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1

AIEEE 2005

MCQ (Single Correct Answer)
An energy source will supply a constant current into the load if its internal resistance is
A
very large as compared to the load resistance
B
equal to the resistance of the load
C
non-zero but less than the resistance of the load
D
zero

Explanation

$$I = {E \over {R + r}},\,$$ Internal resistance $$\left( r \right)$$ is

zero, $$I = {E \over R} = $$ constant.
2

AIEEE 2005

MCQ (Single Correct Answer)
The resistance of hot tungsten filament is about $$10$$ times the cold resistance. What will be resistance of $$100$$ $$W$$ and $$200$$ $$V$$ lamp when not in use ?
A
$$20\Omega $$
B
$$40\Omega $$
C
$$200\Omega $$
D
$$400\Omega $$

Explanation

$$P = Vi = {{{V_2}} \over R}$$

$${R_{hot}} = {{{V^2}} \over P} = {{200 \times 200} \over {100}} = 400\Omega $$

$${R_{cold}} = {{400} \over {10}} = 40\Omega $$
3

AIEEE 2005

MCQ (Single Correct Answer)
Two voltmeters, one of copper and another of silver, are joined in parallel. When a total charge $$q$$ flows through the voltmeters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are $${Z_1}$$ and $${Z_2}$$ respectively the charge which flows through the silver voltmeter is
A
$${q \over {1 + {{{Z_2}} \over {{Z_1}}}}}$$
B
$${q \over {1 + {{{Z_1}} \over {{Z_2}}}}}$$
C
$$q{{{Z_2}} \over {{Z_1}}}$$
D
$$q{{{Z_1}} \over {{Z_2}}}$$

Explanation

Mass deposited

$$m = Zq \Rightarrow Z \propto {1 \over q} \Rightarrow {{{Z_1}} \over {{Z_2}}} = {{{q_2}} \over {{q_1}}}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Also $$q = {q_1} + {q_2}\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$ \Rightarrow {q \over {{q_2}}} = {{{q_1}} \over {{q_2}}} + 1\,\,\,\,\,\,\,\,\,\,$$ $$(\,Dividing\,\,\left( {ii} \right)$$ by $$\left. {{q_2}\,} \right)$$

$$ \Rightarrow {q_2} = {q \over {1 + {{{q_1}} \over {{q_2}}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$

From equations $$(i)$$ and $$(iii),$$ $${q_2} = {q \over {1 + {{{Z_2}} \over {{z_1}}}}}$$
4

AIEEE 2005

MCQ (Single Correct Answer)
A moving coil galvanometer has $$150$$ equal divisions. Its current sensitivity is $$10$$- divisions per milliampere and voltage sensitivity is $$2$$ divisions per millivolt. In order that each division reads $$1$$ volt, the resistance in $$ohms$$ needed to be connected in series with the coil will be -
A
$${10^5}$$
B
$${10^3}$$
C
$$9995$$
D
$$99995$$

Explanation

KEY CONCEPT : Resistance of Galvanometer,

$$G = {{Current\,\,\,sensitivity} \over {Voltage\,\,\,sensityvity}} \Rightarrow G = {{10} \over 2} = 5\Omega $$

Here $${i_g} = $$ Full scale deflection current $$ = {{150} \over {10}} = 15\,\,mA$$

$$V=$$ voltage to be measured $$=150$$ volts

(such that each division reads $$1$$ volt)

$$ \Rightarrow R = {{150} \over {15 \times {{10}^{ - 3}}}} - 5 = 9995\Omega $$

Questions Asked from Current Electricity

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