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PHXII03:CURRENT ELECTRICITY

357350 In Wheatstone's network $P = 2 Ω, Q = 2 Ω, R = 2 Ω a n d S = 3 Ω$ .The resistance with which $S$ is to shunted in order that the bridge may be balanced is

1

2 Ω

2

6 Ω

3

1 Ω

4

4 Ω

Explanation:

Let $X$ be the resistance with which $S$ is to be shunted for the bridge to be balanced.
supporting img
Then, for balanced Wheatstone's bridge
$\frac{P}{Q} = \frac{R}{\frac{S X}{S + X}}$ ( $∵$ $S$ and $X$ are in parallel)
Substituting the given values, we get
$\frac{2}{2} = \frac{2}{\frac{3 X}{3 + X}}$
or, $3 X = 6 + 2 X$
$∴ X = 6 Ω$

KCET - 2015

PHXII03:CURRENT ELECTRICITY

357351 Find the value of $R$ so that no deflection is noticed in the galvanometer when the switch $S$ is closed or open
supporting img

1

6 Ω

2

8 Ω

3

4 Ω

4 None of these

Explanation:

It is a Wheatstone bridge, therefore
$\frac{3}{3 + 3} = \frac{\frac{1 + 1}{8 R}}{8 + R} \Rightarrow R = 8 Ω$

PHXII03:CURRENT ELECTRICITY

357352 Three resistance $P$ , $Q$ , $R$ each of $2 Ω$ and an unknown resistance $S$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 Ω$ is connected in parallel to $S$ the bridge gets balanced. What is the value of $S$ ?

1

2 Ω

2

6 Ω

3

3 Ω

4

1 Ω

Explanation:

As resistance $S$ and $6 Ω$ are in parallel their effective resistance is $\frac{6 S}{6 + S} Ω$
As the bridge is balanced, hence it is balanced Wheatstone bridge. For balancing condition,
$\frac{P}{Q} = \frac{R}{(\frac{6 S}{6 + S})} \Rightarrow \frac{2}{2} = \frac{2 (6 + S)}{6 S} \Rightarrow S = 3 Ω$

PHXII03:CURRENT ELECTRICITY

357353 The value of unknown resistance $(x)$ for which the potential differnce between $B$ and $D$ will be zero in the arrangement shown, is
supporting img

1

42 Ω

2

9 Ω

3

6 Ω

4

3 Ω

Explanation:

The given circuit can be drawn as
supporting img

Given that, $V_{B} = V_{D}$ , thus the Wheatstone bridge is balanced.
$∴ \frac{P}{Q} = \frac{R}{S}$
$or \frac{12}{6 + x} = \frac{0.5}{0.5}$
$\Rightarrow 6 = 3 + 0.5 x \Rightarrow x = 6 Ω$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

357350 In Wheatstone's network $P = 2 Ω, Q = 2 Ω, R = 2 Ω a n d S = 3 Ω$ .The resistance with which $S$ is to shunted in order that the bridge may be balanced is

1

2 Ω

2

6 Ω

3

1 Ω

4

4 Ω

Explanation:

Let $X$ be the resistance with which $S$ is to be shunted for the bridge to be balanced.
supporting img
Then, for balanced Wheatstone's bridge
$\frac{P}{Q} = \frac{R}{\frac{S X}{S + X}}$ ( $∵$ $S$ and $X$ are in parallel)
Substituting the given values, we get
$\frac{2}{2} = \frac{2}{\frac{3 X}{3 + X}}$
or, $3 X = 6 + 2 X$
$∴ X = 6 Ω$

KCET - 2015

PHXII03:CURRENT ELECTRICITY

357351 Find the value of $R$ so that no deflection is noticed in the galvanometer when the switch $S$ is closed or open
supporting img

1

6 Ω

2

8 Ω

3

4 Ω

4 None of these

Explanation:

It is a Wheatstone bridge, therefore
$\frac{3}{3 + 3} = \frac{\frac{1 + 1}{8 R}}{8 + R} \Rightarrow R = 8 Ω$

PHXII03:CURRENT ELECTRICITY

357352 Three resistance $P$ , $Q$ , $R$ each of $2 Ω$ and an unknown resistance $S$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 Ω$ is connected in parallel to $S$ the bridge gets balanced. What is the value of $S$ ?

1

2 Ω

2

6 Ω

3

3 Ω

4

1 Ω

Explanation:

As resistance $S$ and $6 Ω$ are in parallel their effective resistance is $\frac{6 S}{6 + S} Ω$
As the bridge is balanced, hence it is balanced Wheatstone bridge. For balancing condition,
$\frac{P}{Q} = \frac{R}{(\frac{6 S}{6 + S})} \Rightarrow \frac{2}{2} = \frac{2 (6 + S)}{6 S} \Rightarrow S = 3 Ω$

PHXII03:CURRENT ELECTRICITY

357353 The value of unknown resistance $(x)$ for which the potential differnce between $B$ and $D$ will be zero in the arrangement shown, is
supporting img

1

42 Ω

2

9 Ω

3

6 Ω

4

3 Ω

Explanation:

The given circuit can be drawn as
supporting img

Given that, $V_{B} = V_{D}$ , thus the Wheatstone bridge is balanced.
$∴ \frac{P}{Q} = \frac{R}{S}$
$or \frac{12}{6 + x} = \frac{0.5}{0.5}$
$\Rightarrow 6 = 3 + 0.5 x \Rightarrow x = 6 Ω$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

