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Current Electricity 
For JEE Advanced 
   
Match the Columns 
Q 1.  For the circuit shown in figure, match the two columns. 
 
Column I Column II 
(a) current in wire ae (p) 1A 
(b) current is wire be (q) 2 A 
(c) current in wire ce (r) 0.5 A 
(d) current in wire de (s) None of these 
Q 2.  Current i is flowing through a wire of non-uniform cross section as shown. Match the following 
two columns. 
 
Column I Column II 
(a) Current density  (p) is more at 1 
(b) Electric field  (q) is more at 2 
(c) Resistance per unit length  (r) is same at both sections 1 and 2 
(d) Potential difference per unit length  (s) data insufficient 
Q 3.  In the circuit shown in figure, after dosing the switch S. match the following two columns. 
 
Column I Column II 
(a) current through R
1
 (p) will increase 
(b) current through R
2
 (q) will decrease 
Page 2


Current Electricity 
For JEE Advanced 
   
Match the Columns 
Q 1.  For the circuit shown in figure, match the two columns. 
 
Column I Column II 
(a) current in wire ae (p) 1A 
(b) current is wire be (q) 2 A 
(c) current in wire ce (r) 0.5 A 
(d) current in wire de (s) None of these 
Q 2.  Current i is flowing through a wire of non-uniform cross section as shown. Match the following 
two columns. 
 
Column I Column II 
(a) Current density  (p) is more at 1 
(b) Electric field  (q) is more at 2 
(c) Resistance per unit length  (r) is same at both sections 1 and 2 
(d) Potential difference per unit length  (s) data insufficient 
Q 3.  In the circuit shown in figure, after dosing the switch S. match the following two columns. 
 
Column I Column II 
(a) current through R
1
 (p) will increase 
(b) current through R
2
 (q) will decrease 
(c) potential difference across R
1
 (r) will remain same 
(d) potential difference across R
2
 (s) data insufficient 
Q 4.  Match the following two columns. 
Column I Column II 
(a) Electrical resistance (p) [MLT
-2
A
2
] 
(b) Electric potential (q) [ML
2
T
-3
A
-2
] 
(c) Specific resistance (r) [ML
2
T
-3
A
-1
] 
(d) Specific conductance (s) None of these 
Q 5.  In the circuit shown in figure, match the following two columns : 
 
Column I Column II 
(In SI units) 
(a) potential difference across battery A (p) zero 
(b) potential difference across battery B (q) 1 
(c) net power supplied / consumed by A (r) 2 
(d) net power supplied / consumed by B (s) 3 
Answers 
1. (a) ? q (b) ? s (c) ? q (d) ? s 
2. (a) ? p (b) ? p (c) ? p (d) ? p 
3. (a) ? q (b) ? p (c) ? q (d) ? p 
4. (a) ? q (b) ? r  (c) ? s  (d) ? s 
5. (a) ? s (b) ? r  (c) ? s  (d) ? r 
 
Solutions 
1.  Let potential of point e is V volts. Then. 
    I
ae
 + I
be
 + I
ce
 + I
de
 = 0 
  ?  
2 V 4 V 6 V 4 V
0
1 2 1 2
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
  
  or  V = 4 V 
  Now current through any wire can be obtained by the equation, 
   
PD
I
R
? 
2.  i
1
 = i
2
 or i is same at both sections. 
Page 3


Current Electricity 
For JEE Advanced 
   
Match the Columns 
Q 1.  For the circuit shown in figure, match the two columns. 
 
Column I Column II 
(a) current in wire ae (p) 1A 
(b) current is wire be (q) 2 A 
(c) current in wire ce (r) 0.5 A 
(d) current in wire de (s) None of these 
Q 2.  Current i is flowing through a wire of non-uniform cross section as shown. Match the following 
two columns. 
 
Column I Column II 
(a) Current density  (p) is more at 1 
(b) Electric field  (q) is more at 2 
(c) Resistance per unit length  (r) is same at both sections 1 and 2 
(d) Potential difference per unit length  (s) data insufficient 
Q 3.  In the circuit shown in figure, after dosing the switch S. match the following two columns. 
 
