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NEET-XII-Physics

32: Electric Current in Conductors

with Solutions - page 5

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  • #11-a
    How many electrons are transferred per second between the supply and the wire at one end?
    Ans : Let q be the charge transferred per second and n be the number of electrons transferred per second.
    We know:
    `` i=\frac{V}{R}``
    `` \Rightarrow i=\frac{20\,\mathrm{\,V\,}}{{10}^{3}\,\mathrm{\,\Omega \,}}``
    `` \Rightarrow i=20\times {10}^{-3}=2\times {10}^{-2}\,\mathrm{\,A\,}``
    `` q\mathit{=}it``
    `` \Rightarrow q=2\times {10}^{-2}\times 1``
    `` \Rightarrow q=2\times {10}^{-2}\,\mathrm{\,C\,}``
    Also,
    q = ne
    `` \Rightarrow n=\frac{q}{e}=\frac{2\times {10}^{-2}}{1.6\times {10}^{-19}}``
    `` \Rightarrow n=1.25\times {10}^{17}``
  • #11-b
    Write down the current density in the wire.
    Ans : Current density of a wire,
    `` j=\frac{i}{A}``
    `` \Rightarrow j=\frac{2\times {10}^{-2}}{3.14\times {10}^{-8}}``
    `` \Rightarrow j=6.37\times {10}^{5}\,\mathrm{\,A\,}/{\,\mathrm{\,m\,}}^{2}``
    Page No 198:
  • Qstn #12
    Calculate the electric field in a copper wire of cross-sectional area 2.0 mm2 carrying a current of 1 A.
    The resistivity of copper = 1.7 × 10-8 Ω m
    Ans : Given:
    Area of cross-section, A = 2 × 10-6 m2
    Current through the wire, i = 1 A
    Resistivity of copper, ρ = 1.7 × 10-8 Ωm
    Resistance of a wire,
    `` R=\rho \frac{l}{A}``
    Also from Ohm's Law, voltage across a wire,
    `` V=iR=\frac{i\rho l}{A}``
    The electric field of the wire,
    `` E=\frac{V}{l}``
    `` \Rightarrow E=\frac{i\rho l}{Al}=\frac{i\rho }{A}``
    `` \Rightarrow E=\frac{1\times 1.7\times {10}^{-8}}{2\times {10}^{-6}}``
    `` \Rightarrow E=8.5\,\mathrm{\,V\,}/\,\mathrm{\,m\,}``
    Page No 198:
  • Qstn #13
    A wire has a length of 2.0 m and a resistance of 5.0 Ω. Find the electric field existing inside the wire if it carries a current of 10 A.
    Ans : Given:
    Length of the wire, l = 2 m
    Resistance of the wire, R = 5 Ω
    Current through the wire, i = 10 A
    From Ohm's Law, the potential difference across the wire,
    V = iR
    ∴ V= 10 × 5 = 50 V
    Electric field,
    `` \,\mathrm{\,E\,}=\frac{\,\mathrm{\,V\,}}{l}``
    `` =\frac{50}{2}``
    `` =25\,\mathrm{\,V\,}/\,\mathrm{\,m\,}``
    Page No 198:
  • Qstn #14
    The resistance of an iron wire and a copper wire at 20°C are 3.9 Ω and 4.1 Ω, respectively. At what temperature will the resistance be equal? Temperature coefficient of resistivity for iron is 5.0 × 10-3 K-1 and for copper, it is 4.0 × 10-3 K-1. Neglect any thermal expansion.
    Ans : Given:
    Resistance RFe of the iron wire at 20°C = 3.9 Ω
    Resistance RCu of the copper wire at 20°C = 4.1 Ω
    Temperature coefficient αFe for iron = 5.0 × 10-3 K-1
    Temperature coefficient αCu for copper = 4.0 × 10-3 K-1
    Let
    The temperature at which the resistance be equal = T
    Resistance of iron wire at T °C = RFe'
    Resistance of copper wire at T °C = RCu'
    We know:
    `` R={R}_{0}(1+\alpha ∆T)``
    Here, ΔT = T - 20
    `` \Rightarrow {R}_{\,\mathrm{\,Fe\,}}\text{'}={R}_{\,\mathrm{\,Fe\,}}\left[1+{\alpha }_{\,\mathrm{\,Fe\,}}\left(T-20\right)\right]``
    `` {R}_{\,\mathrm{\,Cu\,}}\text{'}={R}_{\,\mathrm{\,Cu\,}}\left[1+{\alpha }_{\,\mathrm{\,Cu\,}}\left(T-20\right)\right]``
    `` {R}_{\,\mathrm{\,Fe\,}}\mathit{\text{'}}\mathit{=}{R}_{\,\mathrm{\,Cu\,}}\text{'}``
    `` \mathit{\Rightarrow }{R}_{\,\mathrm{\,Fe\,}}\left[1+{\alpha }_{\,\mathrm{\,Fe\,}}\left(T-20\right)\right]={R}_{\,\mathrm{\,Cu\,}}\left[1+{\alpha }_{\,\mathrm{\,Cu\,}}\left(T-20\right)\right]``
