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PHXII03:CURRENT ELECTRICITY

356949 Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thicker wire is $10 Ω$ .The total resistance of the combination will be

1

20 Ω

2

10 Ω

3

100 Ω

4

40 Ω

Explanation:

Length of each wire $= ℓ;$ Area of thick wire $(A_{1}) = 3 A;$ Area of thin wire $(A_{2}) = A$ and resistance of thick wire $(R_{1}) = 10 Ω$ Resistance
$(R) = ρ \frac{ℓ}{A}$
$\frac{R_{1}}{R_{2}} = \frac{A_{2}}{A_{1}} = \frac{A}{3 A} = \frac{1}{3}$
$\Rightarrow R_{2} = 3 R_{1} = 3 \times 10 = 30 Ω$
The equivalent resistance of these two resistors in series $= R_{1} + R_{2} = 30 + 10 = 40 Ω .$

PHXII03:CURRENT ELECTRICITY

356950 Three resistances of one ohm each are connected in parallel. Such connection is again connected with $\frac{5}{3} Ω$ resistor in series. The resultant resistance will be

1

2 Ω

2

5 Ω

3

8 Ω

4

10 Ω

Explanation:

Resistance of 1 ohm group $= \frac{R}{n} = \frac{1}{3} Ω$
This is in series with $\frac{5}{3} Ω$ resistor.
$∴$ Total resistance $= \frac{5}{3} + \frac{1}{3} = \frac{6}{3} Ω = 2 Ω$

PHXII03:CURRENT ELECTRICITY

356951 A uniform wire of resistance $R$ and length $L$ is cut into four equal parts, each of length $L / 4$ , which are then connected in parallel combination. The effective resistance of the combination will be

1

R

2

4 R

3

\frac{R}{4}

4

\frac{R}{16}

Explanation:

Resistance, $R = \frac{ρ l}{A}$ or $R \propto l$
$∴ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{L}{L / 4} = 4$
$\Rightarrow R_{2} = \frac{R}{4} (∵ R_{1} = R)$
In parallel combination of such four
resistances,
$\frac{1}{R^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}}$
$or \frac{1}{R^{'}} = \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4}$
$\Rightarrow \frac{1}{R^{'}} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R}$
$∴ \frac{1}{R^{'}} = \frac{16}{R} \Rightarrow R^{'} = \frac{R}{16}$

PHXII03:CURRENT ELECTRICITY

356952 Effective resistance between $A$ and $B$ in the following circuit
supporting img

1

10 Ω

2

20 Ω

3

5 Ω

4

\frac{20}{3} Ω

Explanation:

The given circuit can be redrawn as:
This is balanced Wheatstone bridge.
$∴$ The two $10 Ω$ resistances at the middle are inactive
So, $R_{A B} = \frac{20 \times 20}{20 + 20} = 10 Ω$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

356953 The lowest resistance which can be obtained by connecting 10 resistors each of 1/10 $o h m$ is

1

1 / 250 Ω

2

1 / 200 Ω

3

1 / 100 Ω

4

1 / 10 Ω

Explanation:

Lowest resistance will be in the case when all the resistors are connected in parallel.
$R = \frac{1}{100} Ω$

PHXII03:CURRENT ELECTRICITY

356949 Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thicker wire is $10 Ω$ .The total resistance of the combination will be

1

20 Ω

2

10 Ω

3

100 Ω

4

40 Ω

Explanation:

Length of each wire $= ℓ;$ Area of thick wire $(A_{1}) = 3 A;$ Area of thin wire $(A_{2}) = A$ and resistance of thick wire $(R_{1}) = 10 Ω$ Resistance
$(R) = ρ \frac{ℓ}{A}$
$\frac{R_{1}}{R_{2}} = \frac{A_{2}}{A_{1}} = \frac{A}{3 A} = \frac{1}{3}$
$\Rightarrow R_{2} = 3 R_{1} = 3 \times 10 = 30 Ω$
The equivalent resistance of these two resistors in series $= R_{1} + R_{2} = 30 + 10 = 40 Ω .$

