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PHXII03:CURRENT ELECTRICITY

356941 The equivalent resistance between the points $A$ and $B$ is
supporting img

1

\frac{5 R}{9}

2

\frac{2 R}{3}

3

R

4 None of these

Explanation:

Let $C$ is a point on the circuit. The points having identical potentials are marked with $A$ , $B$ & $C$
supporting img
The circuit can be redrawn as

$R_{A B} = \frac{5 R}{9} .$

PHXII03:CURRENT ELECTRICITY

356942 Each resistance shown in figure is $2 Ω$ . The equivalent resistance between $A$ and $B$ is
supporting img

1

2 Ω

2

4 Ω

3

8 Ω

4

1 Ω

Explanation:

supporting img

Resistance of upper $arm = 2 Ω + 2 Ω = 4 Ω)$ is in parallel with the diagonal resistance $(2 Ω)$ and the resistance of lower arm ( $= 2 Ω + 2 Ω = 4 Ω$ ).
$∴$ The equivalent resistance between $A$ and $B$ is
$\frac{1}{R_{e q}} = \frac{1}{4 Ω} + \frac{1}{2 Ω} + \frac{1}{4 Ω} \Rightarrow R_{e q} = \frac{4 Ω}{4} = 1 Ω$

PHXII03:CURRENT ELECTRICITY

356943 The equivalent resistance between $A$ and $B$ is
supporting img

1

25 Ω

2

18 Ω

3

27 Ω

4

19 Ω

Explanation:

The equivalent circuit is as shown below:
supporting img

$R_{e q} = 6 + 5 + 8 = 19 Ω$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

356944 The equivalent resistance between the points $A$ and $B$ is
supporting img

1

\frac{36}{7} Ω

2

10 Ω

3

\frac{85}{7} Ω

4

\frac{7}{85} Ω

Explanation:

Here, wheat stone bridge is in balanced condition so no current will flow from $10 Ω$ resistance and circuit can be reduce by
supporting img

PHXII03:CURRENT ELECTRICITY

356941 The equivalent resistance between the points $A$ and $B$ is
supporting img

1

\frac{5 R}{9}

2

\frac{2 R}{3}

3

R

4 None of these

Explanation:

Let $C$ is a point on the circuit. The points having identical potentials are marked with $A$ , $B$ & $C$
supporting img
The circuit can be redrawn as

$R_{A B} = \frac{5 R}{9} .$

PHXII03:CURRENT ELECTRICITY

356942 Each resistance shown in figure is $2 Ω$ . The equivalent resistance between $A$ and $B$ is
supporting img

1

2 Ω

2

4 Ω

3

8 Ω

4

1 Ω

Explanation:

supporting img

Resistance of upper $arm = 2 Ω + 2 Ω = 4 Ω)$ is in parallel with the diagonal resistance $(2 Ω)$ and the resistance of lower arm ( $= 2 Ω + 2 Ω = 4 Ω$ ).
$∴$ The equivalent resistance between $A$ and $B$ is
$\frac{1}{R_{e q}} = \frac{1}{4 Ω} + \frac{1}{2 Ω} + \frac{1}{4 Ω} \Rightarrow R_{e q} = \frac{4 Ω}{4} = 1 Ω$

PHXII03:CURRENT ELECTRICITY

356943 The equivalent resistance between $A$ and $B$ is
supporting img

1

25 Ω

2

18 Ω

3

27 Ω

4

19 Ω

Explanation:

The equivalent circuit is as shown below:
supporting img

$R_{e q} = 6 + 5 + 8 = 19 Ω$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

356944 The equivalent resistance between the points $A$ and $B$ is
supporting img

1

\frac{36}{7} Ω

2

10 Ω

3

\frac{85}{7} Ω

4

\frac{7}{85} Ω

Explanation:

Here, wheat stone bridge is in balanced condition so no current will flow from $10 Ω$ resistance and circuit can be reduce by
supporting img

