Electric Resistance and Simple Circuits | Physics JEE previous year questions with Solutions

Q 1 :

The electric current through a wire varies with time as $I = I_{0} + β t$ where $I_{0} = 20 A$ and $β = 3 A / s .$ The amount of electric charge crossed through a section of the wire in 20 sec is [2024]

80 C
1000 C
800 C
1600 C

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(2)

Given that, current $I = I_{0} + β t$

Also, $I_{0} = 20 A$ and $β = 3 A / s$

Hence, $d q / d t = 20 + 3 t \Rightarrow d q = (20 + 3 t) d t$

$\int_{0}^{q} d q = \int_{0}^{20} (20 + 3 t) d t$

$\Rightarrow q = {[20 t + \frac{3 t^{2}}{2}]}_{0}^{20} = 1000 C$

Q 2 :

The current in a conductor is expressed as $I = 3 t^{2} + 4 t^{3}$ , where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t = 1 s$ to $t = 2 s$ is _________ C. [2024]

0%

(22) $I = 3 t^{2} + 4 t^{3}$

$\int d Q = \int I d t$

$Q = \int_{1}^{2} (3 t^{2} + 4 t^{3}) d t$

$= {[t^{3} + t^{4}]}_{1}^{2}$

$= (8 + 16) - (2) = 22 C$

Q 3 :

Which of the following resistivity ( $ρ$ ) v/s temperature (T) curve is most suitable to be used in wire bound standard resistors? [2025]

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(1)

For bound standard resistors the resistivity is independent of temperature and remain nearly constant.

Q 4 :

The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have [2025]

low thermal conductivity and low electrical conductivity
high thermal conductivity and high electrical conductivity
low thermal conductivity and high electrical conductivity
high thermal conductivity and low electrical conductivity

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(3)

Material should have low thermal conductivity and high electrical conductivity.

Q 5 :

Current passing through a wire as function of time is given as $I$ (t) = 0.02 t + 0.01 A. The charge that will flow through the wire from t = 1s to t = 2s is: [2025]

0.06 C
0.02 C
0.07 C
0.04 C

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(4)

$q = \int i d t = \int_{1}^{2} (0.02 t + 0.01) d t$

$q = {[0.02 \frac{t^{2}}{2} + 0.01 t]}_{1}^{2}$

$= \frac{0.02}{2} \times (2^{2} - 1^{2}) + 0.01 \times 1 = 0.04 C$

Q 6 :

A uniform metallic wire carries a current of 2 A, when a 3.4 V battery is connected across it. The mass of the uniform metallic wire is $8.92 \times 10^{- 3} kg$ , density is $8.92 \times 10^{3} {kg/m}^{3}$ and resistivity is $1.7 \times 10^{- 8} Ω -m$ . The length of the wire is [2023]

$l = 10 m$
$l = 100 m$
$l = 5 m$
$l = 6.8 m$

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(1)

$mass = 8.92 \times 10^{- 3} kg$

[IMAGE 174]--------------

$Density, d = 8.92 \times 10^{3} {kg/m}^{3}$

$ρ = 1.7 \times 10^{- 8} Ω - m$

$R = \frac{V}{i} = \frac{3.4}{2} = 1.7 Ω$

$R = \frac{ρ l}{A} = \frac{ρ l^{2}}{vol} = \frac{ρ l^{2} d}{mass} (∵ d = \frac{mass}{vol})$

$l = \sqrt{\frac{R m}{ρ d}} = \sqrt{\frac{1.7 \times 8.92 \times 10^{- 3}}{1.7 \times 10^{- 8} \times 8.92 \times 10^{3}}} = 10 m$

Q 7 :

The resistance of a wire is $5 Ω$ . Its new resistance in ohm, if stretched to 5 times of its original length, will be [2023]

125
5
25
625

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(1)

[IMAGE 175]

