For Questions number 8 and 9, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A): There are generally a greater number of individuals at the lower trophic levels of an ecosystem. Reason (R) : The greatest number is of the producers.
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(B) / Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(a) Define volt, the unit of potential difference. (b) Calculate the work done required to move an electron between two points A and B located at a potential difference of $100\,\text{V}$ in an electric field.
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1 Volt is the potential difference between two points in a current carrying conductor when one Joule work is done to move a charge of 1Coulomb from one point to the other. / $1V = \frac{1J}{1C}$ $V = \frac{W}{Q}$ $W=QxV$ $W = 1.6 \times 10^{-19} \times 100$ $W = 1.6 \times 10^{-17} J$
In an electric circuit carrying current $I$, a $2\ \Omega$ resistor is replaced by an $8\ \Omega$ resistor without changing the connected (ideal) battery. Then :
(a)current becomes $\frac{I}{4}$ and potential difference $\frac{V}{4}$
(b)current becomes $\frac{I}{4}$ and potential difference remains $V$
(c)current becomes $4I$ and potential difference remains $V$
(d)current becomes $4I$ and potential difference $\frac{V}{4}$
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(B)/ current becomes $\frac{1}{4}$ and potential difference remains V.
In which case will the readings of voltmeter and ammeter be maximum in the above experiment ? What conclusion can be drawn from the straight line graph between ‘V' and 'I'?
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(A) The plug key was removed to prevent overheating of circuit components /to avoid accidental burns / to stop the battery from draining. (any one) (1 Mark) (B) $\bullet$ Voltmeter reading increases. ($\frac{1}{2}$ Mark) $\bullet$ because larger potential difference is applied. ($\frac{1}{2}$ Mark) (C) $\bullet$ When all four cells were connected in series in the circuit. (1 Mark) $\bullet$ Current is directly proportional to potential difference applied. (1 Mark) OR $\bullet$ I or current will change. (1 Mark) $\bullet$ Wires made of different material have different resistance. (1 Mark)
Suppose this experiment is carried out using only one cell but wires of same length & thickness but made of different materials like nichrome wire, fuse wire, copper wire and iron wire which are replaced one by one. Out of V and I which will change and why?
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(A) The plug key was removed to prevent overheating of circuit components /to avoid accidental burns / to stop the battery from draining. (any one) (1 Mark) (B) $\bullet$ Voltmeter reading increases. ($\frac{1}{2}$ Mark) $\bullet$ because larger potential difference is applied. ($\frac{1}{2}$ Mark) (C) $\bullet$ When all four cells were connected in series in the circuit. (1 Mark) $\bullet$ Current is directly proportional to potential difference applied. (1 Mark) OR $\bullet$ I or current will change. (1 Mark) $\bullet$ Wires made of different material have different resistance. (1 Mark)
Three students Shweta, Ayesha and Samridhi were performing an experiment to understand the factors on which the resistance of a conductor depends. Each one of them completed electric circuit with the help of a cell, an ammeter, a plug key and wire. Shweta put nichrome wire of length ‘l’ in the circuit and after plugging the key, noted current in the ammeter. Ayesha put nichrome wire of same thickness but twice the length i.e. ‘2l’ in the circuit and after plugging the key, noted current in the ammeter. Samridhi took copper wire of length ‘l’ and same thickness in the circuit and after plugging the key, noted current in the ammeter. (a) If the ammeter reading is X ampere with nichrome wire of length ‘l’, then what will be the ammeter reading if the length of nichrome wire is doubled with same area of cross-section ? (b) What happens to the ammeter reading if the area of cross-section of nichrome wire is doubled, keeping the length of wire ‘l’ the same ? (c) Define ‘resistivity’. Write its SI unit. Compare the resistivity of an alloy with its constituents metals.
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(a) Ammeter reading becomes $\frac{X}{2}$ / halved (1 Mark) (b) Ammeter reading becomes $2X$ / doubled (1 Mark) (c) • Resistivity is equal to electrical resistance of a conductor of unit length and unit area of cross section. (1 Mark) • SI unit = $\Omega\text{ m}$ / ohm metre (½ Mark) • Resistivity of an alloy is higher than its constituent metals. (½ Mark) OR (c) (i) It has high melting point. (1 Mark) (ii) The resistivity of an alloy is generally higher than that of its constituent metals. / Alloys do not oxidise (burn) readily at high temperatures. (1 Mark)
(a) (i) Due to change in length and area of cross-section of a conductor, resistance of conductor changes while resistivity does not change. Why? (ii) Conductors of electric toasters and electric iron are made of an alloy rather than a pure metal. Why ? (iii) Define the S.I. unit of electric current.
