12th Grade > Physics
CIRCUIT NETWORKS MCQs
Total Questions : 30
| Page 1 of 3 pages
Answer: Option A. -> 50 cm
:
A
XY=20100−l=2080=14Y=4X4XY=l′100−l′;l′=50cm
:
A
XY=20100−l=2080=14Y=4X4XY=l′100−l′;l′=50cm
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
Question 3. Eels are able to generate current with biological cells called electroplaques. The electroplaques in an eel are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000 electroplaques. The arrangement is suggestively shown below. Each electroplaques has an emf of 0.15 V and internal resistance of 0.25Ω. The water surrounding the eel completes a circuit between the head and its tail. If the water surrounding it has a resistance of 500 ohm, the current an eel can produce in water is about
Answer: Option A. -> 1.5 A
:
A
Given problem is the case of mixed grouping of cells
So total current produced i=nER+nrm
Here m = 100, n = 5000, R = 500Ω
E = 0.15V and r = 0.25Ω
⇒i=5000×0.15500+5000×0.25100=750512.5≈1.5A
:
A
Given problem is the case of mixed grouping of cells
So total current produced i=nER+nrm
Here m = 100, n = 5000, R = 500Ω
E = 0.15V and r = 0.25Ω
⇒i=5000×0.15500+5000×0.25100=750512.5≈1.5A
Answer: Option C. -> 0.5 ohm
:
C
iSC=Er⇒3=1.5r⇒r=0.5Ω
:
C
iSC=Er⇒3=1.5r⇒r=0.5Ω
Question 6. 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 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
Question 9. In an experiment for calibration of voltmeter, a standard cell of emf 1.5V is balanced at 375cm length of potentiometer wire. The potential difference across a resistance in the circuit is balanced at 1.25 m. If a voltmeter is connected across the same resistance, it reads 0.6V. The error in the voltmeter is
Answer: Option A. -> 0.1V
:
A
iρ=El=1.53.75VmV1=(iρ)l=1.53.75×1.25V1=0.5VError=V−V1=0.6−0.5=0.1V
:
A
iρ=El=1.53.75VmV1=(iρ)l=1.53.75×1.25V1=0.5VError=V−V1=0.6−0.5=0.1V
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