VITEEE 2014 :: Physics :: General questions

• Physics - General questions

A circuit has a self-inductance of 1H and carries a current of 2A . To prevent sparking, when the circuit is switched off, a capacitor that can withstand 400V is used. The least capacitance of capacitor connected across the switch must be equal to 

Answer:

Explanation:

Energy stored in capacitor =energy stored in inductance

i.e, $\frac{1}{2}CV^{2}=\frac{1}{2}LI^{2}$

$\Rightarrow$ $C=\frac{LI^{2}}{V^{2}}= \frac{1 \times (2)^{2}}{(400)^{2}}=25 \mu F$

Four resistance of $10 \Omega,60\Omega,100 \Omega$ and $200 \Omega$ respectively taken in order are used to form a Wheatstone's bridge. A 15 V battery is connected to the ends of a $200 \Omega$ resistance, the current through it will be 

Answer:

Explanation:

Resistance $10 \Omega$,60$\Omega$ and 100$\Omega$ are in series and they together are in parallel to $200 \Omega$ resistance . When a potential difference of 15 V is applied across $200 \Omega$  then current through it 

$I=\frac{15}{200}=7.5 \times 10^{-2} A$

The denial cell is balanced on the 125 cm length of a potentiometer. Now, the cell is short-circuited by the resistance of $2 \Omega$ and the balance is obtained at 100 cm . The internal  resistance  of the denial cell is 

Answer:

Explanation:

Here,    $r=R\left(\frac{I_{1}}{I_{2}}-1\right)=2\left(\frac{125}{100}-1\right)$

$=2\left(\frac{5}{4}-1\right)=2\times\frac{1}{4}=0.05 \Omega$

A galvanometer has a current range of 15 mA and voltage range of 750mV. To convert this galvanometer into an ammeter of range 25 A, the required shunt is 

Answer:

Explanation:

Given : V=$750 \times 10^{-3}$V;

$I_{g}=15 \times 10^{-3} A$ and I=25 A

Using  , $a= \frac{y}{I_{g}}$

                 $= \frac{750 \times 10^{-3}}{15 \times 10^{-3}}=50 \Omega$

 $I_{g}=\frac{S}{S+a} \times I$

 $15 \times 10^{-3}=\left(\frac{S}{S+50}\right)\times25$

$\therefore$    $S=0.03 \Omega$

A coil of resistance 10Ω and an inductance 5H is connected to a 100 V battery. The energy stored in the coil is 

Answer:

Explanation:

Current $I=\frac{V}{R}=\frac{100}{10}=10A$

Energy , $E= \frac{1}{2}LI^{2}$

     $= \frac{1}{2}\times 5 \times (10)^{2}=250 J$

• Physics - General questions