Electricity Exemplar Problems Short Answer Questions Part 2 Class 10 Science

Electricity

Exemplar Problems

Short Answer Questions 2

Question 25. What is the commercial unit of electrical energy? Represent it in terms of joules.

Answer: The kilowatt hour is the commercial unit of electrical energy.

1 kilowatt = 1000 watt

⇒`1kW=(1000J)/text(second)`

Therefore,`1kWh=1000 J/text(second)xx60x60` second

⇒`1 kWh=3600000` joule

⇒`1 kWh=3.6 xx10^6` joule


Question 26. A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

Answer: Given, Electric current through the circuit = 1 A

Potential difference = 10V

Resistance of conductor = 5Ω

Resistance of electric lamp, R = ?

When resistance of 10Ω is connected with parallel with given series connection of electric lamp and conductor, then electric current flowing through 5Ω of the conductor =?

Since, conductor and electric lamp are connected in series, thus, total effective resistance of the circuit

`= 5Ω + R`

We know that, potential difference, V = IR

Therefore,`10V=1A xx(5Ω+R)`

⇒`10V=5ΩA+R A`

⇒`R=(10-5)Ω=5Ω`

Thus, resistance of electric `lA = 5Ω

Now, when a resistance of 10Ω is connected in parallel with the electric lamp and conductor connected in series.

Therefore, Total electric current through the conductors connected in parallel

= electric current through the circuit/2

Because resistance in parallel and total resistance in series are equal.

Thus, electric current through each of the resistance in parallel = 0.5A

Since, there is no division of electric current in series connection, thus electric current through the conductor of 5Ω = 0.5A

Thus, resistance of electric lamp = 5Ω

Electric current through the conductor of 5Ω = 0.5A


Question 27. Why is parallel arrangement used in domestic wiring?

Answer: Parallel arrangement used in domestic connection because:

Question 28. B1, B2 and B3 are three identical bulbs connected as shown in this figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

electricity exemplar problems question circuit diagram 10

(i) What happens to the glow of the other two bulbs when the bulb B1 gets fused?

Answer: Because there is no division of voltage in parallel circuit, thus glow of other two bulbs remain same.


(ii) What happens to the reading of A1, A2, A3 and A when the bulb B2 gets fused?

Answer: Since, the total electric current is equal to the sum of electric current through each of the electrical components.

As all the three bulbs are identical, thus electric current draws from each of the individual bulb will be same.

The electric current through the electric circuit is equal to 3A when all the three bubs are functioning.

Thus, after getting fuse of B2, the total electric current through the circuit will be equal to 2A.

Therefore,

(iii) How much power is dissipated in the circuit when all the three bulbs glow together?

Answer: Given, potential difference, V = 4.5V

Electric current, I = 3A

Thus, Resistance of the circuit, P =?

We know that, power, `P = VI`

Thus, `P = 4.5 xx 3A = 13.5` Watt

Thus, power dissipated in the circuit = 13.5 W



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