PHXII03:CURRENT ELECTRICITY

356950 Three resistances of one ohm each are connected in parallel. Such connection is again connected with $\frac{5}{3} Ω$ resistor in series. The resultant resistance will be

1

2 Ω

2

5 Ω

3

8 Ω

4

10 Ω

Explanation:

Resistance of 1 ohm group $= \frac{R}{n} = \frac{1}{3} Ω$
This is in series with $\frac{5}{3} Ω$ resistor.
$∴$ Total resistance $= \frac{5}{3} + \frac{1}{3} = \frac{6}{3} Ω = 2 Ω$

PHXII03:CURRENT ELECTRICITY

356951 A uniform wire of resistance $R$ and length $L$ is cut into four equal parts, each of length $L / 4$ , which are then connected in parallel combination. The effective resistance of the combination will be

1

R

2

4 R

3

\frac{R}{4}

4

\frac{R}{16}

Explanation:

Resistance, $R = \frac{ρ l}{A}$ or $R \propto l$
$∴ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{L}{L / 4} = 4$
$\Rightarrow R_{2} = \frac{R}{4} (∵ R_{1} = R)$
In parallel combination of such four
resistances,
$\frac{1}{R^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}}$
$or \frac{1}{R^{'}} = \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4}$
$\Rightarrow \frac{1}{R^{'}} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R}$
$∴ \frac{1}{R^{'}} = \frac{16}{R} \Rightarrow R^{'} = \frac{R}{16}$

PHXII03:CURRENT ELECTRICITY

356952 Effective resistance between $A$ and $B$ in the following circuit
supporting img

1

10 Ω

2

20 Ω

3

5 Ω

4

\frac{20}{3} Ω

Explanation:

The given circuit can be redrawn as:
This is balanced Wheatstone bridge.
$∴$ The two $10 Ω$ resistances at the middle are inactive
So, $R_{A B} = \frac{20 \times 20}{20 + 20} = 10 Ω$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

356953 The lowest resistance which can be obtained by connecting 10 resistors each of 1/10 $o h m$ is

1

1 / 250 Ω

2

1 / 200 Ω

3

1 / 100 Ω

4

1 / 10 Ω

Explanation:

Lowest resistance will be in the case when all the resistors are connected in parallel.
$R = \frac{1}{100} Ω$

PHXII03:CURRENT ELECTRICITY

356949 Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thicker wire is $10 Ω$ .The total resistance of the combination will be

1

20 Ω

2

10 Ω

3

100 Ω

4

40 Ω

Explanation:

Length of each wire $= ℓ;$ Area of thick wire $(A_{1}) = 3 A;$ Area of thin wire $(A_{2}) = A$ and resistance of thick wire $(R_{1}) = 10 Ω$ Resistance
$(R) = ρ \frac{ℓ}{A}$
$\frac{R_{1}}{R_{2}} = \frac{A_{2}}{A_{1}} = \frac{A}{3 A} = \frac{1}{3}$
$\Rightarrow R_{2} = 3 R_{1} = 3 \times 10 = 30 Ω$
The equivalent resistance of these two resistors in series $= R_{1} + R_{2} = 30 + 10 = 40 Ω .$

PHXII03:CURRENT ELECTRICITY

356950 Three resistances of one ohm each are connected in parallel. Such connection is again connected with $\frac{5}{3} Ω$ resistor in series. The resultant resistance will be

1

2 Ω

2

5 Ω

3

8 Ω

4

10 Ω

Explanation:

Resistance of 1 ohm group $= \frac{R}{n} = \frac{1}{3} Ω$
This is in series with $\frac{5}{3} Ω$ resistor.
$∴$ Total resistance $= \frac{5}{3} + \frac{1}{3} = \frac{6}{3} Ω = 2 Ω$

PHXII03:CURRENT ELECTRICITY

356951 A uniform wire of resistance $R$ and length $L$ is cut into four equal parts, each of length $L / 4$ , which are then connected in parallel combination. The effective resistance of the combination will be

