JEE Advanced PYQ Physics – Combination of Resistances Questions with Solutions

Q 1 :

In an aluminium (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are $2.7 \times 10^{- 8} Ω m$ and $1.0 \times 10^{- 7} Ω m$ , respectively. The electrical resistance between the two faces P and Q of the composite bar is [2015]

[IMAGE 794]

$\frac{2475}{64} μ Ω$
$\frac{1875}{64} μ Ω$
$\frac{1875}{49} μ Ω$
$\frac{2475}{132} μ Ω$

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(2)

As resistance of a wire, $R = \int ρ \frac{l}{A}$

For iron,

$R_{Fe} = \frac{ρ_{Fe} \times l_{Fe}}{A_{Fe}} = \frac{10^{- 7} \times 50 \times 10^{- 3}}{4 \times 10^{- 6}}$ $= \frac{25}{2} \times 10^{- 4} Ω$

For aluminium,

$R_{Al} = \frac{ρ_{Al} \times l_{Al}}{A_{Al}} = \frac{2.7 \times 10^{- 8} \times 50 \times 10^{- 3}}{(49 - 4) \times 10^{- 6}}$ $= \frac{2.7 \times 50}{45} \times 10^{- 5}$

$= 0.3 \times 10^{- 4} Ω$

As the potential difference across both resistors is the same, they are in parallel combination.

Therefore,

$R_{P Q} = \frac{R_{Fe} \times R_{Al}}{R_{Fe} + R_{Al}}$ $= \frac{12.5 \times 10^{- 4} \times 0.3 \times 10^{- 4}}{12.8 \times 10^{- 4}}$

$= \frac{1875}{64} μ Ω$

Q 2 :

Find out the value of current through $2 Ω$ resistance for the given circuit. [2005]

[IMAGE 795]

zero
2 A
5 A
4 A

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4

(1)

The current in $2 Ω$ resistor = 0 because it is not a part of any closed loop.

[IMAGE 796]

Q 3 :

Six identical resistors are connected as shown in the figure. The equivalent resistance will be: [2004]

[IMAGE 797]

Maximum between P and R
Maximum between Q and R
Maximum between P and Q
All are equal

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(3)

Given resistance of each resistor R

$R_{P Q} = \frac{5}{11} R, R_{Q R} = \frac{4}{11} R, R_{P R} = \frac{3}{11} R$

$∴$ $R_{P Q} is maximum.$

Q 4 :

The effective resistance between points P and Q of the electrical circuit shown in the figure is [2002]

[IMAGE 798]

$\frac{2 R r}{R + r}$
$\frac{8 R (R + r)}{3 R + r}$
$2 r + 4 R$
$\frac{5 R}{2} + 2 r$

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(1)

The circuit above and below the axis POQ is symmetrical and represents a balanced Wheatstone bridge. Hence the central resistance 2R is ineffective.

[IMAGE 799]

Therefore the equivalent circuit is redrawn as follows.

$∴$ $\frac{1}{R_{P Q}} = \frac{1}{4 R} + \frac{1}{4 R} + \frac{1}{2 r} = \frac{r + r + 2 R}{4 R r}$

$∴$ $R_{P Q} = \frac{2 R r}{R + r}$

Topic Question Set

Q 2 :

Find out the value of current through $2 Ω$ resistance for the given circuit. [2005]

[IMAGE 795]

Q 3 :

Six identical resistors are connected as shown in the figure. The equivalent resistance will be: [2004]

[IMAGE 797]

Q 4 :

The effective resistance between points P and Q of the electrical circuit shown in the figure is [2002]

[IMAGE 798]

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Topic Question Set

Q 2 : Find out the value of current through 2Ω resistance for the given circuit. [2005] [IMAGE 795]

Q 3 : Six identical resistors are connected as shown in the figure. The equivalent resistance will be: [2004] [IMAGE 797]

Q 4 : The effective resistance between points P and Q of the electrical circuit shown in the figure is [2002] [IMAGE 798]

Want Us to Email you About Special Offers & Updates?

Q 2 :

Find out the value of current through $2 Ω$ resistance for the given circuit. [2005]

[IMAGE 795]

Q 3 :

Six identical resistors are connected as shown in the figure. The equivalent resistance will be: [2004]

[IMAGE 797]

Q 4 :

The effective resistance between points P and Q of the electrical circuit shown in the figure is [2002]

[IMAGE 798]