Electrical is the branch of Physics dealing with electricity, electronics and electromagnetism. Electrical formulas play a great role in finding the parameter value in any electrical circuits. Most commonly used electrical formulas are formulas related to voltage, current, power, resistance etc.

Volt is a unit of electrical potential or motive force – the potential is required to send one ampere of current through one ohm of resistance. Watt is a unit of electrical energy or power – one watt is the product of one ampere and one volt – one ampere of current flowing under the force of one volt gives one watt of energy

Below are given some commonly used electrical formulas which may be helpful for you.

Quantity Formula Unit
Charge Q = C
\(\begin{array}{l}\times\end{array} \)
V		 Coulomb (C)
Capacitance C =
\(\begin{array}{l}\frac{Q}{V}\end{array} \)
 Farad (F)
Inductance VL = – L
\(\begin{array}{l}\frac{di}{dt}\end{array} \)
 Henry (L or H)
Voltage V = I R Volt (V)
Current I =
\(\begin{array}{l}\frac{V}{R}\end{array} \)
 Ampere (A)
Resistance R =
\(\begin{array}{l}\frac{V}{I}\end{array} \)
 ohm (
\(\begin{array}{l}\omega\end{array} \)
)
Power P = VI Watt (W)
Conductance G =
\(\begin{array}{l}\frac{1}{R}\end{array} \)
 mho (
\(\begin{array}{l}mho\end{array} \)
)
Impedance Z2 = R2 + (xL – xc)2

 

 ohm (
\(\begin{array}{l}\omega\end{array} \)
)
Resonant Frequency f =
\(\begin{array}{l}\frac{1}{2 \pi \sqrt{LC}}\end{array} \)
 Hertz (Hz)

Electrical Formulas helps us to calculate the parameters related to electricals in any electrical components.

Solved Examples

Example 1: A wire carrying a current of 4 Amperes is having a resistance of 5

\(\begin{array}{l}\omega\end{array} \)
. Calculate the potential difference across its ends.
Solution:
Given: Current I = 4 A,
Resistance R = 5
\(\begin{array}{l}\omega\end{array} \)
The Potential difference is given by V = IR
= 4 A
\(\begin{array}{l}\times\end{array} \)
5
\(\begin{array}{l}\omega\end{array} \)
= 20 V.

Example 2: Calculate the charge across the capacitor 5mF and the voltage applied is 25 V.
Solution:
Given: Capacitance of the capacitor C = 5 mF,
Voltage applied V = 25 V,
The Charge across the capacitor is given by Q = CV
= 5 mF

\(\begin{array}{l}\times\end{array} \)
25 V
= 125
\(\begin{array}{l}\times\end{array} \)
10-3 C
= 0.125 C.

Stay tuned with BYJU’S for more such interesting articles. Also, register to “BYJU’S – The Learning App” for loads of interactive, engaging Physics-related videos and an unlimited academic assist.

Comments

Leave a Comment

Your Mobile number and Email id will not be published.

*

*

  1.  Elizah Micah
    February 24, 2021 at 4:13 am

    Happy to join your organization

    Reply