    `` \Rightarrow 3.9\left[1+5\times {10}^{-3}\left(T-20\right)\right]=4.1\left[1+4\times {10}^{-3}\left(T-20\right)\right]``
    `` \Rightarrow 3.9+3.9\times 5\times {10}^{-3}\left(T-20\right)=4.1+4.1\times 4\times {10}^{-3}\left(T-20\right)``
    `` \Rightarrow 4.1\times 4\times {10}^{-3}\left(T-20\right)-3.9\times 5\times {10}^{-3}\left(T-20\right)=-4.1+3.9``
    `` \Rightarrow 16.4\left(T-20\right)-19.5\left(T-20\right)=-0.2\times {10}^{3}``
    `` \Rightarrow \left(T-20\right)\left(-3.1\right)=-0.2\times {10}^{3}``
    `` \Rightarrow T-20=64.5``
    `` \Rightarrow T=84.5°\,\mathrm{\,C\,}``
    `` ``
    `` ``
    Page No 198:
  • Qstn #15
    The current in a conductor and the potential difference across its ends are measured by an ammeter and a voltmeter. The meters draw negligible currents. The ammeter is accurate but the voltmeter has a zero error (that is, it does not read zero when no potential difference is applied). Calculate the zero error if the readings for two different conditions are 1.75 A, 14.4 V and 2.75 A, 22.4 V.
    Ans : Let the magnitude of zero error in the voltmeter reading be V.
    We need to subtract the zero error from the readings obtained under the two given conditions to obtain the true value of potential difference.
    Under both the conditions ,the resistance of the wire will not change.
    `` \Rightarrow {R}_{1}={R}_{2}``
    `` \Rightarrow \frac{{V}_{1}}{{I}_{1}}=\frac{{V}_{2}}{{I}_{2}}``
    `` \frac{{I}_{1}R}{{I}_{2}R}=\frac{{V}_{1}}{{V}_{2}}``
    `` \Rightarrow \frac{1.75}{2.75}=\frac{14.4-V}{22.4-V}``
    `` \Rightarrow \frac{0.35}{0.55}=\frac{14.4-V}{22.4-V}``
    `` \Rightarrow \frac{7}{11}=\frac{14.4-V}{22.4-V}``
    `` \Rightarrow 7\times \left(22.4-V\right)=11\left(14.4-V\right)``
    `` \Rightarrow 156.8-7V=158.4-11V``
    `` \Rightarrow \left(7-11\right)V=156.8-158.4``
    `` \Rightarrow -4V=-1.6``
    `` \Rightarrow \,\mathrm{\,V\,}=0.4\,\mathrm{\,V\,}``
    Magnitude of zero error, V = 0.4 V, which can either be positive or negative. Positive or negative zero error just indicates that the needle of the voltmeter is to the right or left of the zero marked on the device if zero voltage is applied across the voltmeter.
    Page No 198:
  • Qstn #16
    Figure (32-E2) shows an arrangement to measure the emf ε and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed, the voltmeter reading drops to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery.
    Figure 32-E2
    Ans : (a)
    When the switch S is opened, loop 2 will be open and no current will pass through the ammeter. Loop 1 will be closed; so, a current will flow through it. But since the voltmeter has very high resistance, compared to the internal resistance of the battery, the voltage-drop across the internal resistance can be ignored, compared to the voltage drop across the voltmeter. So, the voltage appearing across the voltmeter will be almost equal to the emf of the battery.
    ∴ ε = 1.52 V
    (b) When the switch is closed, current will pass through the circuit in loop 2. In this case, there will be a voltage drop across r due to current i flowing through it.
    Applying the loop rule, we get:
    ε - ir = 1.45
    ⇒ 1.52 - ir = 1.45
    ⇒ ir = 0.07
    ⇒ 1.r = 0.07
    r = 0.07 Ω
    Page No 198:
  • Qstn #17
    The potential difference between the terminals of a battery of emf 6.0 V and internal resistance 1 Ω drops to 5.8 V when connected across an external resistor. Find the resistance of the external resistor.
    Ans : Given:
    Emf of the battery, E = 6 V
    Internal resistance, r = 1 Ω
    Potential difference, V = 5.8 V
    Let R be the resistance of the external resistor.