Column I Column II 
(a) current through R
1
 (p) will increase 
(b) current through R
2
 (q) will decrease 
(c) potential difference across R
1
 (r) will remain same 
(d) potential difference across R
2
 (s) data insufficient 
Q 4.  Match the following two columns. 
Column I Column II 
(a) Electrical resistance (p) [MLT
-2
A
2
] 
(b) Electric potential (q) [ML
2
T
-3
A
-2
] 
(c) Specific resistance (r) [ML
2
T
-3
A
-1
] 
(d) Specific conductance (s) None of these 
Q 5.  In the circuit shown in figure, match the following two columns : 
 
Column I Column II 
(In SI units) 
(a) potential difference across battery A (p) zero 
(b) potential difference across battery B (q) 1 
(c) net power supplied / consumed by A (r) 2 
(d) net power supplied / consumed by B (s) 3 
Answers 
1. (a) ? q (b) ? s (c) ? q (d) ? s 
2. (a) ? p (b) ? p (c) ? p (d) ? p 
3. (a) ? q (b) ? p (c) ? q (d) ? p 
4. (a) ? q (b) ? r  (c) ? s  (d) ? s 
5. (a) ? s (b) ? r  (c) ? s  (d) ? r 
 
Solutions 
1.  Let potential of point e is V volts. Then. 
    I
ae
 + I
be
 + I
ce
 + I
de
 = 0 
  ?  
2 V 4 V 6 V 4 V
0
1 2 1 2
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
  
  or  V = 4 V 
  Now current through any wire can be obtained by the equation, 
   
PD
I
R
? 
2.  i
1
 = i
2
 or i is same at both sections. 
   A
1 
< A
2
 
  (a) Current density 
i1
AA
??
 
 
  (c) 
Resis tan ce 1
lenght A A
?
??
 
 
  (d) and (b) E or potential difference per unit length = (i) (Resistance
 
per unit length) 
   
1
(i)
AA
? ??
??
??
??
 
3.  By introducing parallel resistance R
3
 in the circuit, total resistance of the circuit will decrease. 
Hence main current i will increase. 
Now, 
12
R R 2
V E V E iR ? ? ? ?  
Since i is increasing, so 
2
R
V will increase. Hence 
1
R
V or current passing through R
1
 will decrease. 
4.  (a) 
2
H
R
it
?
 
 
  [R] = [ML
2
T
-2
]/[A
2
T] = [ML
2
T
-3
A
-2
] 
  (b) V = iR 
   [V] = [A][ML
2
T
-3
A
-2
] = [ML
-2
T
-3
A
-1
] 
  (c) 
RA
l
??
 
 
    
2 3 2 2
[ML T A ][L ]
[]
[L]
??
??  
    = [ML
3
T
-3
A
-2
]  
  (d) 
1
[]
??
??
??
?
??
 
= [M
-1
L
-3
T
3
A
2
] 
5.  
41
i 1A
111
?
??
?? 
(anticlockwise) 
  (a) V
A 
= E - ir = 4- 1 x 1 = 3V 
  (b) V
B
 = E + ir = 1 + 1 x 1 = 2 V 
  (c) |P
A
| = Ei - i
2
r = (4 × 1) = (1)
2
(1) = 3W 
  (d) |P
B
| = Ei + i
2
r = (1)(1) + (1)
2
(1)=2W 
 
Subjective Questions 
Q 1.  Find the equivalent resistance of the triangular bipyramid between the points. 
Page 4


Current Electricity 
For JEE Advanced 
   
Match the Columns 
Q 1.  For the circuit shown in figure, match the two columns. 
 
Column I Column II 
(a) current in wire ae (p) 1A 
(b) current is wire be (q) 2 A 
(c) current in wire ce (r) 0.5 A 
(d) current in wire de (s) None of these 
Q 2.  Current i is flowing through a wire of non-uniform cross section as shown. Match the following 
two columns. 
 
Column I Column II 
(a) Current density  (p) is more at 1 
(b) Electric field  (q) is more at 2 
(c) Resistance per unit length  (r) is same at both sections 1 and 2 
(d) Potential difference per unit length  (s) data insufficient 
Q 3.  In the circuit shown in figure, after dosing the switch S. match the following two columns. 
 
