CHAPTER 11 Electricity KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS

CHAPTER 11 Electricity KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS

 

CHAPTER 11 Electricity KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS,  English medium Karnataka state board,the Answers Are Prepared By Our Teachers Which Are Simple ,Pointwise,Easy To Read And Remember.

 

CHAPTER 11 Electricity KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS

 

  1. What does an electric circuit mean?
    • An electric circuit is a continuous and closed path through which electric current flows.
    • It consists of various components such as wires, switches, bulbs, and power sources like batteries or cells.
    • When the circuit is complete and unbroken, the electric current can flow freely, enabling devices like bulbs to function.
    • If the circuit is broken or interrupted, the flow of current stops, causing devices to cease operation.

 

  1. Define the unit of current.
  • The unit of electric current is the ampere (A).
  • It is named after the French scientist Andre-Marie Ampere.
  • One ampere is equivalent to the flow of one coulomb of charge per second.
  • It is represented by the symbol ‘A’ in equations and measurements.
  • Smaller units of current include milliampere (mA), which is 10^{-3} ampere, and microampere (μA), which is  10^{-6}

 

  1. Calculate the number of electrons constituting one coulomb of charge.

Given:
– Charge of one electron (\(e\)) = \(1.6 \times 10^{-19}\) coulombs

Formula:
– Number of electrons \(N = \frac{\text{Total charge}}{\text{Charge of one electron}}\)

    \[ N = \frac{1 \text{ coulomb}}{1.6 \times 10^{-19} \text{ coulombs/electron}} \] \[ N = \frac{1}{1.6 \times 10^{-19}} \] \[ N = \frac{1}{1.6} \times 10^{19} \] \[ N = 0.625 \times 10^{19} \] \[ N = 6.25 \times 10^{18} \]

Therefore, approximately \(6.25 \times 10^{18}\) electrons constitute one coulomb of charge.

 

  1. Name a device that helps to maintain a potential difference across a conductor.
  • A battery or an electric cell helps to maintain a potential difference across a conductor.
  • Chemical action within the cell generates the potential difference.
  • The potential difference produced by the battery sets the charges in motion, creating an electric current.

 

  1. What is meant by saying that the potential difference between two points is 1 V?
  • A potential difference of 1 volt (V) between two points means that 1 joule of work is done to move 1 coulomb of charge from one point to the other.
  • It signifies the amount of energy transferred per unit charge.
  • In other words, if 1 joule of energy is expended to move 1 coulomb of charge across the points, then the potential difference is 1 volt.

 

3.How much energy is given to each coulomb of charge passing through a 6 V battery?

  • Given that the potential difference (V) of the battery is 6 volts.
  • Using the formula W = VQ , where W is the work done,  V  is the potential difference, and Q  is the charge.
  • For a 6 V battery, when 1 coulomb of charge passes through, the work done is W = 6 V * 1 C= 6 J.
  • Therefore, each coulomb of charge passing through a 6 V battery receives 6 joules of energy.

 

 

  1. On what factors does the resistance of a conductor depend?

– The resistance of a conductor depends on three main factors:

  • Length of the conductor (l): Resistance is directly proportional to the length of the conductor.
  • Area of cross-section of the conductor (A): Resistance is inversely proportional to the area of cross-section.
  • Nature of material (resistivity, ρ): Different materials have different resistivities, which determine their resistance.

 

  1. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
    • Current will flow more easily through a thick wire compared to a thin wire of the same material when connected to the same source.
    • This is because the resistance of a wire is inversely proportional to its cross-sectional area, according to Ohm’s law.
    • A thick wire has a larger cross-sectional area, resulting in lower resistance, allowing for easier flow of current.

 

  1. Let the resistance of an electrical component remain 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?
  • According to Ohm’s law (V = IR), if the potential difference (V) across an electrical component decreases, and the resistance (R) remains constant, the current (I) through it will also decrease.
  • Halving the potential difference will halve the current, assuming the resistance remains the same.

