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Engineering
64 Views   |

In a large building, there are 15 bulbs of   40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW.  The voltage of the electric mains is 220 V.  The minimum capacity of the main fuse of the building will be :

• Option 1)

8 A

• Option 2)

10 A

• Option 3)

12 A

• Option 4)

14 A

As we have learned

In parallel Grouping -

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\cdot \cdot \cdot +\frac{1}{R_{n}}$

-

Resistance of each bulb of 40 W $= \frac{v^2 }{P } = \frac{220 ^2 }{40 }= 1210 \Omega$

Resistance of 15 bulbs = 1210/15 = 242/3 $\Omega$

Resistance of 100 W bulb = $\frac{v^2 }{P }= \frac{220 ^2 }{100} = 484 \Omega$

Resistance of 5 such bulbs = 485/5 $\Omega$

Resistance of fan =  $\frac{v^2 }{P }= \frac{220 ^2 }{80}$   =  605 $\Omega$

Resistance of 5 such fan  = 605 /5 $\Omega$

= 121 $\Omega$

Resistance of heater = $\frac{220 ^2 }{1000} = 242 /5\Omega$

Equivalent resistance

$\frac{1}{R} = \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3}+ \frac{1}{R_4}\\ = \frac{3}{242} + \frac{5}{484}+ \frac{5}{605}+\frac{5}{242} \\$

$= \frac{6+5+4+10}{484} = \frac{25 }{484}$

R = 484 / 25

$I \geq \frac{V}{}R = \frac{220 }{\frac{484}{25}} = \frac{125 }{11} = 11.36 A$

$I^{min } = 12 A$

Option 1)

8 A

Option 2)

10 A

Option 3)

12 A

Option 4)

14 A

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Engineering
43 Views   |

In the circuit shown, current (in A) through the 50 V and 30 V batteries are, respectively :

• Option 1)

2.5 and 3

• Option 2)

3.5 and 2

• Option 3)

4.5 and 1

• Option 4)

3  and 2.5

As we discussed in

Cell -

The device which converts Chemical energy into electrical energy is known as electric cell.

Option 1)

2.5 and 3

Option 2)

3.5 and 2

Option 3)

4.5 and 1

Option 4)

3  and 2.5

Engineering
45 Views   |

A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1. The battery terminals are connected to an external resistance ‘R’. The minimum value of ‘R’, so that a current passes through the battery to charge it is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

zero

$r=\left ( \frac{\varepsilon -v}{v} \right )R$

$1=\left ( \frac{20}{220} \right )R$

$\Rightarrow R =11\Omega$

Option 1)

Option 2)

Option 3)

Option 4)

zero

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Engineering
111 Views   |

In the circuit shown, the current in the 1 resistor is :

• Option 1)

1.3 A ,from P to Q

• Option 2)

0 A

• Option 3)

0.13 A ,from Q to P

• Option 4)

0.13 A ,from P to Q

answer is option 3)0.13A from Q  to P

•  Let, Current i1 from 6v battery passes through 3 ohm resistance . At Q current breaks and current i2 goes through 3 ohm and ( i1-i2 ) goes through 1 ohm ,.
• Applying, kirchoffs law in 1st and 2nd loop,we get two eqns as follows
• 4i1-i2=6
• 6i2-i1=9
• solving ,,  i1-i2 =0.13 A
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Engineering
83 Views   |

In the electric network shown, when no current flows through the resistor in the arm EB, the potential difference between the points A and D will be :

• Option 1)

3V

• Option 2)

4V

• Option 3)

5V

• Option 4)

6V

The potential difference can be calculated via following method:

$-4+3+V_{cd}+0=0$

$V_{cd}=1 Volt$

$\therefore V_{A}-V_{D}=9-3-1=5V$

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Engineering
74 Views   |

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

• Option 1)

10.04 Volt

• Option 2)

Zero Volt

• Option 3)

2.9 Volt

• Option 4)

13.3 Volt

Need details on this question

Engineering
63 Views   |

Resistance of a given wire is obtained  by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is

• Option 1)

6%

• Option 2)

zero

• Option 3)

1%

• Option 4)

3%

As discussed in

Ohm's Law -

Current flowing through the conductor is directly proportional to the Potential difference accross two ends .

- wherein

unit -   $\frac{m^{2}}{volt\cdot sec}$

$V$ $\alpha$  $I$

$V=IR$

$R-$ Electric Resistance

$\dpi{100} R=\frac{V}{I}$

$\therefore \; \; \frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}$

Percentage error in $\dpi{100} R$ is

$\dpi{100} \frac{\Delta R}{R}\times 100=\frac{\Delta V}{V}\times 100+\frac{\Delta I}{I}\times 100$

$\dpi{100} =3\%+3\%=6\%$

Option 1)

6%

This option is correct.

Option 2)

zero

This option is incorrect.

Option 3)

1%

This option is incorrect.

Option 4)

3%

This option is incorrect.