Column I Column II 
(a) current through R
1
 (p) will increase 
(b) current through R
2
 (q) will decrease 
(c) potential difference across R
1
 (r) will remain same 
(d) potential difference across R
2
 (s) data insufficient 
Q 4.  Match the following two columns. 
Column I Column II 
(a) Electrical resistance (p) [MLT
-2
A
2
] 
(b) Electric potential (q) [ML
2
T
-3
A
-2
] 
(c) Specific resistance (r) [ML
2
T
-3
A
-1
] 
(d) Specific conductance (s) None of these 
Q 5.  In the circuit shown in figure, match the following two columns : 
 
Column I Column II 
(In SI units) 
(a) potential difference across battery A (p) zero 
(b) potential difference across battery B (q) 1 
(c) net power supplied / consumed by A (r) 2 
(d) net power supplied / consumed by B (s) 3 
Answers 
1. (a) ? q (b) ? s (c) ? q (d) ? s 
2. (a) ? p (b) ? p (c) ? p (d) ? p 
3. (a) ? q (b) ? p (c) ? q (d) ? p 
4. (a) ? q (b) ? r  (c) ? s  (d) ? s 
5. (a) ? s (b) ? r  (c) ? s  (d) ? r 
 
Solutions 
1.  Let potential of point e is V volts. Then. 
    I
ae
 + I
be
 + I
ce
 + I
de
 = 0 
  ?  
2 V 4 V 6 V 4 V
0
1 2 1 2
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
  
  or  V = 4 V 
  Now current through any wire can be obtained by the equation, 
   
PD
I
R
? 
2.  i
1
 = i
2
 or i is same at both sections. 
   A
1 
< A
2
 
  (a) Current density 
i1
AA
??
 
 
  (c) 
Resis tan ce 1
lenght A A
?
??
 
 
  (d) and (b) E or potential difference per unit length = (i) (Resistance
 
per unit length) 
   
1
(i)
AA
? ??
??
??
??
 
3.  By introducing parallel resistance R
3
 in the circuit, total resistance of the circuit will decrease. 
Hence main current i will increase. 
Now, 
12
R R 2
V E V E iR ? ? ? ?  
Since i is increasing, so 
2
R
V will increase. Hence 
1
R
V or current passing through R
1
 will decrease. 
4.  (a) 
2
H
R
it
?
 
 
  [R] = [ML
2
T
-2
]/[A
2
T] = [ML
2
T
-3
A
-2
] 
  (b) V = iR 
   [V] = [A][ML
2
T
-3
A
-2
] = [ML
-2
T
-3
A
-1
] 
  (c) 
RA
l
??
 
 
    
2 3 2 2
[ML T A ][L ]
[]
[L]
??
??  
    = [ML
3
T
-3
A
-2
]  
  (d) 
1
[]
??
??
??
?
??
 
= [M
-1
L
-3
T
3
A
2
] 
5.  
41
i 1A
111
?
??
?? 
(anticlockwise) 
  (a) V
A 
= E - ir = 4- 1 x 1 = 3V 
  (b) V
B
 = E + ir = 1 + 1 x 1 = 2 V 
  (c) |P
A
| = Ei - i
2
r = (4 × 1) = (1)
2
(1) = 3W 
  (d) |P
B
| = Ei + i
2
r = (1)(1) + (1)
2
(1)=2W 
 
Subjective Questions 
Q 1.  Find the equivalent resistance of the triangular bipyramid between the points. 
 
  (a) A and C   (b) D and E 
  Assume the resistance of each branch to be R. 
Q 2.  Nine wires each of resistance r are connected to make a prism as shown in figure. Find the 
equivalent resistance of the arrangement
 
across 
 
  (a) AD   (b) AB 
Q 3.  Draw the circuit for experimental verification of Ohm's law using a source of variable D.C. 
voltage, a main resistance of 100 ?, two galvanometer and two resistance of values 10
6
 ? and 10
-3
 
?, respectively. Clearly show the positions of the voltmeter and the ammeter. 
Q 4.  The figure shows part of certain circuit, find : 
 
  (a) Power dissipated in 5 ? resistance. (b) Potential difference V
C
 - V
B
. 
  (c) Which battery is being charged. 
Q 5.  A 6 V battery of negligible internal resistance is connected across a uniform wire AB of length 
100 cm. The positive terminal of another battery of emf 4 V and internal resistance 1 ? is joined to 
the point A as shown in figure. Take the potential at B to be zero. 
 
  (a) What are the potentials at the points A and C ? 
  (b) At which point D of the wire AB, the potential is equal to the potential at C? 
  (c) If the points C and D are connected by a wire, what will be the current through it ?   
  (d) If the 4V battery is replaced by 7.5 V battery, what would be the answers of parts (a) and (b)? 
Page 5


Current Electricity 
For JEE Advanced 
   
Match the Columns 
Q 1.  For the circuit shown in figure, match the two columns. 
 