 

  1. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
  • Coils of electric toasters and electric irons are made of alloys rather than pure metals because alloys have higher resistivities compared to pure metals.
  • Higher resistivity ensures that the coil reaches a high temperature when current flows through it, allowing the toaster or iron to function effectively.

 

  1. Use the data in Table 12.2 to answer the following:

– (a) Which among iron and mercury is a better conductor?

  • Iron has a lower resistivity (10.0 × 10^-8 Ω m) compared to mercury (94.0 × 10^-8 Ω m), indicating that iron is a better conductor than mercury.

– (b) Which material is the best conductor?

  • Among the listed materials, silver has the lowest resistivity (1.60 × 10^-8 Ω m) , making it the best conductor among the materials listed in the table.

 

  1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Draw a schematic diagram of a circuit consisting of:

– Three cells of 2 V each (Battery)

– A 5 Ω resistor (Resistor 1)

– An 8 Ω resistor (Resistor 2)

– A 12 Ω resistor (Resistor 3)

– A plug key, all connected in series.

 

 

  1. 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 Ω resistor. What would be the readings in the ammeter and the voltmeter

 

Readings:

Ammeter: The same current flows through all components in series, so the reading in the ammeter will be the same as the current through each resistor.

Voltmeter: The potential difference across the 12 Ω resistor can be calculated using Ohm’s law (V = IR), where V is the potential difference, I is the current, and R is the resistance. Since the current through the circuit is the same, the potential difference across the 12 Ω resistor will be V = IR = (Current) × 12 Ω.

 

 

  1. Judge the equivalent resistance when the following are connected in parallel –

(a) 1 Ω and 106 Ω,

(b) 1 Ω and 103 Ω, and 106 Ω.

Let’s tackle these problems one by one:

. Equivalent Resistance in Parallel

**Formula:**

    \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots \]

(a) 1 Ω and 106 Ω:

    \[ \frac{1}{R_{\text{eq}}} = \frac{1}{1} + \frac{1}{106} \] \[ \frac{1}{R_{\text{eq}}} = 1 + \frac{1}{106} = 1 + 0.0094 \approx 1.0094 \] \[ R_{\text{eq}} \approx \frac{1}{1.0094} \approx 0.9907 \, \Omega \]

(b) 1 Ω, 103 Ω, and 106 Ω:

    \[ \frac{1}{R_{\text{eq}}} = \frac{1}{1} + \frac{1}{103} + \frac{1}{106} \] \[ \frac{1}{R_{\text{eq}}} = 1 + \frac{1}{103} + \frac{1}{106} \approx 1 + 0.0097 + 0.0094 \approx 1.0191 \] \[ R_{\text{eq}} \approx \frac{1}{1.0191} \approx 0.9813 \, \Omega \]

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?

Given:
– Electric lamp: \(R_1 = 100 \, \Omega\)
– Toaster: \(R_2 = 50 \, \Omega\)
– Water filter: \(R_3 = 500 \, \Omega\)
– Voltage: \(V = 220 \, V\)

Total Resistance of Parallel Connection:

    \[ \frac{1}{R_{\text{eq}}} = \frac{1}{100} + \frac{1}{50} + \frac{1}{500} \] \[ \frac{1}{R_{\text{eq}}} = 0.01 + 0.02 + 0.002 = 0.032 \] \[ R_{\text{eq}} = \frac{1}{0.032} \approx 31.25 \, \Omega \]

Total Current:

    \[ I = \frac{V}{R_{\text{eq}}} = \frac{220}{31.25} \approx 7.04 \, A \]

Resistance of Electric Iron:
The electric iron takes the same current as all three appliances combined (7.04 A) at the same voltage (220 V).

    \[ R_{\text{iron}} = \frac{V}{I} = \frac{220}{7.04} \approx 31.25 \, \Omega \]

Current through the Iron:

    \[ I_{\text{iron}} = 7.04 \, A \]

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

ANSWER;-

1. Independent Operation: Each device can be operated independently without affecting others.
2. Consistent Voltage: All devices receive the same voltage as the source.
3. Fault Tolerance: If one device fails, others can still operate.

