A potential difference of 1V is applied across the other diagonal. What will be the current in the resistor along the diagonal?
0.5 A
1 A
5 A
Zero
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 :
8 A
10 A
12 A
14 A
As we have learned
In parallel Grouping 

Resistance of each bulb of 40 W
Resistance of 15 bulbs = 1210/15 = 242/3
Resistance of 100 W bulb =
Resistance of 5 such bulbs = 485/5
Resistance of fan = = 605
Resistance of 5 such fan = 605 /5
= 121
Resistance of heater =
Equivalent resistance
R = 484 / 25
Option 1)
8 A
Option 2)
10 A
Option 3)
12 A
Option 4)
14 A
In the circuit shown, current (in A) through the 50 V and 30 V batteries are, respectively :
2.5 and 3
3.5 and 2
4.5 and 1
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
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 :
zero
Option 1)
Option 2)
Option 3)
Option 4)
zero
In the circuit shown, the current in the 1 resistor is :
1.3 A ,from P to Q
0 A
0.13 A ,from Q to P
0.13 A ,from P to Q
answer is option 3)0.13A from Q to P
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 :
3V
4V
5V
6V
The potential difference can be calculated via following method:
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?
10.04 Volt
Zero Volt
2.9 Volt
13.3 Volt
Need details on this question
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
6%
zero
1%
3%
As discussed in
Ohm's Law 
Current flowing through the conductor is directly proportional to the Potential difference accross two ends .
 wherein
unit 
Electric Resistance
Percentage error in is
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.
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.
Statement  I is false, Statement  II is true.
Statement  I is true, Statement  II is true, Statement  II is the correct explanation of Statement  I.
Statement  I is true, Statement  II is true, Statement  II is not the correct explanation of Statement  I.
Statement  I is true, Statement  II is false.
As discussed in
Required shunt 
 wherein
Current through galvanometer
The answer follows from basic Principle of stunt and its use to make an ammeter:
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.
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 :
11.9 V
12.5 V
13.1 V
24.5 V
As we discussed in
In closed loop 
 wherein
Current in the circuit
Reading of the voltmeter
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
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
as discussed in
Parallel Grouping 
Potential  Same
Current  Different
 wherein
Option 1)
This option is incorrect.
Option 2)
This option is correct.
Option 3)
This option is incorrect.
Option 4)
This option is incorrect.
Two electric bulbs marked 25 W220 V and 100 W  220 V are connected in series to a 440 V supply .Which of the bulbs will fuse?
both
100W
25 W
neither
As discussed in:
Series Grouping 
Potential  Different
Current  Same
 wherein
Similarly,
The current flowing through the circuit
After solving this, we get
Thus the bulb rated 25W220V will fuse.
both
Option 2)100W
Option 3)25 W
Option 4)neither
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 ?
1.6 A
0.67 A
2.5 A
0.25 A
As we have learned
In closed loop 
 wherein
since in steady state current through capacitor is 0
USing kirchoff's rule
169I 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
Four identical electrical lamps are labelled 1.5V, 0.5A which describes the condition necessary for them to operate at normal brightness. A 12V battery of negligible internal resistance is connected to lamps as shown, then
The value of R for normal brightness of each lamp is
The value of R for normal brightness of each lamp is
Total power dissipated in circuit when all lamps are normally bright is 24W
Power dissipated in R is 21W when all lamps are normally bright
When the rms voltages V_{L}, V_{C} and V_{R} are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio V_{L} : V_{C} : V_{R }= 1 : 2 : 3. If the rms voltage of the AC source is 100 V, then V_{R} is close to :
50 V
70 V
90 V
100 V
Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating wire is decreased to 2/3 of its initial value, then the same amount of water will boil with the same supply voltage in
15 minutes
12 minutes
10 minutes
8 minutes
The three resistance of equal value are arranged in the different combinations shown below. Arrange them in increasing order of power dissipation
I.
II.
III.
IV.
III < II < IV < I
II < III < IV < I
I < IV < III < II
I < III < II < IV
The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected to (220×.8) volt sources, then the actual power would be
100×0.8 watt
100× (0.8)^{2} watt
Between 100×0.8 watt and 100 watt
Between 100× (0.8)^{2} watt and 100×0.8 watt
In the circuit shown in figure, the heat produced in 5 ohm resistance is 10 calories per second. The heat produced in 4 resistance is
1 cal/sec
2 cal/sec
3 cal/sec
4 cal/sec
In the circuit as shown in the figure, the heat produced by 6 ohm resistance due to current flowing in it is 60 calorie per second. The heat generated across 3 ohm resistance per second will be
30 cal
60 cal
120 cal
100 cal