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Page No 216: - 12) ELECTRICITY


Question 1:

Judge the equivalent resistance when the following are connected in parallel − (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω.

Answer:

(a) When 1 Ω and 10Ω are connected in parallel:

Let R be the equivalent resistance.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1185/Chapter%2012_html_m43044efe.gif

Therefore, equivalent resistance https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1185/Chapter%2012_html_178eba96.gif  1 Ω

(b) When 1 Ω, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1185/Chapter%2012_html_m3f4d261e.gif , and https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1185/Chapter%2012_html_188f686b.gif are connected in parallel:

Let R be the equivalent resistance.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1185/Chapter%2012_html_meaebf27.gif

Therefore, equivalent resistance = 0.999 Ω

Question 2:

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer:

Resistance of electric lamp, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_m24896ba6.gif

Resistance of toaster, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_m678d60c9.gif

Resistance of water filter, https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_38d6e7f1.gif

Voltage of the source, V = 220 V

These are connected in parallel, as shown in the following figure.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_m3a360cf0.jpg

Let R be the equivalent resistance of the circuit.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_6d4b6c34.gif

According to Ohm’s law,

IR

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_28602a50.gif

Where,

Current flowing through the circuit = I

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_609240de.gif

7.04 A of current is drawn by all the three given appliances.

Therefore, current drawn by an electric iron connected to the same source of potential 220 V = 7.04 A

Let https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_349dbe4.gif be the resistance of the electric iron. According to Ohm’s law,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_m15b3c08f.gif

Therefore, the resistance of the electric iron ishttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1186/Chapter%2012_html_m3fff058c.gif  and the current flowing through it is 7.04 A.

Question 3:

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer:

There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.

The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

Question 4:

How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Answer:

There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.

(a) The following circuit diagram shows the connection of the three resistors.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1190/Chapter%2012_html_m335cb2b6.jpg

Here, 6 Ω and 3 Ω resistors are connected in parallel.

Therefore, their equivalent resistance will be given by

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1190/Chapter%2012_html_m4752f789.gif

This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.

Therefore, equivalent resistance of the circuit = 2 Ω + 2 Ω = https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1190/Chapter%2012_html_77914a3b.gif

Hence, the total resistance of the circuit is https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1190/Chapter%2012_html_77914a3b.gif .

2. The following circuit diagram shows the connection of the three resistors.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1190/Chapter%2012_html_m381c3a08.jpg

All the resistors are connected in series. Therefore, their equivalent resistance will be given as

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1190/Chapter%2012_html_585d77c4.gif

Therefore, the total resistance of the circuit ishttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1190/Chapter%2012_html_m6191a4c2.gif .

Question 5:

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer:

There are four coils of resistanceshttps://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1192/Chapter%2012_html_77914a3b.gif ,https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1192/Chapter%2012_html_m40378fe3.gif , 12 Ω, and 24 Ω respectively.

(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1192/Chapter%2012_html_bd825e9.gif

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by

https://img-nm.mnimgs.com/img/study_content/curr/1/10/10/156/1192/Chapter%2012_html_m5b73685b.gif

Therefore, 2 Ω is the lowest total resistance.