357350 In Wheatstone's network $P = 2 Ω, Q = 2 Ω, R = 2 Ω a n d S = 3 Ω$ .The resistance with which $S$ is to shunted in order that the bridge may be balanced is

1

2 Ω

2

6 Ω

3

1 Ω

4

4 Ω

Explanation:

Let $X$ be the resistance with which $S$ is to be shunted for the bridge to be balanced.
supporting img
Then, for balanced Wheatstone's bridge
$\frac{P}{Q} = \frac{R}{\frac{S X}{S + X}}$ ( $∵$ $S$ and $X$ are in parallel)
Substituting the given values, we get
$\frac{2}{2} = \frac{2}{\frac{3 X}{3 + X}}$
or, $3 X = 6 + 2 X$
$∴ X = 6 Ω$

KCET - 2015

PHXII03:CURRENT ELECTRICITY

357351 Find the value of $R$ so that no deflection is noticed in the galvanometer when the switch $S$ is closed or open
supporting img

1

6 Ω

2

8 Ω

3

4 Ω

4 None of these

Explanation:

It is a Wheatstone bridge, therefore
$\frac{3}{3 + 3} = \frac{\frac{1 + 1}{8 R}}{8 + R} \Rightarrow R = 8 Ω$

PHXII03:CURRENT ELECTRICITY

357352 Three resistance $P$ , $Q$ , $R$ each of $2 Ω$ and an unknown resistance $S$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 Ω$ is connected in parallel to $S$ the bridge gets balanced. What is the value of $S$ ?

1

2 Ω

2

6 Ω

3

3 Ω

4

1 Ω

Explanation:

As resistance $S$ and $6 Ω$ are in parallel their effective resistance is $\frac{6 S}{6 + S} Ω$
As the bridge is balanced, hence it is balanced Wheatstone bridge. For balancing condition,
$\frac{P}{Q} = \frac{R}{(\frac{6 S}{6 + S})} \Rightarrow \frac{2}{2} = \frac{2 (6 + S)}{6 S} \Rightarrow S = 3 Ω$

PHXII03:CURRENT ELECTRICITY

357353 The value of unknown resistance $(x)$ for which the potential differnce between $B$ and $D$ will be zero in the arrangement shown, is
supporting img

1

42 Ω

2

9 Ω

3

6 Ω

4

3 Ω

Explanation:

The given circuit can be drawn as
supporting img

Given that, $V_{B} = V_{D}$ , thus the Wheatstone bridge is balanced.
$∴ \frac{P}{Q} = \frac{R}{S}$
$or \frac{12}{6 + x} = \frac{0.5}{0.5}$
$\Rightarrow 6 = 3 + 0.5 x \Rightarrow x = 6 Ω$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

357350 In Wheatstone's network $P = 2 Ω, Q = 2 Ω, R = 2 Ω a n d S = 3 Ω$ .The resistance with which $S$ is to shunted in order that the bridge may be balanced is

1

2 Ω

2

6 Ω

3

1 Ω

4

4 Ω

Explanation:

Let $X$ be the resistance with which $S$ is to be shunted for the bridge to be balanced.
supporting img
Then, for balanced Wheatstone's bridge
$\frac{P}{Q} = \frac{R}{\frac{S X}{S + X}}$ ( $∵$ $S$ and $X$ are in parallel)
Substituting the given values, we get
$\frac{2}{2} = \frac{2}{\frac{3 X}{3 + X}}$
or, $3 X = 6 + 2 X$
$∴ X = 6 Ω$

KCET - 2015

PHXII03:CURRENT ELECTRICITY

357351 Find the value of $R$ so that no deflection is noticed in the galvanometer when the switch $S$ is closed or open
supporting img

1

6 Ω

2

8 Ω

3

4 Ω

4 None of these

Explanation:

It is a Wheatstone bridge, therefore
$\frac{3}{3 + 3} = \frac{\frac{1 + 1}{8 R}}{8 + R} \Rightarrow R = 8 Ω$

PHXII03:CURRENT ELECTRICITY

357352 Three resistance $P$ , $Q$ , $R$ each of $2 Ω$ and an unknown resistance $S$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 Ω$ is connected in parallel to $S$ the bridge gets balanced. What is the value of $S$ ?

1

2 Ω

2

6 Ω

3

3 Ω

4

1 Ω

Explanation:

As resistance $S$ and $6 Ω$ are in parallel their effective resistance is $\frac{6 S}{6 + S} Ω$
As the bridge is balanced, hence it is balanced Wheatstone bridge. For balancing condition,
$\frac{P}{Q} = \frac{R}{(\frac{6 S}{6 + S})} \Rightarrow \frac{2}{2} = \frac{2 (6 + S)}{6 S} \Rightarrow S = 3 Ω$

PHXII03:CURRENT ELECTRICITY

357353 The value of unknown resistance $(x)$ for which the potential differnce between $B$ and $D$ will be zero in the arrangement shown, is
supporting img

1

42 Ω

2

9 Ω

3

6 Ω

4

3 Ω

Explanation:

The given circuit can be drawn as
supporting img

Given that, $V_{B} = V_{D}$ , thus the Wheatstone bridge is balanced.
$∴ \frac{P}{Q} = \frac{R}{S}$
$or \frac{12}{6 + x} = \frac{0.5}{0.5}$
$\Rightarrow 6 = 3 + 0.5 x \Rightarrow x = 6 Ω$

JEE - 2024