NCERT Solution for Class 10 Science Chapter 11 Electricity विद्युत
Page No.172
Question 1:
What does an electric circuit mean?
विद्युत परिपथ का क्या अर्थ है?
Answer
A continuous closed path made of electric components through which an electric current flows is known as an electric circuit.
किसी विद्युत धारा के सतत् तथा बंद पथ को विद्युत परिपथ कहते हैं।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 2:
Define the unit of current.
विद्युत धारा के मात्रक की परिभाषा लिखिए।
Answer
The unit of electric current is the ampere.
One ampere is defined as the flow of one coulomb of electric charge passing through a conductor in one second. It is commonly abbreviated as A.
विद्युत धारा का SI मात्रक ऐम्पियर है।
1 ऐम्पियर विद्युत धारा की प्रवाहित मात्रा है, जो 1 कूलॉम आवेश के किसी चालक से 1 सेकंड में प्रवाहित होती है।
1 ऐम्पियर =1 कूलॉम/1सेकंड
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 3:
Calculate the number of electrons constituting one coulomb of charge.
एक कूलॉम आवेश की रचना करने वाले इलेक्ट्रॉनों की संख्या का परिकलन कीजिए।
Answer
Charge on one electron, e = 1.6 x 10-19 C
Total charge, Q = 1 C
Number of electrons, n = \(\frac{Q}{e} = \frac{1C}{1.6 \times 10^{-19}} = 6.25 \times 10^8\)
ज्ञात है, 1.6×10-19 कूलॉम आवेश = 1 इलेक्ट्रॉन
अतः , 1 कूलॉम आवेश = \(\frac{1C}{1.6 \times 10^{-19}} = 6.25 \times 10^8\) इलेक्ट्रॉन
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Page No. 174
Question 1:
Name a device that helps to maintain a potential difference across a conductor.
उस युक्ति का नाम लिखिए, जो किसी चालक के सिरों पर विभवांतर बनाए रखने में सहायता करती है।
Answer
A battery.
सेल अथवा बैटरी।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 2:
What is meant by saying that the potential difference between two points is 1 V?
यह कहने का क्या तात्पर्य है कि दो बिंदुओं के बीच विभवांतर 1 वोल्ट है?
Answer
It means that 1 J of work is done in moving 1 C charge from one point to the other.
1 वोल्ट विभावांतर का अर्थ है कि एक बिंदु से दूसरे बिंदु तक 1 कूलॉम आवेश को ले जाने में 1 जूल कार्य किया जाता है।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 3:
How much energy is given to each coulomb of charge passing through a 6 V battery?
6 वोल्ट बैटरी से गुजरने वाले हर एक कूलॉम आवेश को कितनी ऊर्जा दी जाती है?
Answer
Energy given by battery = charge x potential difference
or W = QV = 1C X 6V = 6J.
विभवांतर, V= 6 वोल्ट; आवेश Q = 1 कूलॉम ऊर्जा
अथवा, कार्य W = VQ= 6×1 = 6 जूल;
अतः V= 6 वोल्ट बैटरी से गुजरने वाले हर एक कूलॉम आवेश को 6 जूल ऊर्जा दी जाती है।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Page No. 181
Question 1:
On what factors does the resistance of a conductor depend?
किसी चालक का प्रतिरोध किन कारकों पर निर्भर करता है?
Answer
The resistance of a conductor depends
(i) on its length (ii) on its area of cross-section and (iii) on the nature of its material.
किसी चालक का प्रतिरोध (R) निम्न कारकों पर निर्भर करता है।
(i) चालक की लम्बाई (l)
(ii) अनुप्रस्थ-काट का क्षेत्रफल (A)
(iii) चालक के पदार्थ की प्रकृति
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 2:
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
समान पदार्थ के दो तारों में यदि एक पतला तथा दूसरा मोटा हो, तो इनमें से किसमें विद्युत धारा आसानी से प्रवाहित होगी, जबकि उन्हें समान विद्युत स्रोत से संयोजित किया जाता है, क्यों?
Answer
The relation between resistance and the area of cross section can be given as:
R \(\propto \frac{1}{A}\).
Resistance is inversely proportional to the area of cross-section of the wire.
As the resistance decreases, the current increases.