Column I Column II 
(a) current in wire ae (p) 1A 
(b) current is wire be (q) 2 A 
(c) current in wire ce (r) 0.5 A 
(d) current in wire de (s) None of these 
Q 2.  Current i is flowing through a wire of non-uniform cross section as shown. Match the following 
two columns. 
 
Column I Column II 
(a) Current density  (p) is more at 1 
(b) Electric field  (q) is more at 2 
(c) Resistance per unit length  (r) is same at both sections 1 and 2 
(d) Potential difference per unit length  (s) data insufficient 
Q 3.  In the circuit shown in figure, after dosing the switch S. match the following two columns. 
 
Column I Column II 
(a) current through R
1
 (p) will increase 
(b) current through R
2
 (q) will decrease 
(c) potential difference across R
1
 (r) will remain same 
(d) potential difference across R
2
 (s) data insufficient 
Q 4.  Match the following two columns. 
Column I Column II 
(a) Electrical resistance (p) [MLT
-2
A
2
] 
(b) Electric potential (q) [ML
2
T
-3
A
-2
] 
(c) Specific resistance (r) [ML
2
T
-3
A
-1
] 
(d) Specific conductance (s) None of these 
Q 5.  In the circuit shown in figure, match the following two columns : 
 
Column I Column II 
(In SI units) 
(a) potential difference across battery A (p) zero 
(b) potential difference across battery B (q) 1 
(c) net power supplied / consumed by A (r) 2 
(d) net power supplied / consumed by B (s) 3 
Answers 
1. (a) ? q (b) ? s (c) ? q (d) ? s 
2. (a) ? p (b) ? p (c) ? p (d) ? p 
3. (a) ? q (b) ? p (c) ? q (d) ? p 
4. (a) ? q (b) ? r  (c) ? s  (d) ? s 
5. (a) ? s (b) ? r  (c) ? s  (d) ? r 
 
Solutions 
1.  Let potential of point e is V volts. Then. 
    I
ae
 + I
be
 + I
ce
 + I
de
 = 0 
  ?  
2 V 4 V 6 V 4 V
0
1 2 1 2
? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
  
  or  V = 4 V 
  Now current through any wire can be obtained by the equation, 
   
PD
I
R
? 
2.  i
1
 = i
2
 or i is same at both sections. 
   A
1 
< A
2
 
  (a) Current density 
i1
AA
??
 
 
  (c) 
Resis tan ce 1
lenght A A
?
??
 
 
  (d) and (b) E or potential difference per unit length = (i) (Resistance
 
per unit length) 
   
1
(i)
AA
? ??
??
??
??
 
3.  By introducing parallel resistance R
3
 in the circuit, total resistance of the circuit will decrease. 
Hence main current i will increase. 
Now, 
12
R R 2
V E V E iR ? ? ? ?  
Since i is increasing, so 
2
R
V will increase. Hence 
1
R
V or current passing through R
1
 will decrease. 
4.  (a) 
2
H
R
it
?
 
 
  [R] = [ML
2
T
-2
]/[A
2
T] = [ML
2
T
-3
A
-2
] 
  (b) V = iR 
   [V] = [A][ML
2
T
-3
A
-2
] = [ML
-2
T
-3
A
-1
] 
  (c) 
RA
l
??
 
 
    
2 3 2 2
[ML T A ][L ]
[]
[L]
??
??  
    = [ML
3
T
-3
A
-2
]  
  (d) 
1
[]
??
??
??
?
??
 
= [M
-1
L
-3
T
3
A
2
] 
5.  
41
i 1A
111
?
??
?? 
(anticlockwise) 
  (a) V
A 
= E - ir = 4- 1 x 1 = 3V 
  (b) V
B
 = E + ir = 1 + 1 x 1 = 2 V 
  (c) |P
A
| = Ei - i
2
r = (4 × 1) = (1)
2
(1) = 3W 
  (d) |P
B
| = Ei + i
2
r = (1)(1) + (1)
2
(1)=2W 
 
Subjective Questions 
Q 1.  Find the equivalent resistance of the triangular bipyramid between the points. 
 
