QBManager

Current Electricity

151821 A wire of resistance $R_{1}$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

1

9 : 1

2

1 : 9

3

4 : 1

4

3 : 1

Explanation:

A Given,
Initial length $= l_{1}$ , Final length $= l_{2} = l_{1} + 2 l_{1} = 3 l_{1}$
Initial cross-section area of wire $= A_{1}$
Final cross-section area $A_{2}$
Initial volume of wire $=$ final volume of wire
$A_{1} l_{1} = A_{2} l_{2}$
$A_{1} \times l_{1} = A_{2} \times 3 l_{1}$
$A_{2} = A_{1} / 3$
Wire Resistance
$∵ R_{1} = ρ \frac{l_{1}}{A_{1}}$
New wire resistance,
$R_{2} = ρ \frac{l_{2}}{A_{2}}$
$R_{2} = ρ \frac{3 l_{1}}{(A_{1} / 3)}$
$R_{2} = ρ \frac{l_{1}}{A_{1}} \times (\frac{9}{1})$
$R_{2} = R_{1} \times \frac{9}{1}$
$\frac{R_{2}}{R_{1}} = \frac{9}{1} or R_{2} : R_{1} = 9 : 1$
The ratio of new resistance to original resistance, $R_{2} : R_{1} = 9 : 1$

JEE Main-28.07.2022

Current Electricity

151822 Given below are two statements:
Statement I : A uniform wire of resistance $80 Ω$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be $5 Ω$ .
Statement II : Two resistance $2 R$ and $3 R$ are connected in parallel in a electric circuit. The value of thermal energy developed in $3 R$ and $2 R$ will be in the ratio $3 : 2$ .
In the light of the above statements, choose the most appropriate answer from the options given below

1 Both statement I and statement II are correct

2 Both statement I and statement II are incorrect

3 Statement I is correct but statement II is incorrect

4 Statement I is incorrect but statement II is correct.

Explanation:

C Given that,
$R = 80 Ω$
Cut into four equal part,
$R_{1} = R_{2} = R_{3} = R_{4} = 20 Ω$
In parallel combination,

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{4}{20}$
$R_{eq} = 5 Ω$
Statement-II:

Thermal energy,
$Power, P = \frac{V^{2}}{R}$
$∴ \frac{P_{1}}{P_{2}} = \frac{V^{2}}{R_{1}} \times \frac{R_{2}}{V^{2}} = \frac{R_{2}}{R_{1}} = \frac{2 R}{3 R} = \frac{2}{3}$
$P_{1} : P_{2} = 2 : 3$
So, statement II is incorrect.

JEE Main-28.07.2022

Current Electricity

151823 What will be the most suitable combination of three resistors
$A = 2 Ω, B = 4 Ω, C = 6 Ω so that (\frac{22}{3}) Ω is$
equivalent resistance of combination?

1 Parallel combination of

A

and

C

connected in series with B.

2 Parallel combination of A and B connected in series with

C

.

3 Series combination of A and B connected in series with B.

4 Series combination of B and C connected in parallel with A.

Explanation:

B When we add resistance $R_{A}$ and $R_{B}$ in parallel combination then we get,
$\frac{1}{R_{1}} = \frac{1}{R_{A}} + \frac{1}{R_{B}}$
$\frac{1}{R_{1}} = \frac{1}{2} + \frac{1}{4}$
$\frac{1}{R_{1}} = \frac{2 + 1}{4} \Rightarrow R_{1} = \frac{4}{3}$
Now, when add $R_{1}$ and $R_{C}$ in series combination then we get-
$R = R_{1} + R_{C}$
$R = \frac{4}{3} + 6$
$R = \frac{22}{3} Ω$
So, the most suitable combination of three resistors (A, $B, C)$ to get equivalent resistance of combination $\frac{22}{3} Ω$ is parallel combination of $A$ and $B$ connected in series with $C$ .

