Electrical Formulas
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 – 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 $\times$ V  Coulomb (C) 
Capacitance 
C = $\frac{Q}{V}$

Farad (F) 
Inductance 
V_{L} = – L $\frac{di}{dt}$

Henry (L or H) 
Voltage  V = I R  Volt (V) 
Current 
I = $\frac{V}{R}$

Ampere (A) 
Resistance 
R = $\frac{V}{I}$

ohm ($\omega$) 
Power  P = VI  Watt (W) 
Conductance 
G = $\frac{1}{R}$

mho ($mho$) 
Impedance  Z^{2} = R^{2} + (x_{L} – x_{c})^{2 }  ohm ($\omega$) 
Resonant Frequency  f = $\frac{1}{2 \pi \sqrt{LC}}$ 
Hertz (Hz) 
Electrical Formulas helps us to calculate the parameters related to electricals in any electrical components.
Electrical Problems
Problem based on Electrical which may be helpful for you are given below.
Solved Examples
Question 1: A wire carrying a current of 4 Amperes is having resistance of 5 $\omega$. Calculate the potential difference across its ends.
Solution:
Given: Current I = 4 A,
Resistance R = 5 $\omega$
The Potential difference is given by V = IR
= 4 A $\times$ 5 $\omega$
= 20 V.
Solution:
Given: Current I = 4 A,
Resistance R = 5 $\omega$
The Potential difference is given by V = IR
= 4 A $\times$ 5 $\omega$
= 20 V.
Question 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 $\times$ 25 V
= 125 $\times$ 10^{3} C
= 0.125 C.
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 $\times$ 25 V
= 125 $\times$ 10^{3} C
= 0.125 C.