Tag: electric current and its effects

Questions Related to electric current and its effects

......... suggested that the slips should be marked with load lines

  1. Samuel Plimsoll

  2. Einstein

  3. Newton

  4. Archimedes


Correct Option: C
Explanation:

Fuse should be connected to live / phase wire o fthe circuit. If it is connected to neutral wire , the fuse will melt when excess current flows, but the appliance will still be connected to high potential through live wire. Thus, if a person touches the appliance, he receives shock. 
Hence, choice is (3) 

Fill in the blank-
An electric fuse works on the ______effect of current.

  1. heating

  2. magnetic

  3. chemical

  4. none of these


Correct Option: A
Explanation:

Electric fuse work on the principal of heating effect of current . fuse wire is a wire of high resistance and low melting point. when very large current will pass through it, heat will be generated which will melt the fuse wire and break the circuit.By using electric fuses in home we can protect the home appliances from damaging by large amount of current.

A fuse wire is generally made of

  1. tin

  2. tin-lead alloy

  3. copper

  4. either (a) or (b) above


Correct Option: D
Explanation:

Fuse wire are used to save the home appliances from damagaing by large currents. so they are made up of such materials which have high resistance and low melting point so that they can melt down when unexpected amount of current is passed through.for this work either tin metal or tin-lead alloy will be appropriate. 

An electric iron draws a current of 15 A from a 220 V supply. What is the cost of using iron for 30 min everyday for 15 days if the cost of unit (1 unit $=$ 1 k W hr) is 2 rupees?

  1. Rs. 49.5

  2. Rs. 60

  3. Rs. 40

  4. Rs. 10


Correct Option: A
Explanation:

$I=15 A, V=220 V, t=30 min$
$Energy=P\times t=VIt$
$=220\times 15\times \frac {30}{60}\times 15$
$=3300\times \frac {1}{2}\times 15 W hr$
$=3.3\times kW\times 7.5 hr$
$=24.75 kW hr$
$Cost=24.75\times 2=Rs. 49.50$

The parameter irrelevant for an electric fuse wire is 

  1. its radius.

  2. its specific resistance.

  3. current flowing through it.

  4. its length.


Correct Option: D
Explanation:

For fuse wire expression for heat produced is given by eq. is  $H = \dfrac{I^2 \rho}{2 \pi ^2r^3}$
Hence, it is independent on length of fuse wire. Thus, the parameter irrelevant for an electric fuse wire is its length.