Current Electricity — Page 6 | JEE Physics

Q51

A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0(I0 > Ig) by connecting a shunt resistance RA to it and (ii) into a voltmeter of range 0 to V (V = GI0) by connecting a series resistance RV to it. Then,

A

{R_A}{R_V} = {G^2}

and

{{{R_A}} \over {{R_V}}} = {{{I_g}} \over {\left( {{I_0} - {I_g}} \right)}}

B

{R_A}{R_V} = {G^2}\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)

and

{{{R_A}} \over {{R_V}}} = {\left( {{{{I_0} - {I_g}} \over {{I_g}}}} \right)^2}

C

{R_A}{R_V} = {G^2}\left( {{{{I_0} - {I_g}} \over {{I_g}}}} \right)

and

{{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}

D

{R_A}{R_V} = {G^2}

and

{{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}

Correct Answer

Option D

Solution

Galvanometer is converted into ammeter of range 0 to I0 : (Io - Ig)RA = IgG $\Rightarrow$

{R_A} = {{{I_g}G} \over {\left( {{I_0} - {I_g}} \right)}}

............(

1) Galvanometer is converted into voltmeter of range 0 to V : V = Ig(RV + G) Also given, V = GI0 $\therefore$ GI0 = Ig(RV + G) $\Rightarrow$

{R_V} = {{G\left( {{I_0} - {I_g}} \right)} \over {{I_g}}}

.........(2) From equation (1) and (2)

{R_A}{R_V} = {G^2}

{{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}

Q52

A uniform metallic wire has a resistance of 18

\Omega

and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is -

A 12

\Omega

B 2

\Omega

C 4

\Omega

D 8

\Omega

Correct Answer

Option C

Solution

Req berween any two vertex will be

{1 \over {{{\mathop{\rm R}\nolimits} _{eq}}}} = {1 \over {12}} + {1 \over 6} \Rightarrow {{\mathop{\rm R}\nolimits} _{eq.}} = 4\Omega

Q53

A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is -

A 0.4 mA

B 20 mA

C 63 mA

D 100 mA

Correct Answer

Option B

Solution

P = i2R. $\therefore$ for imax, R must be minimum from color coding R = 50 $\times$ 102

\Omega

$\therefore$ imax = 20mA

Q54

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is -

A 11

\times

10–5 W

B 11

\times

10–3 W

C 11

\times

105 W

D 11

\times

10–4 W

Correct Answer

Option A

Solution

P = I2R 4.4 = 4 $\times$ 10 $-$ 6 R R = 1.1 $\times$ 106

\Omega

P' =

{{{{11}^2}} \over R}

=

{{{{11}^2}} \over {1.1}}

$\times$ 10 $-$ 6 = 11 $\times$ 10 $-$ 5 W

Q55

A resistor develops 500 J of thermal energy in 20 s when a current of 1.5A is passed through it. If the current is increased from 1.5A to 3A, what will be the energy developed in 20 s.

A 1000 J

B 2000 J

C 1500 J

D 500 J

Correct Answer

Option B

Solution

{H_1} = i_1^2R\Delta t

{H_2} = i_2^2R\Delta t

\Rightarrow {{{H_1}} \over {{H_2}}} = {{i_1^2} \over {i_2^2}}

\Rightarrow {{500} \over {{H_2}}} = {\left( {{1 \over 2}} \right)^2}

\Rightarrow {H_2} = 2000J

Q56

An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and esistance 5

\Omega

. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is :

A 480

\Omega

B 495

\Omega

C 490

\Omega

D 395

\Omega

Correct Answer

Option D

Solution

Let current flowing in the wire is i. $\therefore$ i =

\left( {{4 \over {R + 5}}} \right)A

If resistance of 10 m length of wire is x then x = 0.5

\Omega

= 5 $\times$

{{0.1} \over 1}

\Omega

$\therefore$

\Delta

V = P.d. on wire = i. x 5 $\times$ 10 $-$ 3 =

\left( {{4 \over {R + 5}}} \right)

.(0.5) $\therefore$

{{4 \over {R + 5}}}

= 10 $-$ 2 or R + 5 = 400

\Omega

$\therefore$ R = 395

\Omega

Q57

An electrical power line, having a total resistance of 2

\Omega

, delivers 1 kW at 220 V. The efficiency of the transmission line is approximately :

A 85%

B 96%

C 72%

D 91%

Correct Answer

Option B

Solution

We know,

\eta = {{{P_{out}}} \over {\left( {{P_{out}} + {P_{loss}}} \right)}} \times 100

Given, P=

vi = {10^3}

$\therefore$

i = {{1000} \over {220}}

=

{{50} \over {11}}

A Power

loss = {i^2}R = {\left( {{{50} \over {11}}} \right)^2} \times 2

$\therefore$ efficiency

= {{1000} \over {1000 + {i^2}R}} \times 100 = 96\%

Q58

A galvanometer of resistance G is converted into a voltmeter of range 0 – 1 V by connecting a resistance R1 in series with it. The additional resistance that should be connected in series with R1 to increase the range of the voltmeter to 0 – 2 V will be :

A G

B R1

C R1 + G

D R1 - G

Correct Answer

Option C

Solution

1 = ig(G + R1) ....(1) 2 = ig(R1 + R2 + G) ....(2) Doing (1)

\div

(2), we get

{1 \over 2} = {{{i_g}\left( {G + {R_1}} \right)} \over {{i_g}\left( {G + {R_1} + {R_2}} \right)}}

$\Rightarrow$ R1 + R2 + G = 2R1 + 2G $\Rightarrow$ R2 = R1 + G

Q59

A galvanometer is used in laboratory for detecting the null point in electrical experiments. If, on passing a current of 6 mA it produces a deflection of 2o, its figure of merit is close to :

A 6

\times

10–3 A/div

B 666o A/div

C 3

\times

10–3 A/div

D 333o A/div

Correct Answer

Option C

Solution

figure of merit =

{I \over \theta }

=

{{6 \times {{10}^{ - 3}}} \over 2}

= 3 $\times$ 10–3 A/div

Q60

Consider four conducting materials copper, tungsten, mercury and aluminium with resistivity

\rho

C,

\rho

T,

\rho

M and

\rho

A respectively. Then :

A

\rho

C >

\rho

A >

\rho

T

B

\rho

M >

\rho

A >

\rho

C

\rho

A >

\rho

T >

\rho

C

D

\rho

A >

\rho

M >

\rho

C

Correct Answer

Option B

Solution

ρM = 98 × 10–8 ρA = 2.80 × 10–8 ρC = 1.72 × 10–8 ρT = 5.65 × 10–8 $\therefore$ $\rho$ M > $\rho$ T > $\rho$ A > $\rho$ C