Current Electricity — Page 13 | JEE Physics

Q121

An electric bulb rated

50 \mathrm{~W}-200 \mathrm{~V}

is connected across a

100 \mathrm{~V}

supply. The power dissipation of the bulb is:

A 100 W

B 50 W

C 12.5 W

D 25 W

Correct Answer

Option C

Solution

To find the power dissipation of the bulb when it's connected to a

100 \mathrm{V}

supply instead of its rated

200 \mathrm{V}

supply, we can use the relation between power (P), voltage (V), and resistance (R), which is given by

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

.

The resistance of the bulb can be considered constant in this case, allowing us to calculate the change in power dissipation due to the change in voltage.

First, let's find the resistance of the bulb based on its rated conditions:

P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P}

Substituting the rated values, we get:

R = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \, \Omega

Now, using this resistance, we can find the power dissipation when the bulb is connected to a

100 \mathrm{V}

supply:

P = \frac{V^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \, W

This means the power dissipation of the bulb when connected to a

100 \mathrm{V}

supply is

12.5 \, W

. Therefore, the correct answer is Option C: 12.5 W.

Q122

Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be _________ to __________ times of its initial length if the water is to be boiled in 15 minutes.

A increased,

\dfrac{3}{4}

B increased,

\dfrac{4}{3}

C decreased,

\dfrac{3}{4}

D decreased,

\dfrac{4}{3}

Correct Answer

Option C

Solution

When an electric kettle is used to heat water, the time taken to boil the water depends on the power of the heating element.

The power supplied to the heating element is inversely proportional to the heating time.

If we want to reduce the boiling time, the power needs to be increased.

The power of the heating element is given by:

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

where

P

is the power,

V

is the voltage, and

R

is the resistance of the heating element. The resistance

R

of the heating element is proportional to its length

L

while the material and cross-sectional area remain constant. So, we can write:

R \propto L

To achieve boiling in 15 minutes instead of 20 minutes, the power needs to increase, which implies the resistance must decrease.

Let the initial length of the heating element be

L_0

, and let the new length needed be

L

. The time taken to heat is inversely proportional to the power:

\frac{T_1}{T_2} = \frac{P_2}{P_1}

Given that:

T_1 = 20 \text{ minutes}

T_2 = 15 \text{ minutes}

We need to find the ratio:

\frac{20}{15} = \frac{P_2}{P_1}

Simplifying:

\frac{4}{3} = \frac{P_2}{P_1}

The power is inversely proportional to the resistance:

\frac{P_2}{P_1} = \frac{R_1}{R_2}

Thus:

\frac{4}{3} = \frac{R_1}{R_2}

Since resistance is proportional to length:

\frac{R_1}{R_2} = \frac{L_1}{L_2}

Hence:

\frac{4}{3} = \frac{L_1}{L_2}

Simplifying, we find:

L_2 = \frac{3}{4} L_1

This means the length of the heating element should be decreased to

\frac{3}{4}

of its initial length. Thus, the correct option is: Option C: decreased,

\frac{3}{4}

Q123

A galvanometer of resistance

100 \Omega

when connected in series with

400 \Omega

measures a voltage of upto

10 \mathrm{~V}

. The value of resistance required to convert the galvanometer into ammeter to read upto

10 \mathrm{~A}

is

x \times 10^{-2} \Omega

. The value of

x

is :

A 2

B 20

C 800

D 200

Correct Answer

Option B

Solution

To convert a galvanometer into an ammeter to measure larger currents, a low resistance known as the shunt resistance (

R_{\text{sh}}

) is connected in parallel with the galvanometer.

The value of this shunt resistance can be calculated using the principles of parallel circuits and the desired maximum current the ammeter should read.

The original configuration of the galvanometer allows it to measure up to

10\,\text{V}

, and it has a resistance of

100\,\Omega

. When connected in series with a

400\,\Omega

resistor, the total resistance in the circuit is

100\,\Omega + 400\,\Omega = 500\,\Omega

. Given this configuration measures up to

10\,\text{V}

, we can calculate the maximum current it is designed to measure using Ohm's law: $I = \dfrac{V}{R} = \dfrac{10\,\text{V}}{500\,\Omega} = 0.02\,\text{A}$ Now, to recalibrate the device to measure up to

10\,\text{A}

, we require the calculation of the shunt resistor

R_{\text{sh}}

that needs to be connected in parallel with the galvanometer. The total current

I

will now be

10\,\text{A}

, and the part of this current flowing through the galvanometer (

I_g

) remains

0.02\,\text{A}

(as before, to ensure we do not exceed the device's original maximum measuring capability), leaving the rest to flow through the shunt.

Thus,

I - I_g

flows through the shunt.