JEE Main-24.06.2022

Current Electricity

151824 Two cells of same emf but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with a resistance $R$ . The value of resistance $R$ , for which the potential difference across second cell is zero, is :

1

r_{2} - r_{1}

2

r_{1} - r_{2}

3

r_{1}

4

r_{2}

Explanation:

A Given, $V_{2} = 0$
Two cells having different internal resistance $r_{1}$ and $r_{2}$ . So,

Now, current flowing through the above circuit
$V = IR$
$I = \frac{2 E}{(R + r_{1} + r_{2})}$
Since, potential difference across second cell is given as-
$V_{2} = E - {Ir}_{2}$
$∵ V_{2} = 0$
$∴ E = {Ir}_{2}$
From equation (i) and (ii) -
$I = \frac{2 {Ir}_{2}}{(R + r_{1} + r_{2})}$
$I = \frac{2 r_{2}}{(R + r_{1} + r_{2})}$
$R + r_{1} + r_{2} = 2 r_{2}$
$R = r_{2} - r_{1}$

JEE Main-25.06.2022

Current Electricity

151821 A wire of resistance $R_{1}$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

1

9 : 1

2

1 : 9

3

4 : 1

4

3 : 1

Explanation:

A Given,
Initial length $= l_{1}$ , Final length $= l_{2} = l_{1} + 2 l_{1} = 3 l_{1}$
Initial cross-section area of wire $= A_{1}$
Final cross-section area $A_{2}$
Initial volume of wire $=$ final volume of wire
$A_{1} l_{1} = A_{2} l_{2}$
$A_{1} \times l_{1} = A_{2} \times 3 l_{1}$
$A_{2} = A_{1} / 3$
Wire Resistance
$∵ R_{1} = ρ \frac{l_{1}}{A_{1}}$
New wire resistance,
$R_{2} = ρ \frac{l_{2}}{A_{2}}$
$R_{2} = ρ \frac{3 l_{1}}{(A_{1} / 3)}$
$R_{2} = ρ \frac{l_{1}}{A_{1}} \times (\frac{9}{1})$
$R_{2} = R_{1} \times \frac{9}{1}$
$\frac{R_{2}}{R_{1}} = \frac{9}{1} or R_{2} : R_{1} = 9 : 1$
The ratio of new resistance to original resistance, $R_{2} : R_{1} = 9 : 1$

JEE Main-28.07.2022

Current Electricity

151822 Given below are two statements:
Statement I : A uniform wire of resistance $80 Ω$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be $5 Ω$ .
Statement II : Two resistance $2 R$ and $3 R$ are connected in parallel in a electric circuit. The value of thermal energy developed in $3 R$ and $2 R$ will be in the ratio $3 : 2$ .
In the light of the above statements, choose the most appropriate answer from the options given below

1 Both statement I and statement II are correct

2 Both statement I and statement II are incorrect

3 Statement I is correct but statement II is incorrect

4 Statement I is incorrect but statement II is correct.

Explanation:

C Given that,
$R = 80 Ω$
Cut into four equal part,
$R_{1} = R_{2} = R_{3} = R_{4} = 20 Ω$
In parallel combination,

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{4}{20}$
$R_{eq} = 5 Ω$
Statement-II:

Thermal energy,
$Power, P = \frac{V^{2}}{R}$
$∴ \frac{P_{1}}{P_{2}} = \frac{V^{2}}{R_{1}} \times \frac{R_{2}}{V^{2}} = \frac{R_{2}}{R_{1}} = \frac{2 R}{3 R} = \frac{2}{3}$
$P_{1} : P_{2} = 2 : 3$
So, statement II is incorrect.

JEE Main-28.07.2022

Current Electricity

151823 What will be the most suitable combination of three resistors
$A = 2 Ω, B = 4 Ω, C = 6 Ω so that (\frac{22}{3}) Ω is$
equivalent resistance of combination?

1 Parallel combination of

A

and

C

connected in series with B.

2 Parallel combination of A and B connected in series with

C

.

