Q1: Sketch a circuit diagram of an electric circuit consisting of a cell, an electric bulb, an ammeter, a voltmeter and a plug key.
Ans:
Q2: Differentiate between Resistance and Resistivity.
Ans: Resistance:
Resistivity:
Q3: Distinguish between resistances in series and resistances in parallel.
Ans: Resistances in series:
Resistances in parallel:
Q4: Nichrome wire is used for making the Ideating elements of electrical appliances like iron, geyser, etc. Give reasons.
Ans: Nichrome wire is used for making the heating elements of electrical appliances like iron, geyser, etc. because:
Q5: A copper wire of resistivity 2.6 × 10^{3} Ωm, has a cross sectional area of 30 × 10^{4} cm^{3}. Calculate the length of this wire required to make a 10 Ω coil.
Ans: Given: R = 10Ω, ρ = 2.6 × 10^{8} Ωm,
To find: l = ?
Formula: R = ρlA
Solution: R = ρlA
∴ ρl = RA
Q6: Two coils of resistance R1 = 3Ω and R2 = 9Ω are connected in series across a battery of potential difference 14 V. Draw the circuit diagram. Find the electrical energy consumed in 1 min in each resistance.
Ans: Given: R_{1} = 3Ω, R_{2} = 9Ω
R_{s} = R_{1} + R_{2 }= 9 + 3 = 12 Ω
Q7: State the relation between work, charge and potential difference for an electric circuit. Calculate the potential difference between the two terminals of a battery if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to the other.
Ans: V = W/Q
Here,
V = Potential difference,
W = Work done,
Q = Electric charge
W = 100 J
Q = 20 C
V = WQ=10020 = 5V
Q8: What is an electric circuit? Distinguish between an open and a closed circuit.
Ans:
Q9: With the help of a diagram, derive the formula for the equivalent resistance of three resistances connected in series.
Ans:
(i) If a number of resistances are connected in such a way that the same current flows through each resistance, then the arrangement is called ‘Resistances in Series’.
(ii) Let R_{1}, R_{2} and R_{3} be three resistances connected in a series combination and let R be their equivalent resistance.
Let V_{1}, V_{2} and V_{3} be the potential difference across the resistances R_{1}, R_{2} and R_{3} respectively. Let ‘V’ be the potential differences across the combination. Let ‘I’ be the current flowing through each resistance.
(iii) According to Ohm’s law,
V = IR
Hence, V_{1 }= IR_{1}; V_{2} = IR_{2}; V_{3 }= IR_{3}
(iv) For series combination of resistances,
Hence, the equivalent resistance in series (Rs) is equal to the sum of the individual resistances.
Q10: With the help of a diagram, derive the formula for the equivalent resistance of three resistances connected in parallel.
Ans: 1. If a number of resistances are connected between two common points in such a way that the potential difference across each resistance is same, then the arrangement is called ‘Resistances in Parallel’.
2. Let R_{1}, R_{2} and R_{3} be the three resistances connected in parallel combination between points C and D and let R_{p} be their equivalent resistance.
Let I_{1}, I_{2 }and I_{3} be the currents flowing through resistances R_{1}, R_{2} and R_{3} respectively.
Let I be the current flowing through the circuit and V be the potential difference of the cell.
3. According to Ohm’s law.
I = V/R
Therefore,
Therefore, the reciprocal of the equivalent resistances in parallel combination is equal to the sum of the reciprocals of the individual resistances.
Q11: What is the better way of connecting lights and other electrical appliances in domestic wiring? Why?
Ans: The better way of connecting lights and other electrical appliances in domestic wiring is parallel connection because of the following advantages:
Q12: An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to aft V battery (Fig.).
Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric conductor.
Ans: (a) Resistance of electric lamp, R_{1} = 20 W
Resistance of series conductor, R_{2} = 4 W
Total resistance in the circuit,
R_{s} = R_{1} + R_{2} = 20 Ω + 4 Ω = 24 Ω.
(b) Total potential difference, V = 6 V
By Ohm’s law, the current through the circuit is
(c) Potential difference across the electric lamp,
V_{1} = IR_{1} = 0.25 A × 20 Ω = 5 V.
Potential difference across the conductor is
V_{2} = IR_{2} = 0.25 A × 4 Ω = 1 V.
Q13: A wire has a resistance of 10Ω. It is melted and drawn into a wire of half of its length. Calculate the resistance of the new wire. What is the percentage change in its resistance?
Ans: Given: R_{1} = 10 Ω, l_{2} = l_{1}/2
To find: (a) R_{2}
(b) Percentage change in the resistance (ΔR%).
If volume of the wire remains same in both the cases.
Dividing eq. (iii) by (ii), we get
Q14: If, in Figure R_{1} = 10 ohms, R_{2 }= 40 ohms, R_{3} = 30 ohms, R_{4 }= 20 ohms, R_{g} = 60 ohms and a 12 volt battery is connected to the arrangement, calculate: (a) the total resistance and (b) the total current flowing in the circuit.
Ans: (a) Let R’ be the equivalent resistance of R_{1} and R_{2}. Then,
R’ = 8Ω
Let R” be the equivalent resistance of R_{3}, R_{4} and R_{5}. Then,
R” = 10 Ω
Total Resistance, R = R’ + R” = 8 + 10 = 18 Ω
(b) Current,
Q15: Two lamps, one rated 60 W at 220 V and other 40 W a 220 V, are connected in parallel to an electric supply at 220 V.
(a) Draw the circuit diagram to show the connections.
(b) Calculate the current drawn from the electric supply.
(c) Calculate the total energy consumed by the two lamp together when they operate for one hour.
Ans: (a) The required circuit diagram is shown below:
(b) Total power of the two lamps = 60 + 40 = 100 W
Applied Voltage, V = 220 V
Current drawn from the electric supply,
(c) Total energy consumed by the lamp in one hour = 60W × 1h + 40W × 1h = 100 Wh = 0.1 kWh.
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