Electricity Test

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Electricity Test - 59

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Question 1
1 / -0

A student's $$9.0V, 7.5W$$ portable radio was left on from $$9:00\ P.M.$$ until $$3:00 A.M.$$ How much charge passed through the wires?

A
$$18000C$$

B
$$24000C$$

C
$$6000C$$

D
$$12000C$$

Solution

As we know,
Power, $$P = V\cdot I$$
$$I = \dfrac {P}{V}$$
$$\Rightarrow I = \dfrac {Q}{t} = \dfrac {P}{V}$$
$$\Rightarrow Q = \dfrac {P}{V} \times t = \dfrac {7.5}{9}\times 3600\times 6$$
$$= 18000\ C$$
$$[\because t = 9\ PM$$ to $$3\ AM = 6\ hr = 6\times 3600s]$$
Question 2
1 / -0

The diagram shows identical lamps X and connected in series with a battery. The lamps light with normal brightness.If a third lamp Z is connected in parallel with lamp X, then what will happens to the brightness of the lamp Y?

A
Brighter than normal

B
Normal

C
Dimmer than normal

D
Very dim (cannot be seen)

Solution

The third lamp Z is connected parallel to X, so the current gets divided among X and Z. So both bulbs Xand Z light dimmer and the bulb Y lights brighter than earlier because current through it increases.
Question 3
1 / -0

Observe the given figure and answer the following questions. The bulb will glow when a ______ is placed in between the probes.

A
Crayon

B
Comb

C
Key

D
Piece of stone

Solution

Materials that allow electric current to pass through them are called conductors. Key is a metallic object. When it is placed in between probes then the current can easily flow in the circuit.
Question 4
1 / -0

Fuse is the most important safety device, used for protecting the circuits due to

A
Short circuiting and overloading

B
Overloading and earthing

C
Earthing only

D
Short circuiting, overloading and earthing.

Solution
Question 5
1 / -0

$$4$$ bulbs rated $$100$$W each, operate for $$6$$ hours per day. What is the cost of the energy consumed in $$30$$ days at the rate of Rs. $$5$$/kWh?

A
Rs. $$360$$

B
Rs. $$90$$

C
Rs. $$120$$

D
Rs. $$400$$

Solution

$$\displaystyle E= Power(in kW)\times time=\frac{4\times 100\times 6}{1000}$$kWh$$=2.4$$kWh
Consumed in $$30$$ days$$=30\times 2.4=72$$
Total cost $$=72\times 5=360$$Rs.
Question 6
1 / -0

Two electric bulbs have ratings respectively of $$25 \ W, 220 \ V$$ and $$100 \ W, 220 \ V$$. If the bulbs are connected in series with a supply of $$440\ V$$, which bulb will fuse?

A
$$25\ W$$ bulb

B
$$100\ W$$ bulb

C
Both of these

D
None of these

Solution

Power is given by $$ P = \dfrac{V^2}{R}$$, where $$P$$ is Power, $$V$$ is voltage across bulb and $$R$$ is resistance of bulb.
Hence, $$R = \dfrac{V^2}{P}$$

Resistance of $$25\ W\ bulb = \dfrac {220\times 220}{25} = 1936\ \Omega$$
Using Ohm's law, $$V = iR$$
Value of current that can pass through it $$i= \dfrac{V}{R} = \dfrac {220}{1936} = 0.11\ A$$

Resistance of $$100\ W\ bulb = \dfrac {220\times 220}{100} = 484\ \Omega$$
Value of current that can pass through it $$i = \dfrac{V}{R} = \dfrac {220}{484} = 0.45\ A$$

When connected in series to $$440 \ V$$ supply, the equivalent resistance is given by,
$$R_{equi} = R_1+R_2 = 1936+484=2420 \ \Omega$$
Calculating the value of the current $$i = \dfrac {440}{2420} = 0.18\ A$$

Thus $$0.18 \ A >0.11 \ A$$

The value of current flowing in the final circuit is greater than that allowed by the $$25W \ bulb$$, so it will fuse.
Question 7
1 / -0

The total electrical resistance between the points $$A$$ and $$B$$ of the circuit shown is:

A
$$9.23\Omega $$

B
$$15\Omega $$

C
$$30\Omega $$

D
$$100\Omega $$

Solution

$$\cfrac { 1 }{ Req. } =\cfrac { 1 }{ 30\Omega } +\cfrac { 1 }{ 20\Omega } +\cfrac { 1 }{ 40\Omega } $$
$$Req.=\cfrac { 120 }{ 13 } \Omega =9.23\Omega $$
Question 8
1 / -0

Two bulbs of $$250 \ V$$ and $$100 \ W$$ are first connected in series and then in parallel with a supply of $$250 \ V$$. Total power in each of the case will be respectively

A
$$100 \ W, 50 \ W$$

B
$$50\ W, 100 \ W$$

C
$$200 \ W, 150 \ W$$

D
$$50 \ W, 200 \ W$$
Question 9
1 / -0

A wire of resistance 'R' is cut into 'n' equal parts. These parts are then connected in parallel with each other. The equivalent resistance of the combination is?

A
nR

B
$$R/n$$

C
$$n/R^2$$

D
$$R/n^2$$

Solution
Question 10
1 / -0

Consider four circuits shown in the figure below. In which circuit power dissipated is highest?

A

B

C

D

Solution

We know that $$\rho =\dfrac { { \varepsilon }^{ 2 } }{ R } \\ \rho \propto \dfrac { 1 }{ R } (\varepsilon = constant)$$ (1)
so we calculate $${ R }_{ eq }$$ in all options as shown in fig
we know that $$\rho \propto \dfrac { 1 }{ R } $$
In (a) option resistance is minimum, so power dissipated will be highest