  1. 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?

 

Resistors:\(R_1 = 2 \, \Omega\), \(R_2 = 3 \, \Omega\), \(R_3 = 6 \, \Omega\)

(a) Total resistance of 4 Ω:
Connect \(R_1\) and \(R_2\) in series and then in parallel with

    \(R_3\): \[ R_{\text{series}} = R_1 + R_2 = 2 + 3 = 5 \, \Omega \] \[ \frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_3} = \frac{1}{5} + \frac{1}{6} = \frac{6 + 5}{30} = \frac{11}{30} \] \[ R_{\text{total}} = \frac{30}{11} \approx 2.73 \, \Omega \]

This doesn’t give exactly 4 Ω. Let’s try another combination.

Connect

    \(R_1\) and \(R_3\) in parallel, and then in series with \(R_2\): \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{6} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \] \[ R_{\text{parallel}} = \frac{3}{2} = 1.5 \, \Omega \] \[ R_{\text{total}} = R_{\text{parallel}} + R_2 = 1.5 + 3 = 4.5 \, \Omega \]

Since no exact solution exists for 4 Ω, either approach as a concept check.

(b) Total resistance of 1 Ω:
All in parallel:

    \[ \frac{1}{R_{\text{total}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 \] \[ R_{\text{total}} = 1 \, \Omega \]

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?

 

Resistances: \(4 \, \Omega\), \(8 \, \Omega\), \(12 \, \Omega\), \(24 \, \Omega\)

(a) Highest Total Resistance:
All in series:

    \[ R_{\text{total}} = 4 + 8 + 12 + 24 = 48 \, \Omega \]

(b) Lowest Total Resistance:
All in parallel:

    \[ \frac{1}{R_{\text{total}}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = 0.25 + 0.125 + 0.0833 + 0.0417 \approx 0.5 \] \[ R_{\text{total}} = \frac{1}{0.5} = 2 \, \Omega \]

These calculations give the range of possible resistances for various configurations.

  1. Why does the cord of an electric heater not glow while the heating element does?
  • The heating element of an electric heater is designed to have high resistance, causing it to dissipate a significant amount of energy in the form of heat when current passes through it.
  • The cord of an electric heater, on the other hand, has much lower resistance compared to the heating element.
  • Since the cord has lower resistance, it dissipates less energy and does not reach a high enough temperature to glow.
  • The heating element, with its higher resistance, absorbs more energy and thus heats up to a temperature where it glows.

 

  1. Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.

– Given: Charge (Q) = 96000 coulombs, Time (t) = 1 hour = 3600 seconds, Potential difference (V) = 50 V.

– Using the formula  H = VIt, where H is the heat generated:

–  H = (50  V)(96000 C)(3600 s) )

–  H = 1.728 times 10^8 J

 

  1. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

– Given: Resistance (R) = 20 Ω, Current (I) = 5 A, Time (t) = 30 s.

– Using the formula  H = VIt , where V is the potential difference across the iron:

– V = IR = (5  A)(20  Ω) = 100 V

–  H = (100 V)(5  A)(30 s)

–  H = 15000 J .

 

  1. What determines the rate at which energy is delivered by a current?
  • The rate at which energy is delivered by a current is determined by the power of the circuit.
  • The power of the circuit depends on the product of the potential difference (voltage) across the circuit and the current flowing through it.
  • Mathematically, the power (P) is given by P = VI , where V is the potential difference and I is the current.

 

  1. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

– Given: Current (I) = 5 A, Voltage (V) = 220 V, Time (t) = 2 hours.

  • To find the power (P), we use P = VI .
  • P = (220 V)(5 A) = 1100 W \)
  • The power of the motor is 1100 watts.
  • To find the energy consumed, we use the formula W = P times t .
  • W = (1100 W)(2 h) = 2200 Wh
  • The energy consumed by the motor in 2 hours is 2200 watt-hours.

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