3 Series combination of A and B connected in series with B.

4 Series combination of B and C connected in parallel with A.

Explanation:

B When we add resistance $R_{A}$ and $R_{B}$ in parallel combination then we get,
$\frac{1}{R_{1}} = \frac{1}{R_{A}} + \frac{1}{R_{B}}$
$\frac{1}{R_{1}} = \frac{1}{2} + \frac{1}{4}$
$\frac{1}{R_{1}} = \frac{2 + 1}{4} \Rightarrow R_{1} = \frac{4}{3}$
Now, when add $R_{1}$ and $R_{C}$ in series combination then we get-
$R = R_{1} + R_{C}$
$R = \frac{4}{3} + 6$
$R = \frac{22}{3} Ω$
So, the most suitable combination of three resistors (A, $B, C)$ to get equivalent resistance of combination $\frac{22}{3} Ω$ is parallel combination of $A$ and $B$ connected in series with $C$ .

JEE Main-24.06.2022

Current Electricity

151824 Two cells of same emf but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with a resistance $R$ . The value of resistance $R$ , for which the potential difference across second cell is zero, is :

1

r_{2} - r_{1}

2

r_{1} - r_{2}

3

r_{1}

4

r_{2}

Explanation:

A Given, $V_{2} = 0$
Two cells having different internal resistance $r_{1}$ and $r_{2}$ . So,

Now, current flowing through the above circuit
$V = IR$
$I = \frac{2 E}{(R + r_{1} + r_{2})}$
Since, potential difference across second cell is given as-
$V_{2} = E - {Ir}_{2}$
$∵ V_{2} = 0$
$∴ E = {Ir}_{2}$
From equation (i) and (ii) -
$I = \frac{2 {Ir}_{2}}{(R + r_{1} + r_{2})}$
$I = \frac{2 r_{2}}{(R + r_{1} + r_{2})}$
$R + r_{1} + r_{2} = 2 r_{2}$
$R = r_{2} - r_{1}$

JEE Main-25.06.2022

Current Electricity

151821 A wire of resistance $R_{1}$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

1

9 : 1

2

1 : 9

3

4 : 1

4

3 : 1

Explanation:

A Given,
Initial length $= l_{1}$ , Final length $= l_{2} = l_{1} + 2 l_{1} = 3 l_{1}$
Initial cross-section area of wire $= A_{1}$
Final cross-section area $A_{2}$
Initial volume of wire $=$ final volume of wire
$A_{1} l_{1} = A_{2} l_{2}$
$A_{1} \times l_{1} = A_{2} \times 3 l_{1}$
$A_{2} = A_{1} / 3$
Wire Resistance
$∵ R_{1} = ρ \frac{l_{1}}{A_{1}}$
New wire resistance,
$R_{2} = ρ \frac{l_{2}}{A_{2}}$
$R_{2} = ρ \frac{3 l_{1}}{(A_{1} / 3)}$
$R_{2} = ρ \frac{l_{1}}{A_{1}} \times (\frac{9}{1})$
$R_{2} = R_{1} \times \frac{9}{1}$
$\frac{R_{2}}{R_{1}} = \frac{9}{1} or R_{2} : R_{1} = 9 : 1$
The ratio of new resistance to original resistance, $R_{2} : R_{1} = 9 : 1$

JEE Main-28.07.2022

Current Electricity

151822 Given below are two statements:
Statement I : A uniform wire of resistance $80 Ω$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be $5 Ω$ .
Statement II : Two resistance $2 R$ and $3 R$ are connected in parallel in a electric circuit. The value of thermal energy developed in $3 R$ and $2 R$ will be in the ratio $3 : 2$ .
In the light of the above statements, choose the most appropriate answer from the options given below

1 Both statement I and statement II are correct

2 Both statement I and statement II are incorrect

3 Statement I is correct but statement II is incorrect

4 Statement I is incorrect but statement II is correct.

Explanation:

C Given that,
$R = 80 Ω$
Cut into four equal part,
$R_{1} = R_{2} = R_{3} = R_{4} = 20 Ω$
In parallel combination,

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{4}{20}$
$R_{eq} = 5 Ω$
Statement-II:

Thermal energy,
$Power, P = \frac{V^{2}}{R}$
$∴ \frac{P_{1}}{P_{2}} = \frac{V^{2}}{R_{1}} \times \frac{R_{2}}{V^{2}} = \frac{R_{2}}{R_{1}} = \frac{2 R}{3 R} = \frac{2}{3}$
$P_{1} : P_{2} = 2 : 3$
So, statement II is incorrect.

JEE Main-28.07.2022

Current Electricity

151823 What will be the most suitable combination of three resistors
$A = 2 Ω, B = 4 Ω, C = 6 Ω so that (\frac{22}{3}) Ω is$
equivalent resistance of combination?

1 Parallel combination of

A

and

C

connected in series with B.

2 Parallel combination of A and B connected in series with

C

.

3 Series combination of A and B connected in series with B.

4 Series combination of B and C connected in parallel with A.

Explanation:

B When we add resistance $R_{A}$ and $R_{B}$ in parallel combination then we get,
$\frac{1}{R_{1}} = \frac{1}{R_{A}} + \frac{1}{R_{B}}$
$\frac{1}{R_{1}} = \frac{1}{2} + \frac{1}{4}$
$\frac{1}{R_{1}} = \frac{2 + 1}{4} \Rightarrow R_{1} = \frac{4}{3}$
Now, when add $R_{1}$ and $R_{C}$ in series combination then we get-
$R = R_{1} + R_{C}$
$R = \frac{4}{3} + 6$
$R = \frac{22}{3} Ω$
So, the most suitable combination of three resistors (A, $B, C)$ to get equivalent resistance of combination $\frac{22}{3} Ω$ is parallel combination of $A$ and $B$ connected in series with $C$ .

JEE Main-24.06.2022

Current Electricity

151824 Two cells of same emf but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with a resistance $R$ . The value of resistance $R$ , for which the potential difference across second cell is zero, is :

1

r_{2} - r_{1}

2

r_{1} - r_{2}

3

r_{1}

4

r_{2}

Explanation:

A Given, $V_{2} = 0$
Two cells having different internal resistance $r_{1}$ and $r_{2}$ . So,

Now, current flowing through the above circuit
$V = IR$
$I = \frac{2 E}{(R + r_{1} + r_{2})}$
Since, potential difference across second cell is given as-
$V_{2} = E - {Ir}_{2}$
$∵ V_{2} = 0$
$∴ E = {Ir}_{2}$
From equation (i) and (ii) -
$I = \frac{2 {Ir}_{2}}{(R + r_{1} + r_{2})}$
$I = \frac{2 r_{2}}{(R + r_{1} + r_{2})}$
$R + r_{1} + r_{2} = 2 r_{2}$
$R = r_{2} - r_{1}$

JEE Main-25.06.2022

Current Electricity

151821 A wire of resistance $R_{1}$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

1

9 : 1

2

1 : 9

3

4 : 1

4

3 : 1

Explanation:

A Given,
Initial length $= l_{1}$ , Final length $= l_{2} = l_{1} + 2 l_{1} = 3 l_{1}$
Initial cross-section area of wire $= A_{1}$
Final cross-section area $A_{2}$
Initial volume of wire $=$ final volume of wire
$A_{1} l_{1} = A_{2} l_{2}$
$A_{1} \times l_{1} = A_{2} \times 3 l_{1}$
$A_{2} = A_{1} / 3$
Wire Resistance
$∵ R_{1} = ρ \frac{l_{1}}{A_{1}}$
New wire resistance,
$R_{2} = ρ \frac{l_{2}}{A_{2}}$
$R_{2} = ρ \frac{3 l_{1}}{(A_{1} / 3)}$
$R_{2} = ρ \frac{l_{1}}{A_{1}} \times (\frac{9}{1})$
$R_{2} = R_{1} \times \frac{9}{1}$
$\frac{R_{2}}{R_{1}} = \frac{9}{1} or R_{2} : R_{1} = 9 : 1$
The ratio of new resistance to original resistance, $R_{2} : R_{1} = 9 : 1$

