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]

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

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4

(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]

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.

Q 3 :

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

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]

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

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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.

Therefore the equivalent circuit is redrawn as follows.

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

$∴$ $R_{P Q} = \frac{2 R r}{R + r}$

Q 5 :

In the following circuit, the current through the resistor $R = 2 Ω$ is $I$ amperes. The value of $I$ is [2015]

(1)

In the fig. (a), no current flows through the $8 Ω$ resistance.

Again, the equivalent resistance of the balanced Wheatstone bridge. In Fig. (b), no current flows through the $10 Ω$ resistance.

$∴$ $R_{eq} = \frac{6 \times 18}{24} = \frac{9}{2} Ω$

Therefore, the current through the resistor $R = 2 Ω$ is

$I = \frac{V}{R} = \frac{6.5}{2 + \frac{9}{2}} = \frac{6.5}{6.5} = 1 A$

Q 6 :

For the circuit shown in the figure [2009]

the current $I$ through the battery is $7.5 mA$
the potential difference across $R_{L}$ is 18 V
ratio of powers dissipated in $R_{1}$ and $R_{2}$ is 3
if $R_{1}$ and $R_{2}$ are interchanged, magnitude of the power dissipated in $R_{L}$ will decrease by a factor of 9

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Select one or more options

(1, 4)

$R_{p} = \frac{R_{2} \times R_{L}}{R_{2} + R_{L}} = \frac{6 \times 1.5}{6 + 1.5} = \frac{9}{7.5} k Ω$

$∴$ $R_{total} = R_{1} + R_{p} = 2 + \frac{9}{7.5} = 3.2 k Ω$

$(1)$ $I = \frac{V}{R} = \frac{24}{3.2} mA = 7.5 mA$

$I = I_{R_{1}}$

$I_{R_{2}} = (\frac{R_{L}}{R_{L} + R_{2}}) I = (\frac{1.5}{1.5 + 6}) \times 7.5 = 1.5 mA$

$∴$ $I_{R_{L}} = I_{R_{1}} - I_{R_{2}} = 7.5 - 1.5 = 6 mA$

$(2) Potential difference across load$

$V_{R_{L}} = (I_{R_{L}}) R_{L} = 6 \times 1.5 = 9 V$

$(3) Ratio of powers dissipated in R_{1} and R_{2}$

$\frac{P R_{1}}{P R_{2}} = \frac{{(I_{R_{1}})}^{2} R_{1}}{{(I_{R_{2}})}^{2} R_{2}} = \frac{{(7.5)}^{2} \times 2}{{(1.5)}^{2} \times 6} = \frac{25}{3}$

$(4) When R_{1} and R_{2} are interchanged, then$

$R'_{p} = \frac{R_{2} R_{L}}{R_{2} + R_{L}} = \frac{2 \times 1.5}{2 + 1.5} = \frac{6}{7} k Ω$

$R'_{total} = 6 + R'_{p} = 6 + \frac{6}{7} k Ω$

$Now p.d. across R_{L}$

$V'_{R_{L}} = 24 (\frac{6 / 7}{6 + 6 / 7}) = 3 V$

$i.e., now potential becomes \frac{3}{9} = \frac{1}{3} rd .$

$Therefore power dissipated$ $P = \frac{V^{2}}{R} or P \propto V^{2}$ $will decrease by a factor of 9 .$

Topic Question Set

Q 2 :

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

Q 3 :

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

Q 4 :

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

Q 5 :

In the following circuit, the current through the resistor $R = 2 Ω$ is $I$ amperes. The value of $I$ is [2015]

Q 6 :

For the circuit shown in the figure [2009]

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Topic Question Set

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

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

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

Q 5 : In the following circuit, the current through the resistor R=2Ω is I amperes. The value of I is [2015]

Q 6 : For the circuit shown in the figure [2009]

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Q 2 :

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

Q 3 :

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

Q 4 :

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

Q 5 :

In the following circuit, the current through the resistor $R = 2 Ω$ is $I$ amperes. The value of $I$ is [2015]

Q 6 :

For the circuit shown in the figure [2009]