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(a) (i) As Resistance, $R = \rho \frac{l}{A}$, it changes with change in length and area of cross section of conductor. But resistivity of conductor is the characteristic property of material and hence it does not change. (1) (ii) The resistivity of an alloy is generally higher than that of its constituent metals. / Alloys do not oxidise (burn) readily at high temperatures. (1) (iii) 1 Ampere is constituted by the flow of 1 Coulomb of charge per second. / $1\text{A}=1\text{C}/1\text{s}$ (1)
(B) (i) Define electrical resistivity of a material. (ii) How will the resistivity and resistance of a conducting wire change if the length and area of cross-section of wire both are doubled? Justify your answer. (iii) Calculate the electric current flowing through a $4\ \Omega$ resistor that produces $100\ \text{J}$ of heat every second.
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(i) Resistivity is the resistance offered by a wire of unit length and unit area of cross section in an electric circuit. (1 Mark) (ii) • Resistivity of conducting wire will remain same. (½ Mark) • Resistivity is the characteristic property of the material. (½ Mark) • Resistance will remain same (½ Mark) • $R = \rho \frac{l}{A}$ If $l' = 2l$, $A' = 2A$ $R' = \rho \frac{l'}{A'}$ $R' = \rho \frac{2l}{2A}$ $R' = R$ (½ Mark) (iii) $H = I^2 R t$ (1 Mark) $I = \sqrt{\frac{H}{Rt}}$ $I = \sqrt{\frac{100}{4 \times 1}}$ (1 Mark) $I = 5\text{ A}$
Attempt either subpart (a) or (b) : OR (b) (i) State the relationship of the resistance (R) of a wire with its area of cross-section (A) and its length (l). With the help of this relation, define resistivity of the material of the wire. (ii) In an electric circuit, $\text{2 A}$ electric current flows through a $40 \Omega$ resistor for $\text{50}$ seconds. (I) Calculate the heat produced in the wire. (II) If another $40 \Omega$ resistor is connected in parallel to the given $40 \Omega$ resistor, calculate the heat produced in the circuit in $\text{50}$ seconds.
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(b) (i) • $R = \rho \frac{l}{A}$ (1 Mark) • Resistivity of a conductor is the resistance offered by a conductor of length 1 m and area of cross section 1 m² (1 Mark) (ii) H = $I^2Rt$ (½ Mark) H = $(2)^2 \times 40 \times 50$ (½ Mark) H = $8000$ J (½ Mark) Voltage applied across $40\Omega$, $V=I \times R$ (½ Mark) = $2 \times 40 = 80$ V (½ Mark) Equivalent resistance $\frac{1}{R} = \frac{1}{40} + \frac{1}{40}$ (½ Mark) $R = 20\Omega$ (½ Mark) Current flowing through circuit $I = \frac{V}{R}$ (½ Mark) $I = \frac{80}{20}$ (½ Mark) $I = 4$ A (½ Mark) H = $I^2Rt$ (½ Mark) H = $(4)^2 \times 20 \times 50$ (½ Mark) H = $16000$ J (½ Mark)
(b) (i) What is meant by electrical resistivity of a material ? (ii) A series electric circuit comprising a resistor made of a metallic wire has an ammeter which reads $300\ \text{mA}$. If the length of the wire is made three times, what will be the new reading in the ammeter ? (iii) Calculate the resistance of a $500\ \text{m}$ long aluminium wire of cross-sectional area $1\times10^{-3}\ \text{m}^2$ and resistivity $$\begin{aligned}& 2.65\times10^{-8} \\ & \Omega\text{m}\end{aligned}$$.