Thicker the wire, less current will pass through it whereas thinner the wire, more current will pass.
किसी तार का प्रतिरोध (R) उसके अनुप्रस्थ-काट के क्षेत्रफल (A) के व्युत्क्रमानुपाती होता है। जो तार मोटा है, उसका अनुप्रस्थ-काट का क्षेत्रफल A अधिक है, अतः प्रतिरोध कम है। यही कारण है कि मोटे तार में से विद्युत धारा अधिक सरलता से प्रवाहित होगी।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 3:
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
मान लीजिए किसी वैद्युत अवयव के दो सिरों के बीच विभवांतर को उसके पूर्व के विभवांतर की तुलना में घटा कर आधा कर देने पर भी उसका प्रतिरोध नियत रहता है। तब उस अवयव से प्रवाहित होने वाली विद्युत धारा में क्या परिवर्तन होगा?
Answer
According to Ohm’s law
V = IR
⇒ I = \(\frac{V}{R}\) … (1)
Now Potential difference is decreased to half
∴ New potential difference Vʹ=\(\frac{V}{2}\)
Resistance remains constant
So the new current Iʹ = \(\frac{V’}{R}\)
= \(\frac{\frac{V}{2}}{R}\) (V/2)/R
= \(\frac{1}{2} \frac{V}{R}\)
= \(\frac{1}{2}\) I = \(\frac{l}{2}\)
Therefore, the amount of current flowing through the electrical component is reduced by half.
प्रथम दशा में, यदि दिए गए विद्युत अवयव का प्रतिरोध = R ओम तथा उसके दो सिरों के बीच विभवांतर = V वोल्ट है।
तब विद्युत धारा = ऐम्पियर होगी।
दूसरी दशा में,
प्रतिरोध = R ओम; V’ = \(\frac{V}{2}\) वोल्ट
विद्युत धारा , I‘ = \(\frac{V}{2} \times \frac{1}{2} = \frac{V}{2R}\)
= \(\frac{\frac{V}{2}}{R}\)
= \(\frac{1}{2} \frac{V}{R}\)
= \(\frac{1}{2}\) I = \(\frac{l}{2}\)
र्थात् विभवांतर आधा करने पर विद्युत धारा भी पहले की अपेक्षा आधी हो जाएगी।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 4:
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
विद्युत टोस्टरों तथा विद्युत इस्तरियों के तापन अवयव शुद्ध धातु के न बना कर किसी मिश्रधातु के क्यों बनाए जाते हैं?
Answer
The coils of electric toasters, electric irons and other heating devices are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of a pure metal.
मिश्रधातुओं की प्रतिरोधकता अपनी अवयवी धातुओं की अपेक्षा अधिक होती है। | इसीलिए मिश्रधातुओं का उच्च ताप पर उपचयन (दहन) नहीं होता। अतः इनका उपयोग विद्युत टोस्टरों, इस्तरियों आदि के तापन अवयव के अवयव बनाने हेतु किया जाता है।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 5:
Use the data in Table 11.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
निम्न प्रश्नों के हल सारणी में दिए हुए आँकड़ों के आधार पर कीजिए।
(a) आयरन (Fe) तथा मरकरी (Hg) में कौन अच्छा विद्युत चालक है?
(b) कौन-सा पदार्थ सर्वश्रेष्ठ चालक है?
Answer
(i) Resistivity of iron = 10.0 x 10-8 Ω m
Resistivity of mercury = 94.0 x 10-8 Ω m.
Thus iron is a better conductor because it has lower resistivity than mercury.
(ii) Because silver has the lowest resistivity (= 1.60 x 10-8 Ω m), therefore, silver is the best conductor.
(a) आयरन (Fe) की विद्युत प्रतिरोधकता = 10.0×10-18 ओम-मी
मर्करी (Hg) की विद्युत प्रतिरोधकता = 94.0×10-8 ओम-मी
आयरन मर्करी की अपेक्षा अच्छा विद्युत चालक है क्योंकि इसकी प्रतिरोधकता अपेक्षाकृत कम है।
(b) सिल्वर (प्रतिरोधकता = 1.60×10-8ओम-मी) सर्वश्रेष्ठ चालक है, क्योंकि इसकी प्रतिरोधकता सबसे कम है।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Page No. 185
Question 1:
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 W resistor, an 8 W resistor, and a 12 W resistor, and a plug key, all connected in series.
