Current Electricity Test

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Current Electricity Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

If four resistances are connected as shown in the fig. between A and B, the effective resistance is:

A
$3\Omega$

B
$1\Omega$

C
$2.4\Omega$

D
$2\Omega$

Solution

Simplified, or equivalent circuit diagram is shown in the figures.

From figure, effective resistance:

$R_{eq} = \dfrac{4 \times4}{2 \times 4}$ $= 2 \Omega$
Question 2
1 / -0

Given circuit has $n$ cells attached to it. All the cells are identical. When one cell is reversed, the current decreases to $0.7 A$ . The value of $n$ is

A
$12$

B
$16$

C
$20$

D
$8$

Solution

Case 1

All same polarity

$nE=I\left [ R+nr \right ]$ ----- 1 $nE=0.8(R+nr)$

Case 2

$(n-2)E=I\left [ R+nr \right ]$ ---- 2 $(n-2)E=0.7(R+nr)$

dividing 1 and 2

$\dfrac{0.8}{0.7}$

$\dfrac{n}{n-2}=\dfrac{8}{7}$

$7n=8n-16$

$n=16$ cells
Question 3
1 / -0

The resistance of a conductor is 1.08 $\ \Omega$ . To reduce it to 1 $\ \Omega$ , the resistance that must be connected is:

A
$0.08$ $\Omega$ in series

B
$13.5$ $\Omega$ is parallel

C
$0.08$ $\Omega$ in parallel

D
$13.5$ $\Omega$ in series.

Solution

To reduce the resistance value we should connect another resistance in parallel.
So, $R = \dfrac{R_{1}R_{2}}{R_{1}+R_{2}}$
or, $1 = \dfrac{1.08 \times R_{2}}{1.08 + R_{2}}$
or, $1.08 + R_{2} = 1.08 R_{2}$
or, $1.08 = 0.08 R_{2}$
$\therefore$ $R_{2} = \dfrac{108}{8} = 13.5\ \Omega$
Question 4
1 / -0

The least resistance that one can have from six resistors each of 0.1 ohm resistance is:

A
0.167 $\ \Omega$

B
0.00167 $\Omega$

C
1.67 $\Omega$

D
0.0167 $\Omega$

Solution

Least resistance is achieved when all the resistors are connected in parallel.
$\therefore R_{eq} = \dfrac{R}{6} = \dfrac{0.1}{6} = 0.0167\ \Omega$
Question 5
1 / -0

A battery has six cells in series and each cell has an electromagnetic force $1.5V$ and internal resistance $1 \Omega$ . If an external load of resistance $24$ $\Omega$ is connected to it. The potential drop across the load is

A
$7.2\ V$

B
$0.3\ V$

C
$6.8\ V$

D
$0.4\ V$

Solution

$\Rightarrow 6E=I\left[ 24+6r \right]$

$\dfrac { 6\times 1.5 }{ 30 }=I$

P.D across load $=$ I x 24

$=\dfrac { 3 }{ 10 } \times 24$

$=\dfrac { 3 \times 12 }{ 5 } =7.2V$
Question 6
1 / -0

The equivalent conductance of two wires of resistance 10 ohm and 5 ohm joined together in parallel is:

A
$\dfrac{50}{15}$ mho

B
$\dfrac{35}{50}$ mho

C
$0.3$ mho

D
$\dfrac{2}{3}$ mho

Solution

Effective resistance of the parallel combination:
$\dfrac{1}{R}=\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}$
So, conductance $= \dfrac{3}{10}=0.3$ mho
Question 7
1 / -0

The resistance between A and B is:

A
$9\Omega$

B
$2\Omega$

C
$12\Omega$

D
$8\Omega$

Solution

$$ Req \ of \ AC = \dfrac{6\times6}{12} = 3$$

$Req \ of \ AD = \dfrac{6\times6}{12} = 3$

$Req \ of \ AE = \dfrac{6\times6}{12} = 3$

$R_{AB} = \dfrac{6\times3}{9} = 2\Omega$
Question 8
1 / -0

Five cells of emf 1.5V and internal resistance 0.2 ohm are connected in series. The maximum current that can be delivered is

A
7.5A

B
1.5A

C
4A

D
2A

Solution

${ E }_{ eff }=1.5\times 5$

$=$ 7.5 V

${r }_{ eff }=0.2\times 5$

$=1\Omega$

for maximum current flow load resistance $=0\Omega$

$\Rightarrow { E }_{ eff }=I{ r }_{ eff }$

$\Rightarrow I=\dfrac { 7.5 }{ 1 } =7.5A$
Question 9
1 / -0

A potentiometer having a wire of $4$ $m$ length is connected to the terminals of a battery with a steady voltage. A leclanche cell has a null point at $1$ $m$ . If the length of the potentiometer wire is increased by $1$ $m$ , the position of the null point is

