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PHXII03:CURRENT ELECTRICITY

356975 The effective resistance between $p$ and $q$ in given figure is
supporting img

1

2 Ω

2

3 Ω

3

5 Ω

4

6 Ω

Explanation:

Figure is equivalent to the one shown below. It is a Wheatstone's bridge in which
supporting img
$P = \frac{6 \times 3}{6 + 3} = 2 Ω$
$Q = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 Ω$
$R = \frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2 Ω$
and $S = \frac{20 \times 5}{20 + 5} = 4 Ω$
We find that, $\frac{P}{Q} = \frac{2}{4} = \frac{R}{S}$
$i . e .$ the bridge is balanced and resistance of arm $C D$ is ineffective. Effective resistance between $p$ and $q$
$\begin{aligned} R_{A B} = \frac{(P + Q) (R + S)}{P + Q + R + S} \\ = \frac{(2 + 4) (2 + 4)}{2 + 4 + 2 + 4} = \frac{6 \times 6}{12} = 3 Ω \end{aligned}$
So, correct option is (2)

AIIMS - 2018

PHXII03:CURRENT ELECTRICITY

356976 The equivalent resistance between points $a$ and $b$ of a network shown in the figure is given by
supporting img

1

\frac{5}{4} R

2

\frac{4}{3} R

3

\frac{4}{5} R

4

\frac{3}{4} R

Explanation:

The circuit can be redrawn as
$R_{e q} = R + R / 3 = 4 R / 3$
supporting img

PHXII03:CURRENT ELECTRICITY

356977 A uniform wire of resistance $R$ stretched uniformly to $n$ times and then cut to from five wires of equal length. These wires are arranged as shown in the figure. The effective resistance between points $A$ and $B$ is
supporting img

1

R / (5 n^{2})

2

R / (5 n)

3

n^{2} R / 5

4

n R / 5

Explanation:

$R = \frac{ρ l_{1}}{A_{1}}$ [ $ρ =$ resistivity of the wire]
$l_{2} = n l_{1}, A_{1} l_{1} = A_{2} l_{2}$
$\Rightarrow A_{2} = A_{1} / n$
New resistance of the wire $= R^{'} = \frac{ρ l_{2}}{A_{2}} = n^{2} R$
Resistance of each wire is $n^{2} R / 5$
Since the wires are arranged to form balanced Wheatstone’s bridge
$R_{A B} = \frac{(n^{2} R / 5) + (n^{2} R / 5)}{2} = \frac{n^{2} R}{5}$

PHXII03:CURRENT ELECTRICITY

356978 In the given circuit all resistances are of value of $R$ ohm each. The equivalent resistance between $A$ and $B$ is:
supporting img

1

\frac{5 R}{2}

2

3 R

3

\frac{5 R}{3}

4

2 R

Explanation:

supporting img
If we combine or squeeze all the points $C$ then
the circuit will be reduced to

JEE - 2018

PHXII03:CURRENT ELECTRICITY

356975 The effective resistance between $p$ and $q$ in given figure is
supporting img

1

2 Ω

2

3 Ω

3

5 Ω

4

6 Ω

Explanation:

Figure is equivalent to the one shown below. It is a Wheatstone's bridge in which
supporting img
$P = \frac{6 \times 3}{6 + 3} = 2 Ω$
$Q = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 Ω$
$R = \frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2 Ω$
and $S = \frac{20 \times 5}{20 + 5} = 4 Ω$
We find that, $\frac{P}{Q} = \frac{2}{4} = \frac{R}{S}$
$i . e .$ the bridge is balanced and resistance of arm $C D$ is ineffective. Effective resistance between $p$ and $q$
$\begin{aligned} R_{A B} = \frac{(P + Q) (R + S)}{P + Q + R + S} \\ = \frac{(2 + 4) (2 + 4)}{2 + 4 + 2 + 4} = \frac{6 \times 6}{12} = 3 Ω \end{aligned}$
So, correct option is (2)

AIIMS - 2018

PHXII03:CURRENT ELECTRICITY

356976 The equivalent resistance between points $a$ and $b$ of a network shown in the figure is given by
supporting img

1

\frac{5}{4} R

2

\frac{4}{3} R

3

\frac{4}{5} R

4

\frac{3}{4} R

Explanation:

The circuit can be redrawn as
$R_{e q} = R + R / 3 = 4 R / 3$
supporting img

PHXII03:CURRENT ELECTRICITY

356977 A uniform wire of resistance $R$ stretched uniformly to $n$ times and then cut to from five wires of equal length. These wires are arranged as shown in the figure. The effective resistance between points $A$ and $B$ is
supporting img

1

R / (5 n^{2})

2

R / (5 n)

3

n^{2} R / 5

4

n R / 5

Explanation:

$R = \frac{ρ l_{1}}{A_{1}}$ [ $ρ =$ resistivity of the wire]
$l_{2} = n l_{1}, A_{1} l_{1} = A_{2} l_{2}$
$\Rightarrow A_{2} = A_{1} / n$
New resistance of the wire $= R^{'} = \frac{ρ l_{2}}{A_{2}} = n^{2} R$
Resistance of each wire is $n^{2} R / 5$
Since the wires are arranged to form balanced Wheatstone’s bridge
$R_{A B} = \frac{(n^{2} R / 5) + (n^{2} R / 5)}{2} = \frac{n^{2} R}{5}$

PHXII03:CURRENT ELECTRICITY

356978 In the given circuit all resistances are of value of $R$ ohm each. The equivalent resistance between $A$ and $B$ is:
supporting img

1

\frac{5 R}{2}

2

3 R

3

\frac{5 R}{3}

4

2 R

Explanation:

supporting img
If we combine or squeeze all the points $C$ then
the circuit will be reduced to