किसी विद्युत परिपथ का व्यवस्था आरेख खींचिए, जिसमें 2 वोल्ट के तीन सेलों की बैटरी, जो 52 प्रतिरोध, 8 ओम प्रतिरोध व ओम 12 ओम प्रतिरोध तथा एक प्लग कुँजी सभी श्रेणीक्रम में संयोजित हों।
Answer
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 2:
Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 W resistor. What would be the readings in the ammeter and the voltmeter?
प्रश्न 1 का परिपथ आरेख दोबारा खींचिए तथा इसमें प्रतिरोधकों से प्रवाहित विद्युत धारा को मापने के लिए अमीटर तथा 12 ओम के प्रतिरोधक के सिरों के बीच विभवांतर मापने के लिए वोल्टमीटर लगाइए। अमीटर तथा वोल्टमीटर के क्या पाठ्यांक होंगे?
Answer
Potential difference, V = 6 V
Current flowing through the circuit/resistors = I
Resistance of the circuit, R = 5 + 8 + 12 = 25Ω
I = \(\frac{V}{R}\) = \(\frac{6}{25}\) = 0.24 A
Potential difference across 12 Ω resistor = V1
Current flowing through the 12 Ω resistor, I = 0.24 A
Therefore, using Ohm’s law, we obtain
V1 = IR = 0.24 x 12 = 2.88 V
Therefore, the reading of the ammeter will be 0.24 A.
The reading of the voltmeter will be 2.88 V.
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Page No. 188
Question 1:
Judge the equivalent resistance when the following are connected in parallel –
(a) 1 W and 106 W, (b) 1 W and 103 W, and 106 W.
जब (a) 1 ओम तथा 106 ओम (b) 1 ओम, 103 ओम तथा 106 ओम के प्रतिरोध समांतर क्रम में संयोजित किए जाते हैं, तो इनके तुल्य प्रतिरोध के संबंध में आप क्या निर्णय करेंगे?
Answer
(a) When 1 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.
Therefore, equivalent resistance ≈ 1 Ω
(b) When 1Ω, 103 Ω and 106 Ω are connected in parallel:
Let R be the equivalent resistance.
Therefore, equivalent resistance = 0.999 Ω
जब विभिन्न प्रतिरोधों को समांतर क्रम में संयोजित किया जाता है, तब तुल्य प्रतिरोध सबसे कम प्रतिरोध से भी कम होता है, अतः
(a) तुल्य प्रतिरोध <1 ओम
(b) तुल्य प्रतिरोध >1 ओम
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 2:
An electric lamp of 100 W, a toaster of resistance 50 W, and a water filter of resistance 500 W 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?
100 ओम का एक विद्युत लैम्प, 50 ओम का एक विद्युत टोस्टर तथा 500 ओम का एक जल फिल्टर 220 वोल्ट के विद्युत स्रोत से समांतर क्रम में संयोजित है। उस विद्युत इस्तरी का प्रतिरोध क्या हैं जिसे यदि समान स्रोत के साथ संयोजित कर दें, तो वह भी उतनी ही विद्युत धारा लेती हैं जितनी तीनों युक्तियाँ लेती हैं। यह भी ज्ञात कीजिए कि इस विद्युत इस्तरी से कितनी विद्युत धारा प्रवाहित होती है?
Answer
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 3:
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
श्रेणीक्रम में संयोजित करने के स्थान पर वैद्युत युक्तियों को समांतर क्रम में संयोजित करने के क्या लाभ हैं?
Answer
वैद्युत युक्तियों को समांतर क्रम में संयोजित करने के निम्न लाभ हैं।
(i) समांतर क्रम में प्रत्येक युक्ति में पूर्ण विद्युत विभव प्राप्त होता है, जबकि धारा विभक्त हो जाती है। प्रत्येक युक्ति में धारा उसके प्रतिरोध के अनुसार जाती है।
(ii) यदि एक युक्ति को ऑन/ऑफ करते हैं, तो अन्य युक्तियाँ अपना कार्य सुचारू रूप से करती रहती हैं।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 4:
How can three resistors of resistances 2 Ω, 3 Ω, and Ω W be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
2 ओम, 3 ओम तथा 6 ओम के तीन प्रतिरोधों को किस प्रकार संयोजित करेंगे कि संयोजन का कुल प्रतिरोध (a) 4 ओम (b) 1 ओम हो?
