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The equivalent resistance of two identical resistors in parallel is 10 Ω. The equivalent resistance of the same resistors in series would be

Physics

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Answer

C

Step-by-step solution

9 steps

Here are the solutions for the next set of questions.

Part 15· 2 steps

  1. Determine the resistance of a single resistor from the parallel combination.

    Let RR be the resistance of each identical resistor. For two identical resistors in parallel, the equivalent resistance Req,pR_{eq,p} is given by:

    1Req,p=1R+1R=2R\frac{1}{R_{eq,p}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}

    We are given Req,p=10ΩR_{eq,p} = 10 \Omega.

    110Ω=2R\frac{1}{10 \Omega} = \frac{2}{R} R=2×10Ω=20ΩR = 2 \times 10 \Omega = 20 \Omega
  2. Calculate the equivalent resistance when these resistors are connected in series.

    For resistors in series, the equivalent resistance Req,sR_{eq,s} is the sum of individual resistances:

    Req,s=R+R=2RR_{eq,s} = R + R = 2R Req,s=2×20Ω=40ΩR_{eq,s} = 2 \times 20 \Omega = 40 \Omega

    The correct option is C.

    C. 40Ω\boxed{\text{C. } 40 \Omega}

Part 16· 3 steps

  1. Identify the given capacitances.

    The capacitance in air is Cair=20μFC_{air} = 20 \mu\text{F}.
    The capacitance with the dielectric is Cdielectric=42μFC_{dielectric} = 42 \mu\text{F}.

  2. Use the formula relating capacitance with and without a dielectric.

    The dielectric constant κ\kappa is defined as the ratio of the capacitance with the dielectric to the capacitance in air:

    κ=CdielectricCair\kappa = \frac{C_{dielectric}}{C_{air}}
  3. Calculate the dielectric constant.

    κ=42μF20μF\kappa = \frac{42 \mu\text{F}}{20 \mu\text{F}} κ=2.1\kappa = 2.1

    The correct option is C.

    C. 2.1\boxed{\text{C. } 2.1}

Part 17· 2 steps

  1. Identify the given values and the formula for work done.

    We are given the charge q=2.0 Cq = 2.0 \text{ C} and the potential difference V=12 VV = 12 \text{ V}.
    The work done WW in moving a charge qq against a potential difference VV is given by:

    W=qVW = qV
  2. Substitute the values and calculate the work done.

    W=(2.0 C)×(12 V)W = (2.0 \text{ C}) \times (12 \text{ V}) W=24 JW = 24 \text{ J}

    The correct option is C.

    C. 24 J\boxed{\text{C. } 24 \text{ J}}

Part 18· 2 steps

  1. Identify the given values and the formula for power dissipated.

    We are given the resistance R=50ΩR = 50 \Omega and the current I=2 AI = 2 \text{ A}.
    The power PP dissipated as heat in a resistor is given by:

    P=I2RP = I^2 R
  2. Substitute the values and calculate the power.

    P=(2 A)2×50ΩP = (2 \text{ A})^2 \times 50 \Omega P=4 A2×50ΩP = 4 \text{ A}^2 \times 50 \Omega P=200 WP = 200 \text{ W}

    The correct option is C.

    C. 200 W\boxed{\text{C. } 200 \text{ W}}

Part 19

Kirchhoff's first law, also known as Kirchhoff's Current Law (KCL) or the junction rule, states that the algebraic sum of currents entering a junction (or node) is equal to the algebraic sum of currents leaving the junction. This law is a direct consequence of the conservation of charges.
The correct option is D.

D. charges\boxed{\text{D. charges}}

Part 20

The amount of work done in bringing a unit positive charge from infinity to a point in an electric field is the definition of Electric potential.
The correct option is A.

A. Electric potential\boxed{\text{A. Electric potential}}
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