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(b) (i) Electrical resistivity is the resistance offered by a wire of length 1m and area of cross-section 1m$^2$ (1) (ii) $\text{V=IR}$ $\text{V = 300 \times R}$ -----------(1) $\text{R} \propto \text{l}$ $\text{R1 = 3 \times R}$ ($\frac{1}{2}$) $\text{V = I1 R1}$ $\text{V = I1 \times 3 R}$ -----------(2) $\text{From (1) and (2)}$ $\text{I1 \times 3 R = 300 \times R}$ $\text{I1 = 100 mA (Reading of ammeter)}$ ($\frac{1}{2}$) (iii) $\text{R} = \rho \frac{\text{l}}{\text{A}}$ (1) $\text{R} = \frac{2.65 \times 10^{-8} \times 500}{1 \times 10^{-3}}$ ($\frac{1}{2}$) $\text{R} = 2.65 \times 5 \times 10^{-3} \text{ \Omega}$ $\text{R} = 13.25 \times 10^{-3} \text{ \Omega}$ ($\frac{1}{2}$)
Attempt either (a) or (b) : (a) (i) What is meant by resistance in an electric circuit ? (ii) In a series electric circuit comprising a resistor made of a metallic wire, the ammeter reads $500\ \text{mA}$. If the length of wire is doubled and area of cross-section is made half, what will be the new reading of ammeter ? (iii) How will the resistivity of a wire be affected if both, the length of the wire and its area of cross-section, are doubled ? Justify your answer.
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(a) (i) Resistance is opposition to the flow of electrons. /Property of a conductor due to which it retards the motion of electrons in an electric circuit. (1) (ii) $\text{V=IR}$; $\text{I} = \frac{\text{V}}{\text{R}} = 500\text{mA}$ (½) $\text{R}=\rho \frac{\text{l}}{\text{A}}$ (½) $\text{R}_1=\rho \frac{\text{l}_1}{\text{A}_1}$; $\text{l}_1=2\text{l}$; $\text{A}_1=\text{A}/2$ $\text{R}_1=\rho \frac{2\text{l}}{\text{A}/2} = 4\text{R}$ (½) $\text{I}_1 = \frac{\text{V}}{\text{R}_1} = \frac{\text{V}}{4\text{R}} = \frac{500}{4} = 125 \text{ mA}$ (½) (iii) • No change in resistivity. (1) • It is the property of material only (1)
(b) (i) Name the device which (I) helps to change the electric current in the given electric circuit and (II) breaks the circuit during short-circuiting. (ii) State Ohm's law and write the factors on which the resistance of a given conductor depends. (iii) Find out the length of a wire of resistance $21\ \Omega$, if the diameter of the wire is $0.02\ \text{cm}$ and resistivity of the material is $$\begin{aligned}& 66 \times 10^{-8} \\ & \Omega\text{m}\end{aligned}$$. [Given : $\pi = \frac{22}{7}$]
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Power consumed per second in parallel combination $\text{P} = \frac{\text{V}^2}{\text{R}_\text{p}} = \frac{8^2}{4/3} = 48\text{W}$ (1 Mark) (b) (i) (I) Rheostat or variable resistance ($\frac{1}{2}$ Mark) (II) Fuse ($\frac{1}{2}$ Mark) (ii) • The potential difference applied across the two ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it provided that its temperature remains constant. (1 Mark) • Resistance depends on $\blacktriangleright$ length of conductor $\blacktriangleright$ area of cross-section/thickness of conductor $\blacktriangleright$ Material/Electrical Resistivity of conductor $\blacktriangleright$ Temperature of conductor. ($\frac{1}{2}$ + $\frac{1}{2}$ Mark) (Any two) (iii) $\text{R} = 21 \Omega$ $\text{r} = 0.01 \times 10^{-2} \text{ m}$ $\rho = 66 \times 10^{-8} \Omega\text{m}$ ($\frac{1}{2}$ Mark) $\text{R} = \rho \frac{\text{l}}{\text{A}} = \rho \frac{\text{l}}{\pi \text{r}^2}$ ($\frac{1}{2}$ Mark) $21 = \frac{66 \times 10^{-8} \times \text{l}}{22 \times (0.01 \times 10^{-2})^2}$ ($\frac{1}{2}$ Mark) $21 = \frac{3 \times 10^{-8} \times \text{l} \times 7}{0.01 \times 0.01 \times 10^{-4}}$ $\text{l} = \frac{21 \times 1 \times 10^{-8}}{21 \times 10^{-8}}$ $\text{l} = 1 \text{ m}$ (1 Mark) (5 Marks)
(b) (i) How many bulbs of resistance $8\ \Omega$ each should be connected in parallel combination to draw a current of $2\ \text{A}$ from a battery of $4\ \text{V}$? (ii) Name the device used for measuring electric current. How is it connected in a circuit ? (iii) State Joule's law of heating.