$∴$ $Volume of wire is constant in stretching$

$V_{i} = V_{f}$

$A_{i} l_{i} = A_{f} l_{f}$

$A l = A' (5 l)$

$A' = \frac{A}{5}$

$R_{f} = \frac{ρ l_{f}}{A_{f}} = \frac{ρ (5 l)}{(\frac{A}{5})} = 25 (\frac{ρ l}{A}) = 25 \times 5 = 125 Ω$

Q 8 :

The charge flowing in a conductor changes with time as $Q (t) = α t - β t^{2} + γ t^{3},$ where $α$ , $β$ and $γ$ are constants. Minimum value of current is [2023]

$α - \frac{3 β^{2}}{γ}$
$α - \frac{γ^{2}}{3 β}$
$β - \frac{α^{2}}{3 γ}$
$α - \frac{β^{2}}{3 γ}$

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(4)

$Q = (α t - β t^{2} + γ t^{3})$

$i = \frac{d Q}{d t} = (α - 2 β t + 3 γ t^{2})$

$\frac{d i}{d t} = (3 γ t - 2 β) = 0 \Rightarrow t = \frac{β}{3 γ}$

$i = (α - 2 β t + 3 γ t^{2}) = (α - \frac{β^{2}}{3 γ})$

Q 9 :

The drift velocity of electrons for a conductor connected in an electrical circuit is $V_{d}$ . The conductor is now replaced by another conductor with the same material and the same length but double the area of cross-section. The applied voltage remains the same. The new drift velocity of electrons will be [2023]

$V_{d}$
$\frac{V_{d}}{2}$
$\frac{V_{d}}{4}$
$2 V_{d}$

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(1)

$i = n A V_{d} e$

$i_{1} = (\frac{V}{R}) and i_{2} = (\frac{2 V}{R})$

$So, \frac{i_{1}}{i_{2}} = \frac{1}{2} = \frac{{(A V_{d})}_{1}}{{(A V_{d})}_{2}} = \frac{V_{d}}{{(V_{d})}_{2}} \times (\frac{1}{2})$

$\frac{1}{2} \times \frac{V_{d}}{{(V_{d})}_{2}} = \frac{1}{2} \Rightarrow {(V_{d})}_{2} = V_{d}$

Q 10 :

In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor [2023]

drift from higher potential to lower potential.
move in the curved paths from lower potential to higher potential.
move with the uniform velocity throughout from lower potential to higher potential.
move in the straight line paths in the same direction.

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(2)

$Move in curved path$

$I = n e A V_{d}$

Topic Question Set

Q 1 :

The electric current through a wire varies with time as $I = I_{0} + β t$ where $I_{0} = 20 A$ and $β = 3 A / s .$ The amount of electric charge crossed through a section of the wire in 20 sec is [2024]

Q 2 :

The current in a conductor is expressed as $I = 3 t^{2} + 4 t^{3}$ , where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t = 1 s$ to $t = 2 s$ is _________ C. [2024]

0%

Q 3 :

Which of the following resistivity ( $ρ$ ) v/s temperature (T) curve is most suitable to be used in wire bound standard resistors? [2025]

Q 4 :

The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have [2025]

Q 5 :

Current passing through a wire as function of time is given as $I$ (t) = 0.02 t + 0.01 A. The charge that will flow through the wire from t = 1s to t = 2s is: [2025]

Q 6 :

A uniform metallic wire carries a current of 2 A, when a 3.4 V battery is connected across it. The mass of the uniform metallic wire is $8.92 \times 10^{- 3} kg$ , density is $8.92 \times 10^{3} {kg/m}^{3}$ and resistivity is $1.7 \times 10^{- 8} Ω -m$ . The length of the wire is [2023]

Q 7 :

The resistance of a wire is $5 Ω$ . Its new resistance in ohm, if stretched to 5 times of its original length, will be [2023]

Q 8 :

The charge flowing in a conductor changes with time as $Q (t) = α t - β t^{2} + γ t^{3},$ where $α$ , $β$ and $γ$ are constants. Minimum value of current is [2023]

Q 10 :

In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor [2023]

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