11th And 12th > Physics
CIRCUIT NETWORKS MCQs
Total Questions : 30
| Page 1 of 3 pages
Answer: Option C. ->
4Ω
:
C
In series the current through the resistor R is 2ER+2r
In parallel connection, current through the resistor R is ER+r2
I=2ER+2r=ER+r22R+r=R+2rR=rr=4Ω
:
C
In series the current through the resistor R is 2ER+2r
In parallel connection, current through the resistor R is ER+r2
I=2ER+2r=ER+r22R+r=R+2rR=rr=4Ω
Answer: Option B. ->
150
:
B
Current sensitivity of galvanometer =4×10−4 Amp/div.
So full scale deflection current (ig)=Current sensitivity×Total number of division=4×10−4×25=10−2 A
To convert galvanometer in to voltmeter, resistance to be put in series is
R=Vig−G=2.510−2−100=150Ω
:
B
Current sensitivity of galvanometer =4×10−4 Amp/div.
So full scale deflection current (ig)=Current sensitivity×Total number of division=4×10−4×25=10−2 A
To convert galvanometer in to voltmeter, resistance to be put in series is
R=Vig−G=2.510−2−100=150Ω
Question 4.
36 cells each of e.m.f. 2 volt and internal resistance 0.5 ohm are connected in mixed grouping so as to deliver maximum current to a bulb of resistance 2 ohm. How are they connected. What is the maximum current through the bulb (n=no. of bulbs in series and, m = no. of bulbs in parallel)
Answer: Option A. ->
n=12 ; m=3; 6 A
:
A
To deliver maximum current, the external resistance should be equal to internal resistance.
mn = 36; Rm = r n
mn=14=28=312mn=rR=0.52=14lmax=mnE2nr=3×12×22×12×0.56A→(1)
:
A
To deliver maximum current, the external resistance should be equal to internal resistance.
mn = 36; Rm = r n
mn=14=28=312mn=rR=0.52=14lmax=mnE2nr=3×12×22×12×0.56A→(1)
Answer: Option A. ->
20
:
A
Vx=2VV100=10VI=V100R=10100=0.1AVx=IX2=0.1XX=20Ω
:
A
Vx=2VV100=10VI=V100R=10100=0.1AVx=IX2=0.1XX=20Ω
Answer: Option A. ->
62V, 2Ω
:
A
V= E - Ir
50 = E - 6r → (1)
60 = E - 1r → (2)
(2) - (1) 10 = 5r; r = 2Ω
E = 62V
:
A
V= E - Ir
50 = E - 6r → (1)
60 = E - 1r → (2)
(2) - (1) 10 = 5r; r = 2Ω
E = 62V
Answer: Option B. ->
56V
:
B
E′=(N−2n)E=(24−2×5)E=14E=14×4E′=56V
:
B
E′=(N−2n)E=(24−2×5)E=14E=14×4E′=56V
Answer: Option C. ->
16V,10Ω
:
C
E=I1[R1+r]=I2[R2+r]0.4[30+r]=0.8[10+r]30+r=20+2rr=10E=0.4[30+10]E=16V
:
C
E=I1[R1+r]=I2[R2+r]0.4[30+r]=0.8[10+r]30+r=20+2rr=10E=0.4[30+10]E=16V
Answer: Option A. ->
4.5 ohm
:
A
From the circuit
154=R+3215=2R+6R=92=4.5Ω
:
A
From the circuit
154=R+3215=2R+6R=92=4.5Ω
Answer: Option B. ->
63 cm
:
B
r=R[l1−l2l2]=R1[l1−l12l12]5[84−7070]=3[84−l12l12]l12×1414×3=84−l12l12+l123=844l12=84×3l12=63cm
:
B
r=R[l1−l2l2]=R1[l1−l12l12]5[84−7070]=3[84−l12l12]l12×1414×3=84−l12l12+l123=844l12=84×3l12=63cm