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(b) (i) $V=4\text{V}, I= 2\text{A}$ Resistance of circuit $R = \frac{V}{I}$ $R=\frac{4}{2}$ $R = 2\Omega$ (½) Let 'n' be the number of bulbs $\frac{1}{R} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + ---n$ $\frac{1}{R} = \frac{n}{8}$ $\frac{1}{2} = \frac{n}{8}$ $n=4$ Therefore, 4 bulbs of resistance $8 \Omega$ should be connected in parallel. (1) (ii) • Ammeter (1) • In series (1) (iii) Heat generated through a current carrying conductor is directly proportional to square of current, resistance of conductor and time for which current flows in conductor. (1) / $H=I^2Rt$ (1)
Read the following passage and answer the questions that follow : Swati, a class 10 student, observes that when she passes close to the refrigerator in her kitchen, she feels the heat, although the things kept inside the refrigerator are cool. Describe the cause of heating in the above-mentioned case.
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(a) A part of current is consumed into useful work and rest is expended in heat to raise the temperature of gadget. (any other suitable explanation) (b) $W=V \times Q= VIt = IR \times It$ $H = I^2Rt$ / $H = \frac{V^2}{R}t$
A current $\text{I}$ flows through a resistor of resistance $\text{R}$ when the potential difference across it is $\text{V}$. Applying Ohm's law, write the formula for amount of heat produced by the resistor in time $\text{t}$.
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(a) A part of current is consumed into useful work and rest is expended in heat to raise the temperature of gadget. (any other suitable explanation) (b) $W=V \times Q= VIt = IR \times It$ $H = I^2Rt$ / $H = \frac{V^2}{R}t$
Read the following passage and answer the questions that follow : Swati, a class 10 student, observes that when she passes close to the refrigerator in her kitchen, she feels the heat, although the things kept inside the refrigerator are cool. (a) Describe the cause of heating in the above-mentioned case.
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(a) A part of current is consumed into useful work and rest is expended in heat to raise the temperature of gadget. (1 Mark) (any other suitable explanation) (b) $W=V \times Q= VIt = IR \times It$ $H = I^2Rt$ / $\frac{V^2}{R}t$ (1 Mark) (c) (i) Electric heater, Oven, Electric iron (Any two, Any other) (1+1 Marks) OR (c) (ii) When 1 kilowatt of power is used for 1 hour then energy consumed is 1 kWh (1 Mark) 1 kWh = $3.6 \times 10^6\text{J}$ (1 Mark)
Write any two practical applications of heating effect of electric current.
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(a) A part of current is consumed into useful work and rest is expended in heat to raise the temperature of gadget. (any other suitable explanation) (b) $W=V \times Q= VIt = IR \times It$ $H = I^2Rt$ / $H = \frac{V^2}{R}t$
(a) Name a device which is used to : (i) Maintain a constant potential difference in a circuit. (ii) Change the electric current in an electric circuit. (b) When the potential difference between the terminals of an electric heater is $110\ \text{V}$, a current of $5\ \text{A}$ flows through it. What will be the value of current flowing through it when the potential difference is increased to $220\ \text{V}$ ?
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(a) (i) Battery / Electric cell (ii) Rheostat / Variable resistance (b) Resistance of the heater, $R = \frac{V}{I} = \frac{110}{5} = 22 \Omega$ Current through the heater, $I' = \frac{V'}{R} = \frac{220}{22} = 10 A$ Alternate answer for (b): • According to Ohm's law, potential difference is directly proportional to current. • When the potential difference is doubled, current will also be doubled. So, value of current will be 10A.
Read the following passage and answer the questions that follow : Swati, a class 10 student, observes that when she passes close to the refrigerator in her kitchen, she feels the heat, although the things kept inside the refrigerator are cool. (a) Describe the cause of heating in the above-mentioned case. (b) A current $\text{I}$ flows through a resistor of resistance $\text{R}$ when the potential difference across it is $\text{V}$. Applying Ohm's law, write the formula for amount of heat produced by the resistor in time $\text{t}$. (c) (i) Write any two practical applications of heating effect of electric current. OR (c) (ii) Define the commercial unit of electric energy and express it in Joules (J).
