
Calculate current through the circuit and potential difference across the circuit in the given figure. The drift current of the diode is \[20\mu\]A.
Answer
124.5k+ views
Hint: The current flowing through the diode and resistor are equal. The potential difference can be calculated by using ohm’s law. If the current flows in a closed circuit in the form of an electrical charge then the potential difference does not move or flow it is applied.
Complete step by step solution:
As evident from the above diagram, the diode and the resistor are in series. Therefore, the amount of current flowing through is equal to the drift current.
Given the drift current is\[20\mu A\].
Therefore, the current through the circuit is also\[20\mu A\].
Converting \[\mu A\] TO A, \[20\mu A = 20 \times {10^{ - 6}}\]
Employing ohm’s law to find the potential drop across the resistor.
Now, we know ohm’s law states that the current through the conductor between two pots is directly proportional to the voltage across the two points, given parameters like temperature, pressure being constant.
Therefore, we can say, voltage is proportional to the current flowing, let R be the resistor of the circuit and also the constant of proportionality.
\[V = IR\]
Where:
\[V = \] Potential Difference across two points
\[I = \] Current flowing through the points
\[R = \]Resistor of the circuit
Putting the values, in the above equation, WE GET:
\[V = 20 \times {10^{ - 6}} \times 20\]
Therefore, solving the equation, we obtain:
\[V = 0.4mV\]
The total voltage applied through the battery is \[5.0V\].
Therefore, the potential difference, across the diode, is total voltage applied to the circuit (via battery) minus the potential difference across the resistor:
\[V = (5.0 - 0.4 \times {10^{ - 3}})V\]
Thus, we obtain,
\[V = 4.9996V\]
Thus, is the required answer is:
Current through the circuit is: \[20\mu A\]
Potential Difference, across the resistor is: \[4.9996V\].
Note: Drift current is defined as the current caused by particles being pulled by an electric field. All units must be converted to their SI units, otherwise it may give an erroneous result. The potential difference is only expressed in Volt.
Complete step by step solution:
As evident from the above diagram, the diode and the resistor are in series. Therefore, the amount of current flowing through is equal to the drift current.
Given the drift current is\[20\mu A\].
Therefore, the current through the circuit is also\[20\mu A\].
Converting \[\mu A\] TO A, \[20\mu A = 20 \times {10^{ - 6}}\]
Employing ohm’s law to find the potential drop across the resistor.
Now, we know ohm’s law states that the current through the conductor between two pots is directly proportional to the voltage across the two points, given parameters like temperature, pressure being constant.
Therefore, we can say, voltage is proportional to the current flowing, let R be the resistor of the circuit and also the constant of proportionality.
\[V = IR\]
Where:
\[V = \] Potential Difference across two points
\[I = \] Current flowing through the points
\[R = \]Resistor of the circuit
Putting the values, in the above equation, WE GET:
\[V = 20 \times {10^{ - 6}} \times 20\]
Therefore, solving the equation, we obtain:
\[V = 0.4mV\]
The total voltage applied through the battery is \[5.0V\].
Therefore, the potential difference, across the diode, is total voltage applied to the circuit (via battery) minus the potential difference across the resistor:
\[V = (5.0 - 0.4 \times {10^{ - 3}})V\]
Thus, we obtain,
\[V = 4.9996V\]
Thus, is the required answer is:
Current through the circuit is: \[20\mu A\]
Potential Difference, across the resistor is: \[4.9996V\].
Note: Drift current is defined as the current caused by particles being pulled by an electric field. All units must be converted to their SI units, otherwise it may give an erroneous result. The potential difference is only expressed in Volt.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

Trending doubts
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

The formula of the kinetic mass of a photon is Where class 12 physics JEE_Main

JEE Main Login 2045: Step-by-Step Instructions and Details

Physics Average Value and RMS Value JEE Main 2025

Inertial and Non-Inertial Frame of Reference - JEE Important Topic

Charging and Discharging of Capacitor

Other Pages
Clemmenson and Wolff Kishner Reductions for JEE

JEE Main 2022 June 29 Shift 2 Question Paper with Answer Keys & Solutions

Current Loop as Magnetic Dipole and Its Derivation for JEE

Elastic Collisions in One Dimension - JEE Important Topic

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Keys & Solutions