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
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?
4 ओम, 8 ओम, 12 ओम तथा 24 ओम प्रतिरोध की चार कुंडलियों को किस प्रकार संयोजित करें कि संयोजन से (a) अधिकतम (b) निम्नतम प्रतिरोध प्राप्त हो सके?
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Page No. 190
Question 1:
Why does the cord of an electric heater not glow while the heating element does?
किसी विद्युत हीटर की डोरी क्यों उत्तप्त नहीं होती, जबकि उसका तापन अवयव उत्तप्त हो जाता है?
Answer
The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminium is very law so it does not glow.
तापन अवयव तथा डोरी दोनों में ही समान विद्युत धारा प्रवाहित होती है, मगर डोरी का बाह्य भाग एक कुचालक पदार्थ से बना होता है, जिसकी प्रतिरोधकता बहुत अधिक होती है तथा इसी कारण प्रतिरोध भी। हीटर का तापन अवयव सुचालक धातु (मिश्रधातु) से निर्मित अर्थात् कम प्रतिरोधकता (तथा प्रतिरोध) वाला होता है। इसलिए यह उत्तप्त हो जाता है मगर डोरी उत्तप्त नहीं होती।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 2:
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
एक घंटे में 50 वोल्ट विभवांतर से 96000 कूलॉम आवेश को स्थानान्तरित करने में उत्पन्न ऊष्मा परिकलित कीजिए।
Answer
V = 50 वोल्ट, t = 1 घंटा, Q = 96000 कूलॉम
H = VIt (परन्तु I =Q/t)
अतः
H = V×Q = 50×96000 = 4800000 जूल = 4800 किलोजूल
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 3:
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
20 ओम प्रतिरोध की कोई विद्युत इस्तरी 5 ऐम्पियर विद्युत धारा लेती है। 30 सेकंड में उत्पन्न ऊष्मा को परिकलित कीजिए।
Answer
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current x Resistance = 5 x 20 = 100 V
H = 100 x 5 x 30 = 15000 = 15 K J.
R = 20 ओम, I = 5 ऐम्पियर; t = 30 सेकंड
उत्पन्न ऊष्मा (H) = I2Rt = 5×5×20×30 = 15000 जूल = 15 किलोजूल
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Page No. 192
Question 1:
What determines the rate at which energy is delivered by a current?
विद्युत धारा द्वारा प्रदत्त ऊर्जा की दर का निर्धारण कैसे किया जाता है?
Answer
The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.
विद्युत धारा द्वारा प्रदत्त ऊर्जा की दर को विद्युत शक्ति (P) भी कहते हैं।
विद्युत शक्ति, P = I2R = VI = V2/R = H/t
जहाँ I विद्युत धारा, R प्रतिरोध, V विभवांतर, H ऊष्मा तथा t समय है।
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 2:
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
कोई विद्युत मोटर 220 वोल्ट के विद्युत स्रोत से 5.0 ऐम्पियर विद्युत धारा लेता है। मोटर की शक्ति निर्धारित कीजिए तथा 2 घंटे में मोटर द्वारा उपमुक्त ऊर्जा परिकलित कीजिए।
Answer
Voltage,V = 220 V
Current, I = 5 A
P = VI = 220 x 5 = 1100 W
Energy consumed by the motor = Pt
Where,
Time, t = 2 h = 2 x 60 x 60 = 7200 s
∴ P = 1100 x 7200 = 7.92 x 106 J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 x 106 J
V = 220 वोल्ट, I = 5.0 ऐम्पियर
t = 2 घंटे = 2×60×60 = 7200 सेकंड
(i) मोटर की शक्ति P = V×I = 220×5.0 = 1100 वाट = 1.1 किलोवाट
(ii) मोटर द्वारा उपमुक्त ऊर्जा, H = P×t
= 1100×7200 = 7920000 जूल
= 7.92×106 जूल = 7.9×103 किलोजूल
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Page No. 193
Exercise
Question 1:
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R¢, then the ratio R/R’ is –
प्रतिरोध R के किसी तार के टुकड़े को पाँच बराबर भागों में काटा जाता है। इन टुकड़ों को फिर समांतरक्रम में संयोजित कर देते हैं। यदि संयोजन का तुल्य प्रतिरोध R’ है, तो R/R’ अनुपात का मान क्या है?
