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NCERT Exemplar · Class 10 Science Electricity

35 questions · 35 still being checked

Short Answer Questions 19–28 (part 3 of 4)

  1. Exercise 19

    A child has drawn the electric circuit to study Ohm's law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections. NCERT_Question_Class10_Science_Exemplar_Ch12_Q19

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    NCERT’s answer
    NCERT_Solution_Class10_Science_Exemplar_Ch12_Q19_ncert
    Corrections:
    Ammeter: in series with \(\displaystyle R\), not across it.
    Voltmeter: in parallel across \(\displaystyle R\), not in the main loop.
    Cells: all the same way round, long plate (\(\displaystyle +\)) to short plate (\(\displaystyle -\)).
    Polarity: current leaves the cell \(\displaystyle +\), enters A and V at \(\displaystyle +\), leaves at \(\displaystyle -\).
    Answer: Ammeter in series, voltmeter across \(\displaystyle R\), cells all one way round, meters \(\displaystyle +\) toward the cell \(\displaystyle +\).

  2. Exercise 20

    Three 2Ω\displaystyle 2 \Omega resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors? NCERT_Question_Class10_Science_Exemplar_Ch12_Q20

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    NCERT’s answer
    Maximum current through resistor A = \(\displaystyle \sqrt{\frac{18}{2}}\) A = $\displaystyle 3$ A. Thus the maximum current through resistors B and C each \(\displaystyle 3\times\frac{1}{2}\) A = $\displaystyle 1.5$ A.
    A carries the full current \(\displaystyle I\); B and C share it, \(\displaystyle I/2\) each. \[P = I^2 R \] \[18\,\mathrm{W} = I^2 (2\,\Omega) \Rightarrow I^2 = 9\,\mathrm{A^2} \Rightarrow I = 3\,\mathrm{A} \] \[P_B = \left(\frac{3\,\mathrm{A}}{2}\right)^2 (2\,\Omega) = 4.5\,\mathrm{W} < 18\,\mathrm{W} \] So A sets the limit. Answer: \(\displaystyle I_{max} = 3\,\mathrm{A}\) through A, with \(\displaystyle 1.5\,\mathrm{A}\) through each of B and C.

  3. Exercise 21

    Should the resistance of an ammeter be low or high? Give reason.

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    NCERT’s answer
    Hint— It should be as close to zero as possible. Ideally it should be zero ohm. If it is non-zero and substantial it will affect the true current.
    An ammeter always sits in series, so the full circuit current passes through it. \[R_A \downarrow \;\Rightarrow\; V_{\text{drop},A} \downarrow \;\Rightarrow\; I_{circuit} \approx I_{true} \] A high \(\displaystyle R_A\) would itself add to the circuit resistance and cut down the very current it is meant to measure. Answer: Low, ideally zero, so that the ammeter does not change the current it measures.

  4. Exercise 22

    Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2Ω\displaystyle 2 \Omega in series with a combination of two resistors (4Ω\displaystyle 4 \Omega each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2Ω\displaystyle 2 \Omega resistor be the same as that across the parallel combination of 4Ω\displaystyle 4 \Omega resistors? Give reason.

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    NCERT’s answer
    Hint— Yes. Total resistance of the parallel combination is also $\displaystyle 2$ ohm ($\displaystyle 2$ Ω ). NCERT_Solution_Class10_Science_Exemplar_Ch12_Q22_ncert
    \[\frac{1}{R_p} = \frac{1}{4\,\Omega} + \frac{1}{4\,\Omega} = \frac{1}{2\,\Omega} \Rightarrow R_p = 2\,\Omega \] The \(\displaystyle 2\,\Omega\) resistor and the parallel block are in series, so the same current \(\displaystyle I\) flows through both: \[V_{2\Omega} = I(2\,\Omega), \qquad V_{p} = I R_p = I(2\,\Omega) \] \[\Rightarrow V_{2\Omega} = V_{p} \] Answer: Yes. \(\displaystyle R_p = 2\,\Omega\) and the same current flows through both, so the p.d. is equal.