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Engineering
65 Views   |

This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statemens.

Statement - I : Higher the range, greater is the resistance of ammeter.

Statement - II : To increase the range of ammeter,additional shunt needs to be used across it.

• Option 1)

Statement - I is false, Statement - II is true.

• Option 2)

Statement - I is true, Statement - II is true, Statement - II is the correct explanation of Statement - I.

• Option 3)

Statement - I is true, Statement - II is true, Statement - II is not the correct explanation of Statement - I.

• Option 4)

Statement - I is true, Statement - II is false.

As discussed in-

Required shunt -

$s= \frac{i_{g}G}{(i-i_{g})}$

- wherein

$i_{g}-$ Current through galvanometer

The answer follows from basic Principle of stunt and its use to make an ammeter:

$s=\frac{I_{g}}{I-I_{g}}$

Option 1)

Statement - I is false, Statement - II is true.

This option is correct.

Option 2)

Statement - I is true, Statement - II is true, Statement - II is the correct explanation of Statement - I.

This option is incorrect.

Option 3)

Statement - I is true, Statement - II is true, Statement - II is not the correct explanation of Statement - I.

This option is incorrect.

Option 4)

Statement - I is true, Statement - II is false.

This option is incorrect.

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Engineering
77 Views   |

A 10 V battery with internal resistance and a 15V battery with internal resistance are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to :

• Option 1)

11.9 V

• Option 2)

12.5 V

• Option 3)

13.1 V

• Option 4)

24.5 V

As we discussed in

In closed loop -

$-i_{1}}R_{1} + i_{2}}R_{2} -E_{1}-i_{3}}R_{3}+E_{2}+E_{3}-i_{4}}R_{4}=0$

- wherein

Current in the circuit

$I=\frac{5}{1.6}=\frac{50}{16}=\frac{25}{8}A$

$V=15-\frac{25}{8}\times 0.6$

$V=15-\frac{15}{8}=13.1V$

Option 1)

11.9 V

The option is incorrect

Option 2)

12.5 V

The option is incorrect

Option 3)

13.1 V

The option is correct

Option 4)

24.5 V

The option is incorrect

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Engineering
27 Views   |

A 3 olt battery with  negligible internal resistance is connected in a circuit as shown in the figure. The current I i n the circuit will be

• Option 1)

• Option 2)

• Option 3)

• Option 4)

as discussed in

Parallel Grouping -

Potential - Same

Current - Different

- wherein

$Req=\frac{(3+3)\times3}{(3+3)+3}=\frac{18}{9}=2\Omega$

$I=\frac{V}{R}=\frac{3}{2}=1.5A$

Option 1)

This option is incorrect.

Option 2)

This option is correct.

Option 3)

This option is incorrect.

Option 4)

This option is incorrect.

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Engineering
32 Views   |

Two electric bulbs marked  25 W-220 V and 100 W - 220 V are connected in series to a 440 V supply .Which of the bulbs will fuse?

• Option 1)

both

• Option 2)

100W

• Option 3)

25 W

• Option 4)

neither

As discussed in:

Series Grouping -

Potential - Different

Current - Same

- wherein

$I_{1}=\frac{P_{1}}{V_{1}}$

$I_{1}=\frac{25}{220}Amp$

Similarly,

$I_{2}=\frac{P_{2}}{V_{2}}=\frac{100}{220}Amp$

The current flowing through the circuit $I=\frac{440}{Reff}$

$Reff= R_{1}+R_{2}$

$R_{1}= \frac{V^{2}_{1}}{P_{1}}=\frac{(220)^{2}}{25}$

$R_{2}= \frac{V^{2}_{2}}{P_{1}}=\frac{(220)^{2}}{100}$

$\therefore I=\frac{440}{\frac{(220)^{2}}{25}+\frac{(220)^{2}}{100}}$

After solving this, we get $I=\frac{40}{220}Amp$

$\because I_{1}(=\frac{25}{220}A)< I(=\frac{40}{220}A)

Thus the bulb rated 25W-220V will fuse.

Option 1)

both

Option 2)

100W

Option 3)

25 W

Option 4)

neither

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Engineering
125 Views   |

The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 , 9 and 9 and a capacitor 5.0 F.

How much is the current I in the circuit in steady state ?

• Option 1)

1.6 A

• Option 2)

0.67 A

• Option 3)

2.5 A

• Option 4)

0.25 A

As we have learned

In closed loop -

$-i_{1}}R_{1} + i_{2}}R_{2} -E_{1}-i_{3}}R_{3}+E_{2}+E_{3}-i_{4}}R_{4}=0$

- wherein

$I'_1 = I_2 = I$  since in steady state current through capacitor is  0

USing kirchoff's rule

16-9I -3I -8V = 0   or I = 8/12 A = 2/3 A

= 0.67 A

Option 1)

1.6 A

Option 2)

0.67 A

Option 3)

2.5 A

Option 4)

0.25 A

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