu and 40% Ni) and manganin, , the increase in resistance is negligible over a considerable range of temperature.

iii) insulator – To decrease the resistance of electrolytes, insulators (such as paper, rubber, glass,mica,etc.) and partial conductors such as carbon. Hence, insulators said to have a negative temperarture co-efficient of resistance.

These are the effect of temperature on resistance of metallic conductor, alloys and insulator.

Q.1.b) Find the induced emf in the coil having inductance of 0.15 H when i] current of 10 A in the coil is switched off in 0.01 sec ii] same current is uniformly reversed in 0.01 sec. [06 M]

Ans.

Given :- L= 0.15 H, I= 10 A , t=0.01 sec

To Find :- i] current of 10 A in the coil is switched off in 0.01 sec

ii] same current is uniformly reversed in 0.01 sec.

Solution :-

i] current of 10 A in the coil is switched off in 0.01 sec

As current in the coil is switched off in 0.01 sec , it means di = 10-0 = 10

e=-L \; \frac{di}{dt} e=- 0.15 \; \frac{10}{0.01}=-\; 150 \; V

|e| = 150 V

e= 150 V

ii) same current is uniformly reversed in 0.01 sec.

e=L\times \frac{di}{dt}=0.15\times[10-(-10)]/0.01= 300 V

e= 300 V

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