Current Electricity Test

Result

Current Electricity Test - 36

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

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

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

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

$$ E_{1} = 1.08 V$$