1

R

2

4 R

3

\frac{R}{4}

4

\frac{R}{16}

Explanation:

Resistance, $R = \frac{ρ l}{A}$ or $R \propto l$
$∴ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{L}{L / 4} = 4$
$\Rightarrow R_{2} = \frac{R}{4} (∵ R_{1} = R)$
In parallel combination of such four
resistances,
$\frac{1}{R^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}}$
$or \frac{1}{R^{'}} = \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4}$
$\Rightarrow \frac{1}{R^{'}} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R}$
$∴ \frac{1}{R^{'}} = \frac{16}{R} \Rightarrow R^{'} = \frac{R}{16}$

PHXII03:CURRENT ELECTRICITY

356952 Effective resistance between $A$ and $B$ in the following circuit
supporting img

1

10 Ω

2

20 Ω

3

5 Ω

4

\frac{20}{3} Ω

Explanation:

The given circuit can be redrawn as:
This is balanced Wheatstone bridge.
$∴$ The two $10 Ω$ resistances at the middle are inactive
So, $R_{A B} = \frac{20 \times 20}{20 + 20} = 10 Ω$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

356953 The lowest resistance which can be obtained by connecting 10 resistors each of 1/10 $o h m$ is

1

1 / 250 Ω

2

1 / 200 Ω

3

1 / 100 Ω

4

1 / 10 Ω

Explanation:

Lowest resistance will be in the case when all the resistors are connected in parallel.
$R = \frac{1}{100} Ω$

PHXII03:CURRENT ELECTRICITY

356949 Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thicker wire is $10 Ω$ .The total resistance of the combination will be

1

20 Ω

2

10 Ω

3

100 Ω

4

40 Ω

Explanation:

Length of each wire $= ℓ;$ Area of thick wire $(A_{1}) = 3 A;$ Area of thin wire $(A_{2}) = A$ and resistance of thick wire $(R_{1}) = 10 Ω$ Resistance
$(R) = ρ \frac{ℓ}{A}$
$\frac{R_{1}}{R_{2}} = \frac{A_{2}}{A_{1}} = \frac{A}{3 A} = \frac{1}{3}$
$\Rightarrow R_{2} = 3 R_{1} = 3 \times 10 = 30 Ω$
The equivalent resistance of these two resistors in series $= R_{1} + R_{2} = 30 + 10 = 40 Ω .$

PHXII03:CURRENT ELECTRICITY

356950 Three resistances of one ohm each are connected in parallel. Such connection is again connected with $\frac{5}{3} Ω$ resistor in series. The resultant resistance will be

1

2 Ω

2

5 Ω

3

8 Ω

4

10 Ω

Explanation:

Resistance of 1 ohm group $= \frac{R}{n} = \frac{1}{3} Ω$
This is in series with $\frac{5}{3} Ω$ resistor.
$∴$ Total resistance $= \frac{5}{3} + \frac{1}{3} = \frac{6}{3} Ω = 2 Ω$

PHXII03:CURRENT ELECTRICITY

356951 A uniform wire of resistance $R$ and length $L$ is cut into four equal parts, each of length $L / 4$ , which are then connected in parallel combination. The effective resistance of the combination will be

1

R

2

4 R

3

\frac{R}{4}

4

\frac{R}{16}

Explanation:

Resistance, $R = \frac{ρ l}{A}$ or $R \propto l$
$∴ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{L}{L / 4} = 4$
$\Rightarrow R_{2} = \frac{R}{4} (∵ R_{1} = R)$
In parallel combination of such four
resistances,
$\frac{1}{R^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}}$
$or \frac{1}{R^{'}} = \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4}$
$\Rightarrow \frac{1}{R^{'}} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R}$
$∴ \frac{1}{R^{'}} = \frac{16}{R} \Rightarrow R^{'} = \frac{R}{16}$

PHXII03:CURRENT ELECTRICITY

356952 Effective resistance between $A$ and $B$ in the following circuit
supporting img

1

10 Ω

2

20 Ω

3

5 Ω

4

\frac{20}{3} Ω

Explanation:

The given circuit can be redrawn as:
This is balanced Wheatstone bridge.
$∴$ The two $10 Ω$ resistances at the middle are inactive
So, $R_{A B} = \frac{20 \times 20}{20 + 20} = 10 Ω$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

356953 The lowest resistance which can be obtained by connecting 10 resistors each of 1/10 $o h m$ is

1

1 / 250 Ω

2

1 / 200 Ω

3

1 / 100 Ω

4

1 / 10 Ω

Explanation:

Lowest resistance will be in the case when all the resistors are connected in parallel.
$R = \frac{1}{100} Ω$

PHXII03:CURRENT ELECTRICITY

356949 Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of thicker wire is $10 Ω$ .The total resistance of the combination will be

1

20 Ω

2

10 Ω

3

100 Ω

4

40 Ω

Explanation:

Length of each wire $= ℓ;$ Area of thick wire $(A_{1}) = 3 A;$ Area of thin wire $(A_{2}) = A$ and resistance of thick wire $(R_{1}) = 10 Ω$ Resistance
$(R) = ρ \frac{ℓ}{A}$
$\frac{R_{1}}{R_{2}} = \frac{A_{2}}{A_{1}} = \frac{A}{3 A} = \frac{1}{3}$
$\Rightarrow R_{2} = 3 R_{1} = 3 \times 10 = 30 Ω$
The equivalent resistance of these two resistors in series $= R_{1} + R_{2} = 30 + 10 = 40 Ω .$

PHXII03:CURRENT ELECTRICITY

356950 Three resistances of one ohm each are connected in parallel. Such connection is again connected with $\frac{5}{3} Ω$ resistor in series. The resultant resistance will be

1

2 Ω

2

5 Ω

3

8 Ω

4

10 Ω

Explanation:

Resistance of 1 ohm group $= \frac{R}{n} = \frac{1}{3} Ω$
This is in series with $\frac{5}{3} Ω$ resistor.
$∴$ Total resistance $= \frac{5}{3} + \frac{1}{3} = \frac{6}{3} Ω = 2 Ω$

PHXII03:CURRENT ELECTRICITY

356951 A uniform wire of resistance $R$ and length $L$ is cut into four equal parts, each of length $L / 4$ , which are then connected in parallel combination. The effective resistance of the combination will be

1

R

2

4 R

3

\frac{R}{4}

4

\frac{R}{16}

Explanation:

Resistance, $R = \frac{ρ l}{A}$ or $R \propto l$
$∴ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{L}{L / 4} = 4$
$\Rightarrow R_{2} = \frac{R}{4} (∵ R_{1} = R)$
In parallel combination of such four
resistances,
$\frac{1}{R^{'}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}}$
$or \frac{1}{R^{'}} = \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4} + \frac{1}{R / 4}$
$\Rightarrow \frac{1}{R^{'}} = \frac{4}{R} + \frac{4}{R} + \frac{4}{R} + \frac{4}{R}$
$∴ \frac{1}{R^{'}} = \frac{16}{R} \Rightarrow R^{'} = \frac{R}{16}$

PHXII03:CURRENT ELECTRICITY

356952 Effective resistance between $A$ and $B$ in the following circuit
supporting img

1

10 Ω

2

20 Ω

3

5 Ω

4

\frac{20}{3} Ω

Explanation:

The given circuit can be redrawn as:
This is balanced Wheatstone bridge.
$∴$ The two $10 Ω$ resistances at the middle are inactive
So, $R_{A B} = \frac{20 \times 20}{20 + 20} = 10 Ω$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

356953 The lowest resistance which can be obtained by connecting 10 resistors each of 1/10 $o h m$ is

1

1 / 250 Ω

2

1 / 200 Ω

3

1 / 100 Ω

4

1 / 10 Ω

Explanation:

Lowest resistance will be in the case when all the resistors are connected in parallel.
$R = \frac{1}{100} Ω$

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