  5. Exercise 23

    How does use of a fuse wire protect electrical appliances?

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    NCERT’s answer
    Hint— If a current larger than a specified value flows in a circuit, temperature of fuse wire increases to its melting point. The fuse wire melts and the circuit breaks.
    A fuse wire is a low-melting-point wire placed in series in the live line. \[H = I^2 R t \] On a fault (short circuit or overload): \(\displaystyle I\uparrow \Rightarrow H\uparrow\) sharply. Answer: Excess current heats the fuse wire to its melting point, so it melts and breaks the circuit before the appliance is damaged.

  6. Exercise 24

    What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5\displaystyle 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

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    NCERT’s answer
    Hint— Use the formula \(\displaystyle R = \rho\frac{l}{A}\). Also, V = R I. R is doubled while V remains unchanged. Hence current becomes \(\displaystyle \frac{I}{2}\).
    Resistivity relates resistance to a conductor's dimensions and is a property of the material alone: \[\rho = \frac{RA}{L}, \qquad \text{unit } \Omega\,\mathrm{m} \] For the wire, \(\displaystyle R=\rho L/A\); with \(\displaystyle \rho\) and \(\displaystyle A\) fixed, doubling \(\displaystyle L\) doubles \(\displaystyle R\): \[R' = \rho\frac{2L}{A} = 2R \] The battery voltage \(\displaystyle V\) is unchanged, so by Ohm's law \(\displaystyle I=V/R\) halves: \[I' = \frac{V}{2R} = \frac{I}{2} = \frac{5\,\mathrm{A}}{2} = 2.5\,\mathrm{A} \] Answer: Doubling the length doubles the resistance; at constant voltage the current therefore falls to half.

  7. Exercise 25

    What is the commercial unit of electrical energy? Represent it in terms of joules.

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    NCERT’s answer
    kW h. $\displaystyle 1$ kW h = $\displaystyle 1000$ W × $\displaystyle 60$ × 60s = $\displaystyle 3.6$ × \(\displaystyle 10^{6}\) J
    The commercial (billing) unit of electrical energy is the kilowatt-hour, commonly called a "unit" of electricity. \[1\,\mathrm{kWh} = 1\,\mathrm{kW}\times1\,\mathrm{h} = 1000\,\mathrm{W}\times3600\,\mathrm{s} \] \[1\,\mathrm{kWh} = 3.6\times10^{6}\,\mathrm{J} \] Answer: \(\displaystyle 1\ \mathrm{kWh} = 3.6\times10^{6}\,\mathrm{J}\).

  8. Exercise 26

    A current of 1\displaystyle 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5Ω\displaystyle 5 \Omega when connected to a 10\displaystyle 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10Ω\displaystyle 10 \Omega is connected in parallel with this series combination, what change (if any) in current flowing through 5Ω\displaystyle 5 \Omega conductor and potential difference across the lamp will take place? Give reason.

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    NCERT’s answer
    (i)
    $\displaystyle 5$ Ω (ii) Hint— Calculate the total resistance of the circuit. There
    will be no change in current flowing through $\displaystyle 5$ Ω conductor. Also
    there will be no change in potential difference across the lamp either.
    In the series circuit, total resistance and the lamp's share come from Ohm's law: \[R_{total} = \frac{V}{I} = \frac{10\,\mathrm{V}}{1\,\mathrm{A}} = 10\,\Omega \] \[R_{lamp} = R_{total} - R = 10\,\Omega - 5\,\Omega = 5\,\Omega \] A \(\displaystyle 10\,\Omega\) resistor is now added in parallel with this whole series branch, both across the same $\displaystyle 10$ V battery.The branch still sees the full $\displaystyle 10$ V regardless of the new path, so nothing inside it changes: \[I_{5\Omega}' = \frac{V}{R_{total}} = \frac{10\,\mathrm{V}}{10\,\Omega} = 1\,\mathrm{A} \] \[V_{lamp}' = I_{5\Omega}'\times R_{lamp} = 1\,\mathrm{A}\times5\,\Omega = 5\,\mathrm{V} \] Answer: \(\displaystyle R_{lamp}=5\,\Omega\); both stay unchanged ($\displaystyle 1$ A, $\displaystyle 5$ V) — every parallel branch gets the full battery voltage independently.