(a) 1/25 (b) 1/5 (c) 5 (d) 25
Answer
(d) 25
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 2:
Which of the following terms does not represent electrical power in a circuit?
निम्नलिखित में से कौन-सा पद विद्युत परिपथ में विद्युत शक्ति को निरूपित नहीं। करता?
(a) I2R (b) IR2 (c) VI (d) V2/R
Answer
(b) IR2
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 3:
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
किसी विद्युत बल्ब का अनुमतांक 220 वोल्ट 100 वाट है। जब इसे 110 वोल्ट पर प्रचालित करते हैं, तब इसके द्वारा उपमुक्त शक्ति कितनी होती है?
(a) 100 W (b) 75 W (c) 50 W (d) 25 W
Answer
(d) 25 W
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 4:
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –
दो चालक तार जिनके पदार्थ, लंबाई तथा व्यास समान हैं, किसी विद्युत परिपथ में पहले श्रेणीक्रम में और फिर समांतर क्रम में संयोजित किए जाते हैं। श्रेणीक्रम तथा समांतर क्रम संयोजन में उत्पन्न ऊष्माओं का अनुपात क्या होगा?
(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
Answer
(c) 1:4
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 5:
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer
To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
Question 6:
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 W m. What will be the length of this wire to make its resistance 10 W? How much does the resistance change if the diameter is doubled?
Answer
Area of cross-section of the wire, A =π (d/2)2
Diameter= 0.5 mm = 0.0005 m
Resistance, R = 10 Ω
We know that
Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω.
Question 7:
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –
| V (volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |
| I (amperes ) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
Plot a graph between V and I and calculate the resistance of that resistor.
Answer
The slope of the line gives the value of resistance (R) as,
Slope = 1/R = BC/AC = 2/6.8
R = 6.8/2 = 3.4 ΩTherefore, the resistance of the resistor is 3.4 Ω.
Question 8:
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer
Resistance (R) of a resistor is given by Ohm’s law as,V= IR
R= \(\frac{V}{I}\)
Where,
Potential difference, V= 12 V
Current in the circuit, I= 2.5 mA = 2.5 x 10-3 A
Therefore, the resistance of the resistor is 4.8 kΩ
Question 9:
A battery of 9 V is connected in series with resistors of 0.2 W, 0.3 W, 0.4 W , 0.5 W and 12 W, respectively. How much current would flow through the 12 W resistor?
Answer
There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as
V = IR
I = \(frac{V}{R}\)
Where,
R is the equivalent resistance of resistances 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω.
These are connected in series.
Hence, the sum of the resistances will give the value of R.
R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Potential difference, V= 9 V
I = \(frac{9}{13.4}\) = 0.671 A
Therefore, the current that would flow through the 12 Ω resistor is 0.671 A.
Question 10:
How many 176 W resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer
For x number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law asV= IR
R= V/I
Where,
Supply voltage, V= 220 V
Current, I = 5 A
Equivalent resistance of the combination = R,
\(\frac{1}{R} = x \times \frac{1}{176}\)
R = \(\frac{176}{x}\)
From Ohms’ law,
\(\frac{V}{I}\) = \(\frac{176}{x}\)
x = \(\frac{176 \times I}{V}\) = \(\frac{176 \times 5}{220}\) = 4
Therefore, four resistors of 176 Ω are required to draw the given amount of current.
Question 11:
Show how you would connect three resistors, each of resistance 6 W, so that the combination has a resistance of (i) 9 W, (ii) 4 W.
Answer
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be 6/2 = 3 Ω is also not desired. Hence, we should either connect the two resistors in series or parallel.
(a) Two resistor in parallel
Two 6 Ω resistors are connected in parallel. Their equivalent resistance will be
\(\frac{1}{\frac{1}{6} + \frac{1}{6}}\) = 3 Ω
The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω+ 3 Ω = 9 Ω.