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(c) (i) Electric heater, Oven, Electric iron (Any two, Any other)
A bulb consumed $24\ \text{J}$ of energy per second, when connected to a $12\ \text{V}$ supply. Calculate the power consumed when it is connected to a $6\ \text{V}$ supply.
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P = 24 J/s ($\frac{1}{2}$ Mark) When connected to 12V supply $P = \frac{V^2}{R}$ $R = \frac{12 \times 12}{24}$ $R = 6 \Omega$ (1 Mark) When connected to 6V supply $P = \frac{V^2}{R}$ $P = \frac{6 \times 6}{6}$ $P = 6 W$ (1 Mark)
(i) How much electric current will an electric iron draw from $220\,\text{V}$ source if the resistance of its heating element when hot, is $55\,\Omega$ ? Calculate the power consumed by the electric iron when it is operated at $220\,\text{V}$. (ii) In a house, 3 bulbs of $100\,\text{watt}$ each, are lit for $5\,\text{hours}$ daily and an electric heater of $1.0\,\text{kW}$ is used for half an hour daily. Calculate the total energy consumed in a month of $30\,\text{days}$ and its cost at the rate of ₹ $3.60$ per $\text{kWh}$. (iii) With reason explain, why are alloys commonly used to make elements of electrical heating devices.
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$I = \frac{V}{R}$ $I = \frac{220}{55}$ $I = 4 A$ Power of electric iron $P = VI$ $P = 220 \times 4$ $P = 880 W$ $E=Pxt$ Energy (3 bulbs) $= 3 \times 100 \times 5$ $= 1500$ $= 1.5 \text{ kWh}$ Energy (electric heater) $= 1.0 \times 0.5$ $= 0.5 \text{ kWh}$ Total energy consumed (1 day) $= 1.5 + 0.5 = 2 \text{ kWh}$ Total energy consumed (30 days) $= 30 \times 2$ $= 60 \text{ kWh}$ $= 60 \text{ units}$ Total cost = Units $\times$ Rate $= 60 \times 3.60$ $= 216$ The resistivity of an alloy is generally higher than that of its constituent metals. / Alloys do not oxidise (burn) readily at high temperatures.
(i) Define the SI unit of power. Express the electric power of an electric appliance in terms of potential difference and current flowing through it. (ii) In a house, 3 bulbs of $100\text{ W}$ each, are lit for $5\text{ hours}$ daily and an electric heater of $1.0\text{ kW}$ is used for half an hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ $3.60$ per $\text{kWh}$. (iii) Convert the commercial unit of electric energy into Joule (J).
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(i) • Power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V / If one joule energy is consumed in one second then power of instrument is said to be 1 watt / 1 W = 1 volt $\times$ 1 ampere (1 Mark) • $P = VI$ (1 Mark) (ii) $E = P \times t$ (½ Mark) Energy (3 bulbs) $= 3 \times 100 \times 5 = 1500\text{ Wh} = 1.5\text{ kWh}$ Energy (electric heater) $= 1.0 \times 0.5 = 0.5\text{ kWh}$ Total energy consumed (1 day) $= 1.5 + 0.5 = 2\text{ kWh}$ (½ Mark) Total energy consumed (30 days) $= 30 \times 2 = 60\text{ kWh} = 60\text{ units}$ Total cost = Units $\times$ Rate $= 60 \times 3.60 = \text{\text{Rs} } 216$ (1 Mark) (ii) $1\text{ kW h} = 1000\text{ watt} \times 3600\text{ second} = 3.6 \times 10^6\text{ watt second} = 3.6 \times 10^6\text{ joule (J)}$ (1 Mark)
Attempt either subpart (a) or (b) : (a) (i) In a house, $\text{5}$ fans, rated $\text{100 W}$; $\text{220 V}$ each are used for $\text{4}$ hours daily and an electric heater rated $\text{1000 W}$; $\text{220 V}$ is used for $\text{5}$ hours daily. Calculate the cost of electrical energy consumed by the fans and the electric heater in a month of $\text{30}$ days if the cost of $\text{1}$ unit of electrical energy is ₹5. (ii) You have two fuse wires X and Y with electric current rating $\text{5 A}$ and $\text{10 A}$ respectively. Which of the two would you use with a heater rated $\text{2000 W}$; $\text{220 V}$ ? Justify your answer.