JEE Main-28.07.2022

Current Electricity

151822 Given below are two statements:
Statement I : A uniform wire of resistance $80 Ω$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be $5 Ω$ .
Statement II : Two resistance $2 R$ and $3 R$ are connected in parallel in a electric circuit. The value of thermal energy developed in $3 R$ and $2 R$ will be in the ratio $3 : 2$ .
In the light of the above statements, choose the most appropriate answer from the options given below

1 Both statement I and statement II are correct

2 Both statement I and statement II are incorrect

3 Statement I is correct but statement II is incorrect

4 Statement I is incorrect but statement II is correct.

Explanation:

C Given that,
$R = 80 Ω$
Cut into four equal part,
$R_{1} = R_{2} = R_{3} = R_{4} = 20 Ω$
In parallel combination,

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} = \frac{4}{20}$
$R_{eq} = 5 Ω$
Statement-II:

Thermal energy,
$Power, P = \frac{V^{2}}{R}$
$∴ \frac{P_{1}}{P_{2}} = \frac{V^{2}}{R_{1}} \times \frac{R_{2}}{V^{2}} = \frac{R_{2}}{R_{1}} = \frac{2 R}{3 R} = \frac{2}{3}$
$P_{1} : P_{2} = 2 : 3$
So, statement II is incorrect.

JEE Main-28.07.2022

Current Electricity

151823 What will be the most suitable combination of three resistors
$A = 2 Ω, B = 4 Ω, C = 6 Ω so that (\frac{22}{3}) Ω is$
equivalent resistance of combination?

1 Parallel combination of

A

and

C

connected in series with B.

2 Parallel combination of A and B connected in series with

C

.

3 Series combination of A and B connected in series with B.

4 Series combination of B and C connected in parallel with A.

Explanation:

B When we add resistance $R_{A}$ and $R_{B}$ in parallel combination then we get,
$\frac{1}{R_{1}} = \frac{1}{R_{A}} + \frac{1}{R_{B}}$
$\frac{1}{R_{1}} = \frac{1}{2} + \frac{1}{4}$
$\frac{1}{R_{1}} = \frac{2 + 1}{4} \Rightarrow R_{1} = \frac{4}{3}$
Now, when add $R_{1}$ and $R_{C}$ in series combination then we get-
$R = R_{1} + R_{C}$
$R = \frac{4}{3} + 6$
$R = \frac{22}{3} Ω$
So, the most suitable combination of three resistors (A, $B, C)$ to get equivalent resistance of combination $\frac{22}{3} Ω$ is parallel combination of $A$ and $B$ connected in series with $C$ .

JEE Main-24.06.2022

Current Electricity

151824 Two cells of same emf but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with a resistance $R$ . The value of resistance $R$ , for which the potential difference across second cell is zero, is :

1

r_{2} - r_{1}

2

r_{1} - r_{2}

3

r_{1}

4

r_{2}

Explanation:

A Given, $V_{2} = 0$
Two cells having different internal resistance $r_{1}$ and $r_{2}$ . So,

Now, current flowing through the above circuit
$V = IR$
$I = \frac{2 E}{(R + r_{1} + r_{2})}$
Since, potential difference across second cell is given as-
$V_{2} = E - {Ir}_{2}$
$∵ V_{2} = 0$
$∴ E = {Ir}_{2}$
From equation (i) and (ii) -
$I = \frac{2 {Ir}_{2}}{(R + r_{1} + r_{2})}$
$I = \frac{2 r_{2}}{(R + r_{1} + r_{2})}$
$R + r_{1} + r_{2} = 2 r_{2}$
$R = r_{2} - r_{1}$

JEE Main-25.06.2022

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