A
$1.5$ $m$

B
$1.25$ $m$

C
$10.05$ $m$

D
$1.31$ $m$

Solution

Let the resistance per unit length of potentiometer wire be $r$ and EMF of battery connected across it's length is $E$ .

the initially the resistance $=$ $4r$

let the balancing length initially, be ${ l }_{ 1 }=1m$

${ E }_{ 1 }={ I }_{ 1 }r\times { l }_{ 1 }$

where $\left[ I_{ 1 }(4r)=E \right]$

final resistance $=$ 5r

let the balancing length $={ l }_{2 }$

then ${ E }_{ 2 }={ I }_{ 2 }\times r\times { l }_{ 2 }$

where $\left[ I_{ 2 }(5r)=E \right]$

$\Rightarrow \dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { { I }_{ 1 }l_{ 1 } }{ { I }_{ 2 }l_{ 2 } }$

as $[E_{1} = E_{2}]$

${ \Rightarrow I }_{ 2 }{ l }_{ 2 }={ I }_{ 1 }{ l }_{ 1 }$

${ l }_{ 2 }=\times 1$

${ l }_{ 2 }= 5/4=1.25m$
Question 10
1 / -0

A potentiometer wire is $10m$ long and a potential difference of $6V$ is maintained between its ends. The emf of a cell which balances against a length of $180 cm$ of the potentiometer wire is:

A
$1.8 V$

B
$1.1 V$

C
$1.08 V$

D
$1.2 V$

Solution

formula $\dfrac{E}{E_{1}} = \dfrac{l}{l_{1}}$

E - potential difference across ends $= 6 V$

$E_{1}$ - emf of cell
$l$ - length of wire
$l_{1}$ - balancing length

$\dfrac{6}{E_{1}} = \dfrac{1000}{180}$

$E_{1} = \dfrac{108}{100}$

$E_{1} = 1.08 V$

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Current Electricity Test - 36

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Current Electricity Test - 36

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Question 1

3Ω3\Omega 3Ω

1Ω1\Omega 1Ω

2.4Ω2.4\Omega 2.4Ω

2Ω2\Omega 2Ω

Solution

Question 2

121212

161616

202020

888

Solution

Question 3

0.080.080.08 Ω\Omega Ω in series

13.513.513.5 Ω\Omega Ω is parallel

0.080.080.08 Ω\Omega Ω in parallel

13.513.513.5 Ω\Omega Ω in series.

Solution

Question 4

0.167 Ω\ \Omega Ω

0.00167 Ω\Omega Ω

1.67 Ω\Omega Ω

0.0167 Ω\Omega Ω

Solution

Question 5

7.2 V7.2\ V7.2 V

0.3 V0.3\ V0.3 V

6.8 V6.8\ V6.8 V

0.4 V0.4\ V0.4 V

Solution

Question 6

5015\dfrac{50}{15}1550​ mho

3550\dfrac{35}{50}5035​ mho

0.30.30.3 mho

23\dfrac{2}{3}32​ mho

Solution

Question 7

9Ω9\Omega 9Ω

2Ω2\Omega 2Ω

12Ω12\Omega 12Ω

8Ω8\Omega 8Ω

Solution

Question 8

7.5A

1.5A

4A

2A

Solution

Question 9

1.51.51.5 mmm

1.251.251.25 mmm

10.0510.0510.05 mmm

1.311.311.31 mmm

Solution

Question 10

1.8V1.8 V1.8V

1.1V1.1 V1.1V

1.08V1.08 V1.08V

1.2V1.2 V1.2V

Solution

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$3\Omega$

$1\Omega$

$2.4\Omega$

$2\Omega$

$12$

$16$

$20$

$8$

$0.08$ $\Omega$ in series

$13.5$ $\Omega$ is parallel

$0.08$ $\Omega$ in parallel

$13.5$ $\Omega$ in series.

0.167 $\ \Omega$

0.00167 $\Omega$

1.67 $\Omega$

0.0167 $\Omega$

$7.2\ V$

$0.3\ V$

$6.8\ V$

$0.4\ V$

$\dfrac{50}{15}$ mho

$\dfrac{35}{50}$ mho

$0.3$ mho

$\dfrac{2}{3}$ mho

$9\Omega$

$2\Omega$

$12\Omega$

$8\Omega$

$1.5$ $m$

$1.25$ $m$

$10.05$ $m$

$1.31$ $m$

$1.8 V$

$1.1 V$

$1.08 V$

$1.2 V$