Q 1 :    

A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is            [2025]

  • R16

     

  • R8

     

  • R64

     

  • R32

     

(1)

[IMAGE 173]

Resistance of each part, r=R8

R1=R2=r4=R32

R1 and R2 are in series

   Req=R1+R2=R32+R32=R16



Q 2 :    

A uniform metal wire of length l has 10Ω resistance. Now this wire is stretched to a length 2l and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is              [2024]

  • 10Ω

     

  •  

  • 40Ω

     

  • 20Ω

     

(1)

The volume of wire remains same even after stretching the wire, i.e., V'=V ; A'l'=Al

              A'=All'=Al2l=A2  (∵ l'=2l)

and Resistance of wire is R=ρLA=10Ω

After stretching, resistance of circle becomes

           R'=ρ(2l)(2)A=4R=40Ω

As the wire is bent at its mid-point to form a circle, thus, the effective resistance of semi-circle is =R'2=402=20Ω

   Equivalent resistance across diameter of circle, 

        =20Ω20Ω

          1Req=120+120

   Req=10Ω



Q 3 :    

A wire of length l and resistance 100Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is                [2024]

  • 26 Ω

     

  • 52 Ω

     

  • 55 Ω

     

  • 60 Ω

     

(2)

As the wire having resistance R=100Ω is divided into 10 equal parts, each part will have resistance R10.

Equivalent resistance of series combination of first 5 parts,

R'eq=R1+R2+R3+R4+R5

=R10+R10+R10+R10+R10;  Req'=R2

Equivalent resistance of parallel combination of next 5 parts,

1Req=1R6+1R7+1R8+1R9+1R10=10R+10R+10R+10R+10R=50R

Req=R50

Resistance of final series combination is, Rfinal=R'eq+Req=R2+R50=26R50=26×10050=52Ω

   Rfinal=52 Ω



Q 4 :    

Two heaters A and B have power rating of 1kW and 2kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is          [2024]

  • 1 : 1

     

  • 2 : 9

     

  • 1 : 2

     

  • 2 : 3

     

(2)

Given that, PA=1 kW, PB=2 kW

For the series combination of two heaters A and B of same voltage, equivalent resistance is,

            Rs=RA+RB                                               ...(i)

Power across both heaters, PA=V2RA,  PB=V2RB

   V2P=V2PA+V2PB                                                        (Using (i))

   1Ps=1PA+1PB    Ps=PA·PBPA+PB    Ps=1×21+2=23 kW

For the parallel combination of the two heaters,

1Rp=1RA+1RB; PPV2=PAV2+PBV2  

  PP=PA+PB=3 kW

So the ratio of power output in series to power output in parallel is =23×13=29



Q 5 :    

Three resistors having resistances r1,r2 and r3 are connected as shown in the given circuit. The ratio i3i1 of currents in terms of resistances used in the circuit is    [2021]

[IMAGE 174]
 

  • r2r1+r3

     

  • r1r2+r3

     

  • r2r2+r3

     

  • r1r1+r2

     

(3)

i1=i2+i3                               ...(i)

Voltage is same in r2 and r3 as they are in parallel

      i2r2=i3r3

      i2=i3r3r2                           ...(ii)

[IMAGE 175]-----------------------------

From eqn. (i) and (ii), i1=i3r3r2+i3

or       i1=i3r3+i3r2r2=i3(r2+r3r2)  ;  i3i1=r2r2+r3



Q 6 :    

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25Ω. What will be the effective resistance if they are connected in series?           [2021]
 

  • Ω

     

  • 0.25 Ω

     

  • 0.5 Ω

     

  • Ω

     

(1)

Given that, four wires of same material, equal area of cross-section and equal length.

Resistance of each wire is 'R'.

For parallel combination, Req=0.25Ω

When wires are connected in parallel, 

             1Req=1R1+1R2+1R3+1R4

              1Req=1R+1R+1R+1R ;  10.25=4RR=1 Ω

So, resistance of each wire is 1Ω.

Now, wires are connected in series. Then, equivalent resistance is 

Req=R1+R2+R3+R4

Req=1Ω+1Ω+1Ω+1Ω=4Ω

Hence, the equivalent resistance in series will be 4 Ω.



Q 7 :    

In the circuits shown below, the readings of the voltmeters and the ammeters will be                [2019]

[IMAGE 176]

 

  • V2>V1 and i1>i2

     

  • V2>V1 and i1=i2

     

  • V1=V2 and i1>i2

     

  • V1=V2 and i1=i2

     

(4)

 



Q 8 :    

Six similar bulbs are connected as shown in the figure with a DC source of emf E, and zero internal resistance. The ratio of power consumption by the bulbs when
(i) all are glowing and
(ii) in the situation when two from section A and one from section B are glowing, will be                   [2019]

[IMAGE 177]

  • 2 : 1

     

  • 4 : 9

     

  • 9 : 4

     

  • 1 : 2

     

(3)

[IMAGE 178]-----------------------------

Let R be the resistance of each bulb.

Case (i)

Net resistance of the circuit, R1=R3+R3=2R3

Power consumption by the bulbs = Power supply by the sources

P1=E2(2R/3)=3E22R

[IMAGE 179]-----------------------------

Case (ii)

Net resistance of the circuit,

R2=R2+R=3R2

Power consumption by the bulbs,

P2=E2(3R/2)=23(E2R);  P1P2=32×32=94



Q 9 :    

A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is              [2016]

  • 230 Ω

     

  • 46 Ω

     

  • 26 Ω

     

  • 13 Ω

     

(3)

Resistance of bulb, RB=V2P=(100)2500=20 Ω

Power of the bulb in the circuit,

[IMAGE 180]---------------------------------------------

P=VI;  I=PVB;  I=500100=5A

VR=IR(230-100)=5×R    R=26Ω



Q 10 :    

Two metal wires of identical dimensions are connected in series. If σ1 and σ2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is                    [2015]
 

  • σ1+σ2σ1σ2

     

  • σ1σ2σ1+σ2

     

  • 2σ1σ2σ1+σ2

     

  • σ1+σ22σ1σ2

     

(3)

As both metal wires are of identical dimensions, so their length and area of cross-section will be same. Let them be l and A respectively. Then the resistance of the first wire is

             R1=lσ1A                                                 ...(i)

and that of the second wire is R2=lσ2A              ...(ii)

[IMAGE 181]----------------------------------------------------------------

As they are connected in series, their effective resistance is

          Rs=R1+R2=lσ1A+lσ2A                             (using (i) and (ii))

                =lA(1σ1+1σ2)                           ...(iii)

If σeff is the effective conductivity of the combination, then

                Rs=2lσeffA                                     ...(iv)

Equating eqns. (iii) and (iv), we get

2lσeffA=lA(1σ1+1σ2);   2σeff=σ2+σ1σ1σ2  or  σeff=2σ1σ2σ1+σ2