Current Electricity — Page 9 | NEET Physics

Q81

Two cells, having the same e.m.f. are connected in series through an external resistance R. Cells have internal resistances r 1 and r 2 (r 1 > r 2 ) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is

A r 1 + r 2

B r 1

-

r 2

C

{{{r_1} + {r_2}} \over 2}

D

{{{r_1} - {r_2}} \over 2}

Correct Answer

Option B

Solution

Kirchhoff’s law has to be applied to the whole loop. while

i = {{2E} \over {\left( {{r_1} + {r_2} + R} \right)}}

If through one section (here the first battery) has zero potential difference, current cannot flow.

The question could have been modified.

The statement that when the circuit is closed, the potential difference across the first cell is zero implies that in a series circuit, one part cannot conduct current which is wrong.

Kirchhoff’s law is violated.

Assuming that ir 1 = E as given in the question paper, some students could have found that R = r 1 – r 2 .

They have to be given marks.

Q82

In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will

A flow from B to A

B flow from A to B

C flow in the direction which will be decided by the value of V be zero.

D be zero

Correct Answer

Option A

Solution

Current will flow from B to A Potential drop over the resistance CA will be more due to higher value of resistance.

So potential at A will be less as compared with at B.

Hence, current will flow from B to A.

Q83

For the network shown in the figure the value of the current

i

is

A

{{9V} \over {35}}

B

{{18V} \over 5}

C

{{5V} \over 9}

D

{{5V} \over {18}}

Correct Answer

Option D

Solution

It is a balanced Wheatstone bridge. Hence resistance 4

\Omega

can be eliminated. $\therefore$

{R_{eq}} = {{6 \times 9} \over {6 + 9}} = {{18} \over 5}

\therefore i = {V \over {{R_{eq}}}} = {{5V} \over {18}}

Q84

A 5-ampere fuse wire can withstand a maximum power of 1 watt in the circuit. The resistance of the fuse wire is

A 0.04 ohm

B 0.2 ohm

C 5 ohm

D 0.4 ohm

Correct Answer

Option A

Solution

P = i 2 R or 1 = 25 × R

R = {1 \over {25}} = 0.04\Omega

Q85

When a wire of uniform cross-section

a

length

l

and resistance R is bent into a complete circle. resistance between any two of diametrically opposite points will be

A R/4

B 4R

C R/8

D R/2

Correct Answer

Option A

Solution

{R_{eq}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {{\left( {{R \over 2}.{R \over 2}} \right)} \over {{R \over 2} + {R \over 2}}}

$\therefore$

{R_{eq}} = {R \over 4}

Q86

Two batteries, one of emf 18 volts and internal resistance 2

\Omega

and the other of emf 12 volts and internal resistance 1

\Omega

, are connected as shown. The voltmeter V will record a reading of

A 30 volt

B 18 volt

C 15 volt

D 14 volt

Correct Answer

Option D

Solution

From Kirchhoff’s law, I × 2 + I × 1 = 18 – 12 Current in the circuit,

I = {V \over R} = {6 \over 2} = 2A

Voltage drop across

2\Omega

, V 1 = 2 × 2 = 4 V Voltmeter reading = 18 – 4 = 14 V.

Q87

A 6 volt battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of 100 ohm. The difference of potential between two points on the wire separated by a distance of 50 cm will be

A 2 volt

B 3 volt

C 1 volt

D 1.5 volt

Correct Answer

Option C

Solution

R\,\alpha \,\ell

For 300 cm, R = 100

\Omega

For 50 cm,

R' = {{100} \over {300}} \times 50 = {{50} \over 3}\Omega

$\therefore$ IR = 6

\Rightarrow IR' = {6 \over R} \times R' = {6 \over {100}} \times {{50} \over 3} = 1\,volt.

Q88

In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be

A R

B 2R

C R/4

D R/2

Correct Answer

Option C

Solution

P = {{{V^2}} \over R}

\Rightarrow R = {{{V^2}} \over P} = {{{{\left( {220} \right)}^2}} \over {60}} = {{4 \times {{\left( {110} \right)}^2}} \over {60}}

R' = {{{{\left( {110} \right)}^2}} \over {60}} = {R \over 4}

Q89

Five equal resistances each of resistances R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be

A

{{3V} \over R}

B

{V \over R}

C

{V \over {2R}}

D

{{2V} \over R}

Correct Answer

Option C

Solution

A balanced Wheststone’s bridge exists between A & B.

$\therefore$ R eq = R Current through circuit = V/R Current through AFCEB = V/2R

Q90

The electric resistance of a certain wire of iron is R. If its length and radius are both doubled, then

A The resistance will be doubled and the specific resistance will be halved.

B The resistance will be halved and the specific resistance will remain unchanged.

C The resistance will be halved and the specific resistance will be doubled.

D The resistance and the specific resistance, will both remain unchanged.

Correct Answer

Option B

Solution

Resistance of wire =

\rho {l \over A}

R \propto {l \over A} = {l \over {\pi {r^2}}}

When length and radius are both doubled

{R_1} \propto {{2l} \over {\pi {{\left( {2r} \right)}^2}}} \Rightarrow {R_1} \propto {1 \over 2}R

The specific resistance of wire is independent of geometry of the wire, it only depends on the material of the wire.