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1

AIEEE 2008

MCQ (Single Correct Answer)
Consider a block of conducting material of resistivity $$'\rho '$$ shown in the figure. Current $$'I'$$ enters at $$'A'$$ and leaves from $$'D'$$. We apply superposition principle to find voltage $$'\Delta V'$$ developed between $$'B'$$ and $$'C'$$. The calculation is done in the following steps:
(i) Take current $$'I'$$ entering from $$'A'$$ and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $$E(r)$$ at distance $$'r'$$ from A by using Ohm's law $$E = \rho j,$$ where $$j$$ is the current per unit area at $$'r'$$.
(iii) From the $$'r'$$ dependence of $$E(r)$$, obtain the potential $$V(r)$$ at $$r$$.
(iv) Repeat (i), (ii) and (iii) for current $$'I'$$ leaving $$'D'$$ and superpose results for $$'A'$$ and $$'D'.$$

$$\Delta V$$ measured between $$B$$ and $$C$$ is

A
$${{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$$
B
$${{\rho I} \over a} - {{\rho I} \over {\left( {a + b} \right)}}$$
C
$${{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$$
D
$${{\rho I} \over {2\pi \left( {a - b} \right)}}$$

Explanation

Let $$j$$ be the current density.

Then $$j \times 2\pi {r^2} = I \Rightarrow j = {I \over {2\pi {r^2}}}$$

$$\therefore$$ $$E = \rho j = {{\rho I} \over {2\pi {r^2}}}$$

Now, $$\Delta V{'_{BC}} = $$ $$ - \int\limits_{a + b}^a {\overrightarrow E .\,\overrightarrow {dr} } $$ $$ = - \int\limits_{a + b}^a {{{\rho I} \over {2\pi {r^2}}}} dr$$

$$ = - {{\rho I} \over {2\pi }}\left[ { - {1 \over r}} \right]_{a + b}^a$$

$$ = {{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$$

On applying superposition as mentioned we get

$$\Delta {V_{BC}} = 2 \times \Delta {V_{BC}} = {{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$$
2

AIEEE 2008

MCQ (Single Correct Answer)
Consider a block of conducting material of resistivity $$'\rho '$$ shown in the figure. Current $$'I'$$ enters at $$'A'$$ and leaves from $$'D'$$. We apply superposition principle to find voltage $$'\Delta V'$$ developed between $$'B'$$ and $$'C'$$. The calculation is done in the following steps:
(i) Take current $$'I'$$ entering from $$'A'$$ and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field $$E(r)$$ at distance $$'r'$$ from A by using Ohm's law $$E = \rho j,$$ where $$j$$ is the current per unit area at $$'r'$$.
(iii) From the $$'r'$$ dependence of $$E(r)$$, obtain the potential $$V(r)$$ at $$r$$.
(iv) Repeat (i), (ii) and (iii) for current $$'I'$$ leaving $$'D'$$ and superpose results for $$'A'$$ and $$'D'.$$

For current entering at $$A,$$ the electric field at a distance $$'r'$$ from $$A$$ is

A
$${{\rho I} \over {8\pi {r^2}}}$$
B
$${{\rho I} \over {{r^2}}}$$
C
$${{\rho I} \over {2\pi {r^2}}}$$
D
$${{\rho I} \over {4\pi {r^2}}}$$

Explanation

As shown above $$E = {{\rho I} \over {2\pi {r^2}}}$$
3

AIEEE 2008

MCQ (Single Correct Answer)
Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.

The value of the unknown resister $$R$$ is

A
$$13.75\Omega $$
B
$$220\Omega $$
C
$$110\Omega $$
D
$$55\Omega $$

Explanation

According to the condition of balancing

$${{55} \over {20}} = {R \over {80}} \Rightarrow R = 220\Omega $$
4

AIEEE 2007

MCQ (Single Correct Answer)
The resistance of a wire is $$5$$ ohm at $${50^ \circ }C$$ and $$6$$ ohm at $${100^ \circ }C.$$ The resistance of the wire at $${0^ \circ }C$$ will be
A
$$3$$ ohm
B
$$2$$ ohm
C
$$1$$ ohm
D
$$4$$ ohm

Explanation

KEY CONCEPT : We know that

$${R_t} = R{}_0\left( {1 + \alpha t} \right),$$

where $${R_t}$$ is the resistance of the wire at $${t^ \circ }C,$$

$${R_0}$$ is the resistance of the wire at $${0^ \circ }C$$

and $$\alpha $$ is the temperature coefficient of resistance

$$ \Rightarrow {R_{50}} = {R_0}\left( {1 + 50\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

$${R_{100}} = {R_0}\left( {1 + 100\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$\left( i \right),\,{R_{50}} - {R_0} = 50\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$

From $$\left( {ii} \right),{R_{100}} - {R_0} = 100\alpha {R_0}\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iv} \right)$$

Dividing $$(iii)$$ by $$(iv),$$ we get

$${{{R_{50}} - {R_0}} \over {{R_{100}} - R{}_0}} = {1 \over 2}$$

Here, $${{R_{50}}}$$ $$ = 5\Omega $$ and $${{R_{100}} = 6\Omega }$$

$$\therefore$$ $${{5 - {R_0}} \over {6 - {R_0}}} = {1 \over 2}$$

or, $$6 - {R_0} = 10 - 2{R_0}$$

or, $${R_0} = 4\Omega .$$

Questions Asked from Current Electricity

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