(b) Two resistor in series
Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω.The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be
\(\frac{1}{\frac{1}{12} + \frac{1}{6}}\) = 4 Ω
Therefore, the total resistance is 4 Ω.
Question 12:
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer
Resistance R1 of the bulb is given by the expression,
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of an electric bulb P=10watts
Because R =\(\frac{V^2}{P}\)
R1 = \(\frac{220^2}{10}\) = 4840 Ω
According to Ohms’ law,
V= IR
Let R is the total resistance of the circuit for x number of electric bulbs
R = \(\frac{V}{I}\) = \(\frac{220}{5}\) = 44 Ω
Resistance of each electric bulb, R1 = 4840 Ω
\(\frac{1}{R}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) + ….. upto x times
\(\frac{1}{R}\) = \(\frac{1}{R_1} \times \) x
x = \(\frac{R_1}{R}\) = \(\frac{4840}{R}\) = 110
∴ Number of electric bulbs connected in parallel are 110.
Question 13:
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 W resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer
Supply voltage, V= 220 V
Resistance of one coil, R= 24 Ω
(i) Coils are used separatelyAccording to Ohm’s law,
V= I1R1
Where,
I1 is the current flowing through the coilI1 = V/R1 = 220/24 = 9.166 A
Therefore, 9.16 A current will flow through the coil when used separately.
(ii) Coils are connected in series
Total resistance, R2 = 24 Ω + 24 Ω = 48 ΩAccording to Ohm’s law,V = I2R2
Where,
I2 is the current flowing through the series circuitI2 = V/R2 = 220/48 = 4.58 A
Therefore, 4.58 A current will flow through the circuit when the coils are connected in series.
(iii) Coils are connected in parallel
Total resistance, R3 is given as = \(\frac{1}{1/24 + 1/24}\) = \(\frac{24}{4}\) = 12Ω
According to Ohm’s law,V= I3R3
Where,
I3 is the current flowing through the circuit I3 = V/R3 = 220/12 = 18.33 A
Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.
Question 14:
Compare the power used in the 2 W resistor in each of the following circuits:
(i) a 6 V battery in series with 1 W and 2 W resistors, and (ii) a 4 V battery in parallel with 12 W and 2 W resistors.
Answer
(i) Potential difference, V = 6 V
1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω
According to Ohm’s law,
V = IR
Where,
I is the current through the circuit
I = \(\frac{6}{3}\) = 2 A
This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is 2 A. Power is given by the expression,
P= (I)2R = (2)2 x 2 = 8 W
(ii) Potential difference, V = 4 V
12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be 4 V.
Power consumed by 2 Ω resistor is given by
P= \(\frac{V^2}{R}\) = \(\frac{4^2}{2}\) = 8 W
Therefore, the power used by 2 Ω resistor is 8 W.
Question 15:
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer
Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.
Current drawn by the bulb of rating 100 W is given by, Power = Voltage x Current
Current = Power/Voltage = 60/220 A
Hence, current drawn from the line = 100/220 + 60/220 = 0.727 A
Question 16:
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer
Energy consumed by an electrical appliance is given by the expression,H = Pt
Where,
Power of the appliance = P
Time = t
Energy consumed by a TV set of power 250 W in 1 h = 250 ×3600 = 9 ×105 J
Energy consumed by a toaster of power 1200 W in 10 minutes = 1200 ×600 = 7.2×105 J
Therefore, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.
Question 17:
An electric heater of resistance 8 W draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer
Rate of heat produced by a device is given by the expression for power as, P= I2R
Where,
Resistance of the electric heater, R= 8 Ω
Current drawn, I = 15 A
P = (15)2 x 8 = 1800 J/s
Therefore, heat is produced by the heater at the rate of 1800 J/s.
Question 18:
Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
Answer
The melting point and of Tungsten is an alloy which has very high melting point and very high resistivity so does not burn easily at a high temperature.
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Answer
The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of metals which produces large amount of heat.
(c) Why is the series arrangement not used for domestic circuits?
Answer
In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits.
(d) How does the resistance of a wire vary with its area of cross-section?
Answer
Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e. when area of cross section increases the resistance decreases or vice versa.
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer
Copper and aluminium are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.
NCERT Solution of Class 10 Science Ch-11 Electricity विद्युत