PHXII03:CURRENT ELECTRICITY

356941 The equivalent resistance between the points $A$ and $B$ is
supporting img

1

\frac{5 R}{9}

2

\frac{2 R}{3}

3

R

4 None of these

Explanation:

Let $C$ is a point on the circuit. The points having identical potentials are marked with $A$ , $B$ & $C$
supporting img
The circuit can be redrawn as

$R_{A B} = \frac{5 R}{9} .$

PHXII03:CURRENT ELECTRICITY

356942 Each resistance shown in figure is $2 Ω$ . The equivalent resistance between $A$ and $B$ is
supporting img

1

2 Ω

2

4 Ω

3

8 Ω

4

1 Ω

Explanation:

supporting img

Resistance of upper $arm = 2 Ω + 2 Ω = 4 Ω)$ is in parallel with the diagonal resistance $(2 Ω)$ and the resistance of lower arm ( $= 2 Ω + 2 Ω = 4 Ω$ ).
$∴$ The equivalent resistance between $A$ and $B$ is
$\frac{1}{R_{e q}} = \frac{1}{4 Ω} + \frac{1}{2 Ω} + \frac{1}{4 Ω} \Rightarrow R_{e q} = \frac{4 Ω}{4} = 1 Ω$

PHXII03:CURRENT ELECTRICITY

356943 The equivalent resistance between $A$ and $B$ is
supporting img

1

25 Ω

2

18 Ω

3

27 Ω

4

19 Ω

Explanation:

The equivalent circuit is as shown below:
supporting img

$R_{e q} = 6 + 5 + 8 = 19 Ω$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

356944 The equivalent resistance between the points $A$ and $B$ is
supporting img

1

\frac{36}{7} Ω

2

10 Ω

3

\frac{85}{7} Ω

4

\frac{7}{85} Ω

Explanation:

Here, wheat stone bridge is in balanced condition so no current will flow from $10 Ω$ resistance and circuit can be reduce by
supporting img

PHXII03:CURRENT ELECTRICITY

356941 The equivalent resistance between the points $A$ and $B$ is
supporting img

1

\frac{5 R}{9}

2

\frac{2 R}{3}

3

R

4 None of these

Explanation:

Let $C$ is a point on the circuit. The points having identical potentials are marked with $A$ , $B$ & $C$
supporting img
The circuit can be redrawn as

$R_{A B} = \frac{5 R}{9} .$

PHXII03:CURRENT ELECTRICITY

356942 Each resistance shown in figure is $2 Ω$ . The equivalent resistance between $A$ and $B$ is
supporting img

1

2 Ω

2

4 Ω

3

8 Ω

4

1 Ω

Explanation:

supporting img

Resistance of upper $arm = 2 Ω + 2 Ω = 4 Ω)$ is in parallel with the diagonal resistance $(2 Ω)$ and the resistance of lower arm ( $= 2 Ω + 2 Ω = 4 Ω$ ).
$∴$ The equivalent resistance between $A$ and $B$ is
$\frac{1}{R_{e q}} = \frac{1}{4 Ω} + \frac{1}{2 Ω} + \frac{1}{4 Ω} \Rightarrow R_{e q} = \frac{4 Ω}{4} = 1 Ω$

PHXII03:CURRENT ELECTRICITY

356943 The equivalent resistance between $A$ and $B$ is
supporting img

1

25 Ω

2

18 Ω

3

27 Ω

4

19 Ω

Explanation:

The equivalent circuit is as shown below:
supporting img

$R_{e q} = 6 + 5 + 8 = 19 Ω$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

356944 The equivalent resistance between the points $A$ and $B$ is
supporting img

1

\frac{36}{7} Ω

2

10 Ω

3

\frac{85}{7} Ω

4

\frac{7}{85} Ω

Explanation:

Here, wheat stone bridge is in balanced condition so no current will flow from $10 Ω$ resistance and circuit can be reduce by
supporting img