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A cell of internal resistance 5Ω is connected to a resistor of 3Ω. Find the current flowing through the resistor if the e.m.f. of the battery is 1.2V.

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1. A cell of internal resistance 5Ω5\Omega is connected to a resistor of 3Ω3\Omega. Find the current flowing through the resistor if the e.m.f. of the battery is 1.2V1.2\text{V}.

  1. Calculate the total resistance in the circuit.

    The internal resistance (rr) and the external resistance (RR) are in series.

    Rtotal=R+rR_{\text{total}} = R + r Rtotal=3Ω+5ΩR_{\text{total}} = 3\Omega + 5\Omega Rtotal=8ΩR_{\text{total}} = 8\Omega
  2. Use Ohm's Law to find the current.

    Ohm's Law states that V=IRV = IR, where VV is the e.m.f., II is the current, and RR is the total resistance.

    I=VRtotalI = \frac{V}{R_{\text{total}}} I=1.2V8ΩI = \frac{1.2\text{V}}{8\Omega} I=0.15AI = 0.15\text{A}

    The correct option is (a).

    0.15 A\boxed{\text{0.15 A}}

    2. What is the resistivity of a wire of length 5cm5\text{cm} and cross sectional area of 0.2m20.2\text{m}^2 if it offers a resistance of 10Ω10\Omega to the flow of current?

  3. Convert the length to meters.

    L=5cm=5×10−2m=0.05mL = 5\text{cm} = 5 \times 10^{-2}\text{m} = 0.05\text{m}
  4. Use the formula for resistance to find resistivity.

    The resistance RR of a wire is given by R=ρLAR = \rho \frac{L}{A}, where ρ\rho is the resistivity, LL is the length, and AA is the cross-sectional area.
    Rearrange the formula to solve for resistivity ρ\rho:

    ρ=R×AL\rho = \frac{R \times A}{L} ρ=10Ω×0.2m20.05m\rho = \frac{10\Omega \times 0.2\text{m}^2}{0.05\text{m}} ρ=2Ω⋅m20.05m\rho = \frac{2\Omega \cdot \text{m}^2}{0.05\text{m}} ρ=40Ω⋅m\rho = 40\Omega \cdot \text{m}

    The correct option is (b).

    40 Ω⋅m\boxed{\text{40 }\Omega \cdot \text{m}}

    3. A heating filament with p.d. of 20V20\text{V} across its ends has a current of 2A2\text{A} flowing through it for 3 seconds. Determine the heat energy generated in the filament if 80 percent of the electrical energy produced is converted to heat.

  5. Calculate the total electrical energy produced.

    The electrical energy (EE) produced is given by E=VItE = VIt, where VV is the potential difference, II is the current, and tt is the time.

    E=20V×2A×3sE = 20\text{V} \times 2\text{A} \times 3\text{s} E=120JE = 120\text{J}
  6. Calculate the heat energy generated.

    Only 80%80\% of the electrical energy is converted to heat.

    Heat energy=0.80×E\text{Heat energy} = 0.80 \times E Heat energy=0.80×120J\text{Heat energy} = 0.80 \times 120\text{J} Heat energy=96J\text{Heat energy} = 96\text{J}

    The correct option is (d).

    96 J\boxed{\text{96 J}}

    4. If the current flowing through a conductor is doubled and the resistance is constant, the power consumed by the conductor will increase to ..... times the original amount.

  7. Recall the formula for power.

    Power (PP) consumed by a conductor is given by P=I2RP = I^2R, where II is the current and RR is the resistance.

  8. Analyze the change in power when current is doubled.

    Let the original current be I1I_1 and the original power be P1P_1.

    P1=I12RP_1 = I_1^2 R

    If the current is doubled, the new current I2=2I1I_2 = 2I_1. The resistance RR remains constant.
    The new power P2P_2 will be:

    P2=I22RP_2 = I_2^2 R P2=(2I1)2RP_2 = (2I_1)^2 R P2=4I12RP_2 = 4I_1^2 R
  9. Compare the new power to the original power.

    P2=4(I12R)P_2 = 4 (I_1^2 R)

    Since P1=I12RP_1 = I_1^2 R, we have:

    P2=4P1P_2 = 4 P_1

    The power consumed will increase to 4 times the original amount.
    The correct option is (b).

    4\boxed{\text{4}}

    5. In a series circuit with two resistors of equal value and a total current flow of 12A12\text{A}, the current through each resistor is ............

  10. Understand current in a series circuit.

    In a series circuit, the current is the same at every point. This means the current flowing through each component (including each resistor) is equal to the total current flowing through the circuit.

  11. Determine the current through each resistor.

    Given that the total current flow is 12A12\text{A}, the current through each resistor in the series circuit will also be 12A12\text{A}. The fact that there are two resistors of equal value is relevant for calculating voltage drops or total resistance, but not for the current through individual components in a series circuit.
    The question does not provide options that include 12A12\text{A}. This suggests there might be a misunderstanding in the question or the provided options. However, based on the principles of series circuits, the current through each resistor must be the same as the total current. If this were a parallel circuit, the current would divide. Since it's explicitly stated as a series circuit, the current is constant.

    Given the options (a) 6A6\text{A} and (b) 5A5\text{A}, neither is 12A12\text{A}. If the question intended to ask about a parallel circuit where the total current of 12A12\text{A} splits equally between two identical resistors, then the current through each would be 6A6\text{A}. However, the question clearly states "series circuit".

    Assuming there might be a typo in the question and it meant to ask for current in a parallel circuit with two identical resistors and a total current of 12A12\text{A}:
    If it were a parallel circuit with two equal resistors, the total current would split equally between them.

    Ieach resistor=Itotal2=12A2=6AI_{\text{each resistor}} = \frac{I_{\text{total}}}{2} = \frac{12\text{A}}{2} = 6\text{A}

    If we strictly follow the question as written (series circuit), the answer should be 12A12\text{A}. Since 12A12\text{A} is not an option, and 6A6\text{A} is an option, it's highly probable the question implicitly assumes a parallel circuit or has a typo. Given the options, 6A6\text{A} is the most plausible intended answer if the question meant to imply current division.

    Let's assume the question intended to ask about a parallel circuit, as 6A6\text{A} is an option.
    In a parallel circuit, if two resistors are of equal value, the total current divides equally between them.

    Ieach resistor=Itotal2I_{\text{each resistor}} = \frac{I_{\text{total}}}{2} Ieach resistor=12A2I_{\text{each resistor}} = \frac{12\text{A}}{2} Ieach resistor=6AI_{\text{each resistor}} = 6\text{A}

    The correct option would be (a) under this assumption.

    However, if we strictly adhere to the "series circuit" statement:
    In a series circuit, the current is the same through all components.
    Therefore, the current through each resistor would be 12A12\text{A}. Since 12A12\text{A} is not an option, there is an inconsistency. I will proceed with the assumption that the question implies a scenario where the current divides, which would be a parallel circuit, given the options.

    Assuming the question implies a parallel circuit for the options to make sense:

    6 A\boxed{\text{6 A}}
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