    Applying KVL in the above circuit, we get:
    `` E\mathit{-}i\mathit{(}R\mathit{+}r\mathit{)}=0``
    `` \Rightarrow i\mathit{=}\frac{E}{R\mathit{+}r}``
    `` \mathit{=}\frac{6}{R\mathit{+}1}``
    Also,
    `` V\mathit{=}E\mathit{-}ir``
    `` \Rightarrow 5.8=6-\frac{6}{R+1}\times 1``
    `` \frac{6}{R+1}=0.2``
    `` R+1=\frac{6}{0.2}=30``
    `` R=29\,\mathrm{\,\Omega \,}``
    Page No 199:
  • Qstn #18
    The potential difference between the terminals of a 6.0 V battery is 7.2 V when it is being charged by a current of 2.0 A. What is the internal resistance of the battery?
    Ans : Given:
    Potential difference across the terminals of the battery, V = 7.2 V
    Emf of the battery, E = 6 V
    Current flowing through the circuit, i = 2 A
    As the battery is getting charged, the voltage drop across the terminals of the battery will be equal to the emf of the battery plus the voltage drop across the internal resistance of the battery.
    So, V = E + ir
    ⇒ 7.2 = 6 + 2 × r
    ⇒ 1.2 = 2r
    ⇒ r = 0.6 Ω
    Page No 199:
  • Qstn #19
    The internal resistance of an accumulator battery of emf 6 V is 10 Ω when it is fully discharged. As the battery gets charged up, its internal resistance decreases to 1 Ω.
    The battery in its completely discharged state is connected to a charger that maintains a constant potential difference of 9 V. Find the current through the battery
  • #19-a
    just after the connections are made and
    Ans : When the battery is being charged:
    Emf of the battery, E = 6 V
    Internal resistance of the battery, r = 10 Ω
    Potential difference, V = 9 V
    Net e.m.f. across the resistance while charging = 9 - 6 = 3 V
    `` \therefore \,\mathrm{\,Curre\,}\text{nt}=\frac{3}{10}=0.3\,\mathrm{\,A\,}``
  • #19-b
    after a long time when it is completely charged.
    Ans : When the battery is completely charged:
    Internal resistance, r' = 1 Ω
    `` \therefore \,\mathrm{\,Current\,}=\frac{3}{1}=3\,\mathrm{\,A\,}``
    Page No 199:
  • Qstn #20
    Find the value of i1/i2 in the figure (32-E3) if
  • #20-a
    R = 0.1 Ω
    Ans : For R = 0.1 Ω:
    Applying KVL in the given circuit, we get:

    `` 0.1{i}_{1}+1{i}_{1}-6+1{i}_{1}-6=0``
    `` \Rightarrow 0.1{i}_{1}+1{i}_{1}+1{i}_{1}=12``
    `` \Rightarrow {i}_{1}=\frac{12}{\left(2.1\right)}=5.71\,\mathrm{\,A\,}``
    `` ``
    Now, consider the given circuit.

    Applying KVL in the loop ABCDA, we get:
    `` 0.1{i}_{2}+1i-6=0``
    `` \Rightarrow 0.1{i}_{2}+i=6``
    `` \Rightarrow i=6-0.1{i}_{2}``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2\,\mathrm{\,i\,}-{i}_{2}=0``
    `` \Rightarrow 2\left[6-0.1{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow {i}_{2}=10\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=0.571``
  • #20-b
    R = 1 Ω and
    Ans : For R = 1 Ω:
    Applying KVL in the circuit given in figure 1, we get:
    `` 1{i}_{1}+1.{i}_{1}-6+{i}_{1}-6=0``
    `` \Rightarrow 3{i}_{1}=12``
    `` \Rightarrow {i}_{1}=4``
    `` ``
    Now, for figure 2:
    Applying KVL in ABCDA, we get:
    `` {i}_{2}+i-6=0``
    `` \Rightarrow {i}_{2}+i=6``
    Applying KVL in ADEFA, we get:
    `` i-6+6-\left({i}_{2}-i\right)1=0``
    `` \Rightarrow i-{i}_{2}+i=0``
    `` \Rightarrow 2i-{i}_{2}=0``
    `` \Rightarrow 2\left[6-{i}_{2}\right]-{i}_{2}=0``
    `` \Rightarrow 12-3{i}_{2}=0``
    `` \Rightarrow {i}_{2}=4\,\mathrm{\,A\,}``
    `` \therefore \frac{{i}_{1}}{{i}_{2}}=1``
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