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(a) (i) E=Pxt (½ Mark) E (fan) = $5 \times 100 \times 4$ (½ Mark) = $2000$ Wh (½ Mark) = $2$ kWh (½ Mark) E(heater) = $1000 \times 5$ (½ Mark) = $5000$ Wh (½ Mark) = $5$ kWh (½ Mark) Total energy, E = E(fan) + E(heater) (½ Mark) = $2 + 5 = 7$ kWh (½ Mark) Cost = $7 \times 30 \times 5$= Rs $1050$ (1 Mark) (ii) P=VI (½ Mark) Current required for heater $I = \frac{P}{V}$ (½ Mark) $I = \frac{2000}{220}$ (½ Mark) $I = 9.09$ A (½ Mark) Current rating of Y wire is slightly greater than required value. Hence fuse wire Y will be used. (1 Mark)
Attempt either (a) or (b) : (a) A smart home uses energy monitoring system to track electricity usage of appliances. In a month of 30 days, the following devices are used: 1. An electric heater of $1500\ \text{W}$ used for 4 hours daily. 2. A microwave oven rated at $1.5\ \text{kW}$ used for 3 hours daily. (i) Define electric power and write its SI unit. (ii) Calculate the total energy consumed in joules and $\text{kWh}$ in a whole month. (iii) Find out the total cost of electricity for running both appliances for a month of 30 days if the cost of 1 unit is ₹7.
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(a) (i) $\bullet$ Electric power is defined as the rate at which electrical energy is consumed in an electric circuit. (1) $\bullet$ S.I unit is watt. ($\frac{1}{2}$) (ii) Electrical energy consumed=Power$\times$ time Electrical energy consumed by heater in 1 day $\text{E=1500}\times\text{4=6000Wh=6kWh}$ Electrical energy consumed by heater in 30days=6$\times$30=180kWh ($\frac{1}{2}$) Electrical energy consumed by oven in 1 day=1.5$\times$3=4.5kWh Electrical energy consumed by oven in 30 days=4.5$\times$30=135kWh ($\frac{1}{2}$) Total energy consumed in whole month=180+135=315kWh ($\frac{1}{2}$) Total energy consumed in joules=315$\times 3.6 \times 10^6 = 1134 \times 10^6 \text{J}$ ($\frac{1}{2}$) (iii)Total cost =Total energy consumed $\times$ Cost of 1 unit =315$\times$7= Rs.2205 (1)
(a) (i) Write the definition of the SI unit of power. (ii) Two resistors $\text{R}_1 = 1\ \Omega$ and $\text{R}_2 = 2\ \Omega$ are connected in an electric circuit with a battery of $12\ \text{V}$. Calculate the power consumed, when $\text{R}_1$ and $\text{R}_2$ are connected in (I) Parallel combination, and (II) Series combination. (iii) Calculate the minimum number of identical electric bulbs of rating $4\ \text{V}$, $6\ \text{W}$; that can work safely with their optimum brightness, when connected in series with a $240\ \text{V}$ mains supply.
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1 W is the power consumed by a device that carries 1A current when operated at a potential difference of 1V. / When 1J electrical energy is consumed in 1 s the power consumed is 1W. 1 Parallel combinatioin $\frac{1}{R} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2}$ $R = \frac{2}{3} \Omega$ $P = \frac{V^2}{R}$ $P = \frac{(12)^2}{2/3} = 216W$ 1 Series: $R=1+2=3 \Omega$ $P = \frac{V^2}{R} = \frac{(12)^2}{3} = 48 \text{ W}$ 1 $R = \frac{V^2}{P}$ $R = \frac{(4)^2}{6} = \frac{8}{3} \Omega$ 1/2 $P = VI$ $I = \frac{P}{V} = \frac{6}{4} = \frac{3}{2} A$ 1/2 Since same current will flow in all resistors, hence $R_S = \frac{240}{3/2} = 160 \Omega$ 1/2 $n = \frac{R_S}{R} = \frac{160}{8/3} = 60$ 1/2