Numerical Problems Based on Definition of Electric Current for Class 12 Physics

by
  • Last modified on:2 days ago
  • Reading Time:14Minutes
Home CBSE Class 12 Physics Numerical Problems Definition of Electric Current

Electric Current

The physics of charges at rest is called electrostatics or static electricity. We shall now study the motion or dynamics of charges. As the term current implies some sort of motion, so the motion of electric charges constitutes an electric current.

Understanding Electric current

If two bodies charged to different potentials are connected together by means of a conducting wire, charges begin to flow from one body to another. The charges continue to flow till the potentials of the two bodies become equal.

Tire flow of electric charges through a conductor constitutes an electric current. Quantitatively, electric current in a conductor across an area held perpendicular to the direction of flow of charge is defined as the amount of charge flowing across that area per unit time.

SI Unit:

SI emit of current is ampere (A). If one coulomb of charge crosses an area in one second, then the current through that area is one ampere (A).

Note:

1 milliampere = 1 mA = 10-6 A
1 microampere = 1 μA = 10-6 A

    Numerical Problems on Current Electricity

    Class 12 Physics · Current Electricity
    SQ

    Definition of Electric Current

    Apply $I = q/t$, $q = ne$, and $I = ef$
    Ch 3 · Current Electricity
    1
    One billion electrons pass from a point P towards another point Q in $10^{-3}\text{ s}$. What is the current in ampere? What is its direction?
    Answer $1.6 \times 10^{-7}\text{ A}$, from Q to P 📝
    Detailed Solution

    Number of electrons, $n = 1\text{ billion} = 10^9$

    Time, $t = 10^{-3}\text{ s}$; Elementary charge, $e = 1.6 \times 10^{-19}\text{ C}$

    The magnitude of the current ($I$) is the rate of flow of charge:

    $$I = \frac{q}{t} = \frac{ne}{t}$$
    $$I = \frac{10^9 \times 1.6 \times 10^{-19}}{10^{-3}} = \frac{1.6 \times 10^{-10}}{10^{-3}}$$
    $$I = 1.6 \times 10^{-7}\text{ A}$$

    By convention, the direction of electric current is opposite to the direction of flow of negatively charged electrons. Since electrons flow from P to Q, the current flows from Q to P.

    2
    If $2.25 \times 10^{20}$ electrons pass through a wire in one minute, find the magnitude of the current flowing through the wire. [Punjab 02]
    Answer 0.6 A 📝
    Detailed Solution

    Number of electrons, $n = 2.25 \times 10^{20}$

    Time, $t = 1\text{ minute} = 60\text{ s}$; Elementary charge, $e = 1.6 \times 10^{-19}\text{ C}$

    The current ($I$) is calculated as:

    $$I = \frac{ne}{t}$$
    $$I = \frac{2.25 \times 10^{20} \times 1.6 \times 10^{-19}}{60}$$
    $$I = \frac{3.6 \times 10^1}{60} = \frac{36}{60} = 0.6\text{ A}$$
    3
    A solution of sodium chloride discharges $6.1 \times 10^{16}\text{ Na}^+$ ions and $4.6 \times 10^{16}\text{ Cl}^-$ ions in 2 s. Find the current passing through the solution.
    Answer $8.56 \times 10^{-3}\text{ A}$ 📝
    Detailed Solution

    In an electrolyte, both positive and negative ions contribute to the total current. Since they move in opposite directions, their equivalent charge transfer adds up.