  (a) A and C   (b) D and E 
  Assume the resistance of each branch to be R. 
Q 2.  Nine wires each of resistance r are connected to make a prism as shown in figure. Find the 
equivalent resistance of the arrangement
 
across 
 
  (a) AD   (b) AB 
Q 3.  Draw the circuit for experimental verification of Ohm's law using a source of variable D.C. 
voltage, a main resistance of 100 ?, two galvanometer and two resistance of values 10
6
 ? and 10
-3
 
?, respectively. Clearly show the positions of the voltmeter and the ammeter. 
Q 4.  The figure shows part of certain circuit, find : 
 
  (a) Power dissipated in 5 ? resistance. (b) Potential difference V
C
 - V
B
. 
  (c) Which battery is being charged. 
Q 5.  A 6 V battery of negligible internal resistance is connected across a uniform wire AB of length 
100 cm. The positive terminal of another battery of emf 4 V and internal resistance 1 ? is joined to 
the point A as shown in figure. Take the potential at B to be zero. 
 
  (a) What are the potentials at the points A and C ? 
  (b) At which point D of the wire AB, the potential is equal to the potential at C? 
  (c) If the points C and D are connected by a wire, what will be the current through it ?   
  (d) If the 4V battery is replaced by 7.5 V battery, what would be the answers of parts (a) and (b)? 
Q 6.  A thin uniform wire AB of length 1 m, an unknown resistance X and a resistance of 12 ?, are 
connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a 
sliding jockey connected to it) are also available. Connections are to be made to measure the 
unknown resistance X. Using the principle of Wheatstone bridge answer
 
the following questions :  
 
  (a) Are there positive and negative terminals on the galvanometer ? 
(b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) 
connected at appropriate points. 
(c) After appropriate connections are made, it is found that no deflection takes place in the 
galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the 
value of the resistance X 
Q 7.  A galvanometer (coil resistance 99 ?) is converted into an ammeter using a shunt of 1 ? and 
connected as shown in figure (a). The ammeter reads 3 A. The same galvanometer is converted 
into a voltmeter by connecting a resistance of 101 ? in series. This voltmeter is connected as 
shown in figure (b). Its reading is found to be 4/5 of the full scale reading. Find : 
      
  (a) internal resistances of the cell  (b) range of the ammeter and voltmeter 
  (c) full scale deflection current of the galvanometer. 
Q 8.  In a circuit shown in figure if the internal resistances of the sources are negligible then at what 
value of resistance R will the thermal power generated in it will be the maximum. What is its 
value ? 
 
Q 9.  In the circuit shown in figure, find :  
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FAQs on DC Pandey Solutions (JEE Advance): Current Electricity- 2 - DC Pandey Solutions for JEE Physics

1. What is the significance of current electricity in JEE Advanced?
Ans. Current electricity is an important topic in JEE Advanced as it forms the basis for understanding various electrical phenomena and circuits. It is essential for solving complex problems related to resistors, capacitors, and inductors, which frequently appear in the JEE Advanced exam.
2. How can I prepare effectively for the current electricity section in JEE Advanced?
Ans. To prepare effectively for the current electricity section in JEE Advanced, it is important to understand the fundamental concepts and principles. Start by studying the basic laws such as Ohm's law, Kirchhoff's laws, and the concept of electric power. Practice solving numerical problems and attempt previous year question papers to get familiar with the exam pattern.
3. Are DC Pandey's solutions helpful for JEE Advanced preparation in current electricity?
Ans. Yes, DC Pandey's solutions are highly recommended for JEE Advanced preparation, particularly for the current electricity section. His book provides detailed explanations and step-by-step solutions to help students grasp the concepts effectively. It is a popular resource among JEE aspirants and can greatly enhance your understanding and problem-solving skills.
4. What are some important topics to focus on while studying current electricity for JEE Advanced?
Ans. While studying current electricity for JEE Advanced, it is important to focus on topics such as Ohm's law, Kirchhoff's laws, electrical circuits, resistors and their combinations, capacitors, inductors, and AC circuits. Understanding the concepts and their applications in solving numerical problems is crucial for scoring well in this section.
5. How can I improve my problem-solving skills in the current electricity section for JEE Advanced?
Ans. Improving problem-solving skills in the current electricity section for JEE Advanced requires regular practice. Solve a variety of numerical problems from different sources, including textbooks, practice books, and previous year question papers. Analyze your mistakes and learn from them. Additionally, join online forums or study groups to discuss and solve problems with fellow aspirants, as this can further enhance your understanding and problem-solving abilities.
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