  9. Exercise 27

    Why is parallel arrangement used in domestic wiring?

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    NCERT’s answer
    Hint— Provide the same potential difference across each electrical appliance.
    Appliances in a house are wired in parallel across the mains, not in series: \[I = I_1+I_2+I_3+\cdots, \qquad V_1=V_2=V_3=V \]So each appliance gets the full mains voltage, is switched by its own switch/fuse, and one failing or being turned off leaves the rest unaffected. Answer: Parallel wiring gives every appliance full mains voltage and an independent, individually-switchable path.

  10. Exercise 28

    B1, B2\displaystyle \mathrm{B}_1, \mathrm{~B}_2 and B3\displaystyle \mathrm{B}_3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
    (i)
    What happens to the glow of the other two bulbs when the bulb B1\displaystyle \mathrm{B}_1 gets fused?
    (ii)
    What happens to the reading of A1, A2, A3\displaystyle \mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3 and A when the bulb B2\displaystyle \mathrm{B}_2 gets fused?
    (iii)
    How much power is dissipated in the circuit when all the three bulbs glow together?
    NCERT_Question_Class10_Science_Exemplar_Ch12_Q28

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    NCERT’s answer
    Hint— (i) The glow of the bulbs \(\displaystyle B_{2}\) and \(\displaystyle B_{3}\) will remain the same.
    (ii)
    \(\displaystyle A_{1}\) shows $\displaystyle 1$ ampere, \(\displaystyle A_{2}\) shows zero, \(\displaystyle A_{3}\) shows $\displaystyle 1$ ampere and A
    shows $\displaystyle 2$ ampere
    (iii)
    P = V × I = $\displaystyle 4.5$ × $\displaystyle 3$ = $\displaystyle 13.5$ W
    Identical bulbs on the same $\displaystyle 4.5$ V share the main current equally:
    \[I_1=I_2=I_3=\frac{3\,\mathrm{A}}{3}=1\,\mathrm{A} \]
    (i)
    \(\displaystyle \mathrm{B}_1\) fuses: \(\displaystyle \mathrm{B}_2\) and \(\displaystyle \mathrm{B}_3\) sit on separate parallel branches, so they glow as before.
    (ii)
    \(\displaystyle \mathrm{B}_2\) fuses: its branch opens, so \(\displaystyle \mathrm{A}_2\to0\). \(\displaystyle \mathrm{A}_1\) and \(\displaystyle \mathrm{A}_3\) stay at \(\displaystyle 1\,\mathrm{A}\); the main meter reads only their sum:
    \[I=I_1+I_3=1\,\mathrm{A}+1\,\mathrm{A}=2\,\mathrm{A} \]
    (iii)
    All three bulbs glowing:
    \[P=VI=4.5\,\mathrm{V}\times3\,\mathrm{A}=13.5\,\mathrm{W} \]
    Answer: (i) \(\displaystyle \mathrm{B}_2\), \(\displaystyle \mathrm{B}_3\) unchanged. (ii) \(\displaystyle \mathrm{A}_2=0\); \(\displaystyle \mathrm{A}_1=\mathrm{A}_3=1\,\mathrm{A}\); the main meter falls from \(\displaystyle 3\,\mathrm{A}\) to \(\displaystyle 2\,\mathrm{A}\). (iii) \(\displaystyle P=13.5\,\mathrm{W}\).