Current electricity advance based questions made with a lot of effort

LEVEL UP - JEE ADVANCED 2022
CURRENT ELECTRICITY EDUNITI
1. A network of nine conductors connects six points A, B, 1 6 1 6 35 22
C, D, E and F as shown in figure. The figure denotes      RAB  R
RAB 11R 2 R 11R 22 R 35
resistances in ohms. Find the equivalents resistance
3. Relation between current in conductor and time is shown
between A and D.
in figure.
A
i
2 i0
D
1 1
1 1
t
E 1 F 2 t0
2
B C (a) Find the total charge flow through the conductor
1
(b) Write the expression of current in terms of time
1. From the symmetry of the circuit about AD we can see (c) If the resistance of conductor is R, then the total heat
that VB = VC and VE = VF dissipated across resistance R is
Thus the reduced circuit is as shown in the figure. 3.(a) The total charge flow q is equal to the area under the
given i  t graph.
1 it
 q  i0  t0  0 0
2 2
(b) This graph is of form, y = mx + c
where, m  i0 / t0 and c  i0
Therefore, the expression of current in terms of time is
The equivalent resistance across AD is given by i0  t 
i t  i0  i  i0  1  
1  1 1 t0  t0 
   1  RAD  1
RAD 4 2 4
t 2 t
0
 t  0
 2t t 2 
2. Find the equivalent resistance of the circuit between points (c) H   i 2 Rdt  i02 R  1   dt  i02 R   1   2  dt
0
t0  0
t 0 t0 
A and B shown in figure. The resistance of each branch
is R i02 Rt0

3
C D
4. Find the current through () resistor in the figure
A shown.  
B
O


E F
 

10V
2. Due to symmetry, we observe that
4. Converting the delta BCD into star, we get the resistor
ICO= IOD, IEO = EOF and IAO = IOB.
network as shown in the fig.
The resistances of BD, BC and CD are respectively
2 20
 4  2  .
(), 4 and 2. Their sum is
3 3
So, resistances BO, OD and OC are respectively




We can therefore detatch the branches at centre O and
redraw the circuit as shown in figure. The equivalent
resistance across AB is given by


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Mohit Goenka | Founder of Eduniti | IIT KGP Alumnus

, LEVEL UP - JEE ADVANCED 2022
CURRENT ELECTRICITY EDUNITI




1
4   t 
5 i  i0 sin   , where i0 is peak current.
T 
T T 0
 t  iT  t  2i T
()  4 2 ()  2 1 2 4 6  Q   idt  i0  sin   dt  0 cos    0
 ,   and   0 0 T   T T 
5 5 5
Q
 21/ 5 84 6  i0 
Now, RAO    and ROC   2T
 21/ 5 55 5 Heat generated in resistor R during time t = 0 to T will be
VAO RAO 14
   T T
 t  i2 R 
T
2t 
VOC ROC 6/5 11 H   i 2 Rdt  i02 R  sin 2   dt  0  1  cos dt
0 0 T  2 0 T 
14 28
 VAO   10  V i02 R  sin(2t / T )  i02 RT
T
14  11 5  t  
V 7 2  2 / T  0 2
 I1  AO   A and 2
3  Q  RT 2 Q 2 R
 H    
V 4  2T  2 8T
I 2  AO   A
21/ 5 21/ 5 3
Therefore current through () resistor is 7. A rod of length L and cross–section area A lies along the
I1  I 2    1A x–axis between x = 0 and x = L. The material obeys
ohm’s law and its resistivity varies along the rod according
5. A hemisphere network of radius 'a' is made by using a to   x   0 e  x / L . The potential is V0 at x = 0 and is zero
conducting wire of resistance per unit length r. Find the at x = L.
equivalent resistance across OP. (a) Find the total resistance of the rod and the current in
the rod.
(b) Find the electric potential in the rod as a function of x.
7.(a) Small resistance of an element of thickness dx is,
dx
dR 
A
The resistance of the rod from 0 to x is

5. Here, ROA  ROB  ROC  ROD  ar

and RPA  RPB  RPC  RPD  ar
2
From symmetry, VA  VB  VC  VD
So, the circuit can be drawn as shown in the figure.
x x
0  L L 0 L
R( x)   e dx  0  e x / L   0 (1  e x / L )
0
A A x A
The total resistance of rod is
ar ar ar
ROP    (2  ) 0 L  L 1
4 8 8 R  R ( L)  (1  e  L / L )  0 1  
A A  e
6. A total charge Q flows across a resistor R during a V0 V0 Ae
time interval T in such a way that the current v/s time The current through rod is I  
R 0 L (e  1)
graph for 0 to T is like the loop of a sine curve in the (b) If V(x) is the potential at x, then
range 0 to . What will be the total heat generated in V0 Ae L
the resistor. V0  V ( x)  IR( x )   0 (1  e  x / L )
0 L (e  1) A
6. Since the variation of i vs t for t = 0 to T is a sine curve in
the range 0 to , we have V0 (e  e1 x / L )

e 1

EDUNITI - LEARN LIKE NEVER BEFORE Mohit Goenka | Founder of Eduniti | IIT KGP Alumnus 2