    Total number of charge carriers, $n = n_{\text{Na}^+} + n_{\text{Cl}^-} = 6.1 \times 10^{16} + 4.6 \times 10^{16} = 10.7 \times 10^{16}$

    Total charge, $q = ne = 10.7 \times 10^{16} \times 1.6 \times 10^{-19}\text{ C}$

    Time, $t = 2\text{ s}$

    $$I = \frac{q}{t} = \frac{10.7 \times 10^{16} \times 1.6 \times 10^{-19}}{2}$$
    $$I = \frac{17.12 \times 10^{-3}}{2} = 8.56 \times 10^{-3}\text{ A}$$
    4
    An electric current of $2.0\text{ }\mu\text{A}$ exists in a discharge tube. How much charge flows across a cross-section of the tube in 5 minutes?
    Answer $6.0 \times 10^{-4}\text{ C}$ 📝
    Detailed Solution

    Current, $I = 2.0\text{ }\mu\text{A} = 2.0 \times 10^{-6}\text{ A}$

    Time, $t = 5\text{ minutes} = 5 \times 60 = 300\text{ s}$

    The total charge ($q$) that flows is:

    $$q = I \times t$$
    $$q = 2.0 \times 10^{-6} \times 300$$
    $$q = 600 \times 10^{-6} = 6.0 \times 10^{-4}\text{ C}$$
    5
    In a hydrogen atom, the electron makes about $0.6 \times 10^{16}$ revolutions per second around the nucleus. Determine the average current at any point on the orbit of the electron.
    Answer 0.96 mA 📝
    Detailed Solution

    Frequency of revolution, $f = 0.6 \times 10^{16}\text{ rev/s} = 0.6 \times 10^{16}\text{ Hz}$

    Charge of an electron, $e = 1.6 \times 10^{-19}\text{ C}$

    The revolving electron constitutes a circular current loop. The current is defined as the charge passing a given point per unit time, which is charge times frequency:

    $$I = e \times f$$
    $$I = 1.6 \times 10^{-19} \times 0.6 \times 10^{16}$$
    $$I = 0.96 \times 10^{-3}\text{ A} = 0.96\text{ mA}$$
    6
    An electron moves in a circular orbit of radius 10 cm with a constant speed of $4.0 \times 10^6\text{ ms}^{-1}$. Determine the electric current at a point on the orbit.
    Answer $1.02 \times 10^{-12}\text{ A}$ 📝
    Detailed Solution

    Radius, $r = 10\text{ cm} = 0.1\text{ m}$; Speed, $v = 4.0 \times 10^6\text{ ms}^{-1}$

    Charge of an electron, $e = 1.6 \times 10^{-19}\text{ C}$

    The time period of revolution ($T$) is the distance covered in one orbit divided by speed: $T = \frac{2\pi r}{v}$.

    The current ($I$) is charge divided by the time period:

    $$I = \frac{e}{T} = \frac{ev}{2\pi r}$$
    $$I = \frac{1.6 \times 10^{-19} \times 4.0 \times 10^6}{2 \times 3.1416 \times 0.1} = \frac{6.4 \times 10^{-13}}{0.62832}$$
    $$I \approx 10.18 \times 10^{-13}\text{ A} = 1.018 \times 10^{-12}\text{ A} \approx 1.02 \times 10^{-12}\text{ A}$$
    7
    In a hydrogen discharge tube, the number of protons drifting across a cross-section per second is $1.1 \times 10^{18}$, while the number of electrons drifting in the opposite direction across another cross-section is $3.1 \times 10^{18}$ per second. Find the current flowing in the tube.
    Answer 0.672 A 📝
    Detailed Solution

    Number of protons per second, $n_p = 1.1 \times 10^{18}$

    Number of electrons per second, $n_e = 3.1 \times 10^{18}$

    Protons moving in one direction and electrons moving in the opposite direction both contribute to a conventional current in the same direction (the direction of proton flow). We add their effective charge transfer rates together.

    Total elementary charges passing per second, $n = n_p + n_e$:

    $$n = 1.1 \times 10^{18} + 3.1 \times 10^{18} = 4.2 \times 10^{18}$$

    Total Current ($I$) is:

    $$I = n \times e = 4.2 \times 10^{18} \times 1.6 \times 10^{-19}$$
    $$I = 6.72 \times 10^{-1} = 0.672\text{ A}$$

    Related Posts

    Also check

    Class-wise Contents

    Leave a Reply

    Join Telegram Channel

    Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

    Download Now!

    Discover more from Gurukul of Excellence

    Subscribe now to keep reading and get access to the full archive.

    Continue reading