JEE - 2018

PHXII03:CURRENT ELECTRICITY

356975 The effective resistance between $p$ and $q$ in given figure is
supporting img

1

2 Ω

2

3 Ω

3

5 Ω

4

6 Ω

Explanation:

Figure is equivalent to the one shown below. It is a Wheatstone's bridge in which
supporting img
$P = \frac{6 \times 3}{6 + 3} = 2 Ω$
$Q = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 Ω$
$R = \frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2 Ω$
and $S = \frac{20 \times 5}{20 + 5} = 4 Ω$
We find that, $\frac{P}{Q} = \frac{2}{4} = \frac{R}{S}$
$i . e .$ the bridge is balanced and resistance of arm $C D$ is ineffective. Effective resistance between $p$ and $q$
$\begin{aligned} R_{A B} = \frac{(P + Q) (R + S)}{P + Q + R + S} \\ = \frac{(2 + 4) (2 + 4)}{2 + 4 + 2 + 4} = \frac{6 \times 6}{12} = 3 Ω \end{aligned}$
So, correct option is (2)

AIIMS - 2018

PHXII03:CURRENT ELECTRICITY

356976 The equivalent resistance between points $a$ and $b$ of a network shown in the figure is given by
supporting img

1

\frac{5}{4} R

2

\frac{4}{3} R

3

\frac{4}{5} R

4

\frac{3}{4} R

Explanation:

The circuit can be redrawn as
$R_{e q} = R + R / 3 = 4 R / 3$
supporting img

PHXII03:CURRENT ELECTRICITY

356977 A uniform wire of resistance $R$ stretched uniformly to $n$ times and then cut to from five wires of equal length. These wires are arranged as shown in the figure. The effective resistance between points $A$ and $B$ is
supporting img

1

R / (5 n^{2})

2

R / (5 n)

3

n^{2} R / 5

4

n R / 5

Explanation:

$R = \frac{ρ l_{1}}{A_{1}}$ [ $ρ =$ resistivity of the wire]
$l_{2} = n l_{1}, A_{1} l_{1} = A_{2} l_{2}$
$\Rightarrow A_{2} = A_{1} / n$
New resistance of the wire $= R^{'} = \frac{ρ l_{2}}{A_{2}} = n^{2} R$
Resistance of each wire is $n^{2} R / 5$
Since the wires are arranged to form balanced Wheatstone’s bridge
$R_{A B} = \frac{(n^{2} R / 5) + (n^{2} R / 5)}{2} = \frac{n^{2} R}{5}$

PHXII03:CURRENT ELECTRICITY

356978 In the given circuit all resistances are of value of $R$ ohm each. The equivalent resistance between $A$ and $B$ is:
supporting img

1

\frac{5 R}{2}

2

3 R

3

\frac{5 R}{3}

4

2 R

Explanation:

supporting img
If we combine or squeeze all the points $C$ then
the circuit will be reduced to

JEE - 2018

PHXII03:CURRENT ELECTRICITY

356975 The effective resistance between $p$ and $q$ in given figure is
supporting img

1

2 Ω

2

3 Ω

3

5 Ω

4

6 Ω

Explanation:

Figure is equivalent to the one shown below. It is a Wheatstone's bridge in which
supporting img
$P = \frac{6 \times 3}{6 + 3} = 2 Ω$
$Q = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4 Ω$
$R = \frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2 Ω$
and $S = \frac{20 \times 5}{20 + 5} = 4 Ω$
We find that, $\frac{P}{Q} = \frac{2}{4} = \frac{R}{S}$
$i . e .$ the bridge is balanced and resistance of arm $C D$ is ineffective. Effective resistance between $p$ and $q$
$\begin{aligned} R_{A B} = \frac{(P + Q) (R + S)}{P + Q + R + S} \\ = \frac{(2 + 4) (2 + 4)}{2 + 4 + 2 + 4} = \frac{6 \times 6}{12} = 3 Ω \end{aligned}$
So, correct option is (2)

AIIMS - 2018

PHXII03:CURRENT ELECTRICITY

356976 The equivalent resistance between points $a$ and $b$ of a network shown in the figure is given by
supporting img

1

\frac{5}{4} R

2

\frac{4}{3} R

3

\frac{4}{5} R

4

\frac{3}{4} R

Explanation:

The circuit can be redrawn as
$R_{e q} = R + R / 3 = 4 R / 3$
supporting img

PHXII03:CURRENT ELECTRICITY

356977 A uniform wire of resistance $R$ stretched uniformly to $n$ times and then cut to from five wires of equal length. These wires are arranged as shown in the figure. The effective resistance between points $A$ and $B$ is
supporting img

1

R / (5 n^{2})

2

R / (5 n)

3

n^{2} R / 5

4

n R / 5

Explanation:

$R = \frac{ρ l_{1}}{A_{1}}$ [ $ρ =$ resistivity of the wire]
$l_{2} = n l_{1}, A_{1} l_{1} = A_{2} l_{2}$
$\Rightarrow A_{2} = A_{1} / n$
New resistance of the wire $= R^{'} = \frac{ρ l_{2}}{A_{2}} = n^{2} R$
Resistance of each wire is $n^{2} R / 5$
Since the wires are arranged to form balanced Wheatstone’s bridge
$R_{A B} = \frac{(n^{2} R / 5) + (n^{2} R / 5)}{2} = \frac{n^{2} R}{5}$

PHXII03:CURRENT ELECTRICITY

356978 In the given circuit all resistances are of value of $R$ ohm each. The equivalent resistance between $A$ and $B$ is:
supporting img

1

\frac{5 R}{2}

2

3 R

3

\frac{5 R}{3}

4

2 R

Explanation:

supporting img
If we combine or squeeze all the points $C$ then
the circuit will be reduced to

JEE - 2018

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