Since the voltage drop across both the shunt and the galvanometer must be the same for parallel components, we use Ohm’s Law

V = IR

for both and set up an equation to calculate

R_{\text{sh}}

: $I_g R_g = (I - I_g) R_{\text{sh}}$ Substituting known values (

R_g = 100\,\Omega

,

I = 10\,\text{A}

, and

I_g = 0.02\,\text{A}

): $0.02\,\text{A} \times 100\,\Omega = (10\,\text{A} - 0.02\,\text{A}) R_{\text{sh}}$ This simplifies to: $2\,\text{V} = 9.98\,\text{A} \times R_{\text{sh}}$ Solving for

R_{\text{sh}}

gives: $R_{\text{sh}} = \dfrac{2\,\text{V}}{9.98\,\text{A}} \approx 0.2004\,\Omega$ Expressing this in terms of

\times 10^{-2} \Omega

gives

R_{\text{sh}} \approx 20.04 \times 10^{-2} \Omega

. Therefore, the value of

x

is approximately

20.04

, and rounding it according to the provided options leads to the closest value: Option B: 20

Q124

A galvanometer having a coil of resistance

30 \Omega

need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be

\dfrac{30}{X} \Omega

, where

X

is

A 447

B 298

C 149

D 596

Correct Answer

Option C

Solution

Given: Galvanometer resistance:

R_g = 30 \, \Omega

Full-scale deflection current:

I_g = 20 \, mA = 0.02 \, A

Maximum current to be measured:

I = 3 \, A

When using the galvanometer to measure 3 A, the extra current passing through the shunt resistor ( $R_s$ ) is:

I_s = I - I_g = 3 - 0.02 = 2.98 \, A.

Since the galvanometer and the shunt resistor are connected in parallel, their voltage drops must be equal. Therefore:

I_g R_g = I_s R_s.

Substitute the given values:

0.02 \times 30 = 2.98 \times R_s \quad \Rightarrow \quad R_s = \frac{0.6}{2.98} \approx 0.2013 \, \Omega.

The shunt resistance is represented as:

R_s = \frac{30}{X} \, \Omega.

Equate the two expressions:

\frac{30}{X} = 0.2013.

Solving for $X$ :

X = \frac{30}{0.2013} \approx 149.

Thus, the value of $X$ is approximately

149.

Q125

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into

a

square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

A

8 / 9

B

9 / 8

C

32 / 27

D

27 / 32

Correct Answer

Option C

Solution

We know,

R = \rho {l \over A} \Rightarrow R \propto l

Side length of triangle

= {l \over 3}

Side length of square

= {l \over 4}

So,

\Rightarrow {\left( {{R_{eq}}} \right)_1} = {{{{2R} \over 3} \times {R \over 3}} \over {{{2R} \over 3} + {R \over 3}}}

and

{\left( {{R_{eq}}} \right)_2} = {{{{3R} \over 4} \times {R \over 4}} \over {{{3R} \over 4} + {R \over 4}}}

= {{2{R^2}} \over 9} \times {3 \over {3R}} = {{2R} \over 9}

and

{\left( {{R_{eq}}} \right)_2} = {{3{R^2}} \over {16}} - R = {{3R} \over {16}}

Hence,

{{{{\left( {{R_{eq}}} \right)}_1}} \over {{{\left( {{R_{eq}}} \right)}_2}}} = {{{{2R} \over 9}} \over {{{3R} \over {16}}}} = {{32} \over {27}}

Q126

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged. Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor

\left(R / \sqrt{R^2+\omega^2 L^2}\right)

, where

\omega

is frequency of the supply across resistor

R

and inductor

L

. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage. In the light of the above statements, choose the most appropriate answer from the options given below :

A (A) is true but (R) is false

B Both

(\mathbf{A})

and

(\mathbf{R})

are true and

(\mathbf{R})

is the correct explanation of

(\mathbf{A})

3.

C Both

(\mathbf{A})

and

(\mathbf{R})

are true but

(\mathbf{R})

is not the correct explanation of

(\mathbf{A})

D (A) is false but (R) is true

Correct Answer

Option B

Solution

Assertion (A): A choke coil is simply a coil having large inductance but small resistance.

Choke coils are used with fluorescent mercury-tube fittings.

If household electric power is directly connected to a mercury tube, the tube will be damaged.

A choke (or ballast) does have high inductance and low resistance.

In a fluorescent lamp (or mercury‐vapor lamp) circuit, the choke limits the current through the tube.

If we connect the tube directly to the full line voltage without any current-limiting device, excessive current will flow and damage the tube.

Therefore, (A) is true.

Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor $\displaystyle \dfrac{R}{\sqrt{R^2 + (\omega L)^2}}$ , where $\omega$ is the supply frequency, and $R$ and $L$ are the resistance and inductance in series.

If the choke coil were not used, the voltage across the resistor (tube) would be the same as the applied voltage.

In a simple series R–L circuit (where $R$ represents the lamp’s effective resistance once conducting, and $L$ is the choke’s inductance), the total impedance is $Z = \sqrt{\,R^2 + (\omega L)^2\,}.$ The current is the same through $R$ and $L$ .

The fraction of the total supply voltage appearing across $R$ alone is $\dfrac{V_R}{V_{\text{total}}} \;=\; \dfrac{I\,R}{I\,Z} \;=\; \dfrac{R}{\sqrt{R^2 + (\omega L)^2}}.$ This shows that using a large inductance $L$ (the choke) can substantially reduce (or limit) the voltage—and hence the current—through the lamp.

Without the choke coil, the tube (treated here as a resistor once it starts conducting) would indeed see nearly the full line voltage, causing a large current that could destroy it.

Therefore, (R) is also true.

Does (R) correctly explain (A)?

(A) states why a choke coil is needed (to prevent the lamp from being damaged by excessive current).

(R) shows how the choke coil limits the voltage/current through the lamp—by giving the fraction of the supply voltage that appears across the lamp in a series R–L circuit.

Thus, (R) does provide the correct (physics-based) explanation: the choke’s inductive reactance drops a significant portion of the supply voltage, thereby protecting the lamp.

Hence, the correct option is: $\boxed{\text{Option B: Both (A) and (R) are true and (R) is the correct explanation of (A).}}$

Q127

The battery of a mobile phone is rated as

4.2 \mathrm{~V}, 5800 \mathrm{~mAh}

. How much energy is stored ir it when fully charged?

A 87.7 kJ

B 43.8 kJ

C 48.7 kJ

D 24.4 kJ

Correct Answer

Option A

Solution

To find the energy stored in the battery, follow these steps: Convert mAh to Ah: The battery capacity is given as 5800 mAh.

Since

1\,\text{Ah} = 1000\,\text{mAh},

we have

5800\,\text{mAh} = 5.8\,\text{Ah}.

Calculate energy in watt-hours (Wh): The energy in Wh is given by the product of the voltage and the capacity in Ah.

Thus,

\text{Energy (Wh)} = 4.2\,\text{V} \times 5.8\,\text{Ah} = 24.36\,\text{Wh}.

Convert watt-hours to joules (J): Since

1\,\text{Wh} = 3600\,\text{J},

the total energy in joules is:

24.36\,\text{Wh} \times 3600\,\text{J/Wh} = 87,696\,\text{J} \approx 87.7\,\text{kJ}.

Thus, the energy stored in the battery when fully charged is about 87.7 kJ. The correct answer is Option A: 87.7 kJ.

Q128

A wire of length 25 m and cross-sectional area

5 \mathrm{~mm}^2

having resistivity of

2 \times 10^{-6} \Omega \mathrm{~m}

is bent into a complete circle. The resistance between diametrically opposite points will be

A

100 \Omega

B

2.5 \Omega

C

12.5 \Omega

D

50 \Omega

Correct Answer

Option B

Solution

Length of the wire, $L = 25\rm\,m$ Cross‐sectional area, $A = 5\rm\,mm^2 = 5\times10^{-6}\,m^2$ Resistivity, $\rho = 2\times10^{-6}\,\Omega\cdot m$ Resistance of the entire wire:

R_{\rm total}=\frac{\rho\,L}{A} =\frac{2\times10^{-6}\times25}{5\times10^{-6}} =\frac{50\times10^{-6}}{5\times10^{-6}} =10\ \Omega

When bent into a circle, points diametrically opposite divide the ring into two equal halves.

Each half-circle has length $L/2 = 12.5\rm\,m$ , so its resistance is half of $R_{\rm total}$ :

R_{\rm half}=\frac{R_{\rm total}}{2} =\frac{10}{2} =5\ \Omega

These two $5\,\Omega$ halves are in parallel between the opposite points. Equivalent resistance $R$ is given by:

\frac{1}{R}=\frac{1}{R_{\rm half}}+\frac{1}{R_{\rm half}} =\frac{1}{5}+\frac{1}{5} =\frac{2}{5}

R=\frac{5}{2}=2.5\ \Omega

Answer: 2.5 Ω (Option B)

Q129

There are '

n

' number of identical electric bulbs, each is designed to draw a power

p

independently from the mains supply. They are now joined in series across the mains supply. The total power drawn by the combination is :

A np

B p

C

\dfrac{\mathrm{p}}{\mathrm{n}^2}

D

\dfrac{p}{n}

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3+\ldots \ldots \ldots+\mathrm{R}_{\mathrm{n}} \\ & \frac{\mathrm{~V}^2}{\mathrm{P}_{\mathrm{s}}}=\frac{\mathrm{V}^2}{\mathrm{P}}+\frac{\mathrm{V}^2}{\mathrm{P}}+\ldots \ldots . .+\frac{\mathrm{V}^2}{\mathrm{P}_{\mathrm{n}}} \\ & \mathrm{P}_{\mathrm{s}}=\frac{\mathrm{P}}{\mathrm{n}} \end{aligned}

Q130

A wire when connected to

220

V

mains supply has power dissipation

{P_1}.

Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is

{P_2}.

Then

{P_2}:{P_1}

is

A

1

B

4

C

2

D

3

Correct Answer

Option B

Solution

Case 1 :

{P_1} = {{{V^2}} \over R}

Case 2 : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is

{R \over 2}.

These are connected in parallel $\therefore$

{{\mathop{\rm R}\nolimits} _{eq}} = {{R/2} \over 2} = {R \over 4}

$\therefore$

{P_2} = {{{V^2}} \over {R/4}} = 4\left( {{{{V^2}} \over R}} \right) = 4{P_1}