Ohm’s Law V=IR Solved Examples and Solution | Electric Current

Today’s Physics is having a lot of formulas and explanations, which are also applied in real life. From walking to operating current electricity everything is based on some principle or laws. There had been many scientist and researchers who had made many difficult things possible to us. Ohm’s Law is also one such law which defines many current electricity related theories. Being a very basic and important theorem, it is very much widely used in physics.

Ohm’s Law is one of the fundamental formula which is widely used in Electrical Circuits. Ohm’s Law defines that the Electric Current is directly dependent or proportional to voltage or potential difference across the circuit or the conductor. V=IR Law was first introduced while experimenting the behaviors of electricity conducted by the metals. It was developed by the German Physicist in 1827 (Georg Ohm). Here we will discuss about some of the popular V=IR Solved Examples.

Ohm’s Law Equation

Ohm’s Law basically defined the relationship between three parameters i.e resistance, voltage and current.


  • where V is the Voltage or potential difference in volts (V)
  • I is the current applied in amperes (I)
  • R is the resistance in Ω (Omega)

So, solver can rearrange the parameters as per her or her own need. You can get more familiar to ohm’s law by solving the V=IR Solved Examples.

For example if current needs to be calculated than:

I= \frac{V}{R}

if Resistance(R) needs to be calculated than:

R= \frac{V}{I}

V=IR Ohm’s Law Solved Examples

Below you can find some types of V=IR Solved Examples and solution, which you can practice and have the idea.

Ques 1. In a circuit, Voltage across the circuit is 10V. Resistance applied to the circuit is 5 Ohms. What is the Current flowing through the circuit?

Ans. As we know V=IR; So rearranging to calculate Current (I) we will have:

I= \frac{V}{R} = \frac{10 V}{5 \Omega } = 2 ASo, 2 Ampere current is flowing in the circuit.

Ques 2. Voltage applied across the circuit is 220 Volts while the current flowing is 5 A. Find the Resistance applied in the circuit.

Ans. Applying V=IR, we get

R= \frac{V}{I} = \frac{220 V}{5 A} = 44 \Omega

Ques 3. A circuit with 11 KΩ resistance has 750 μA current flowing inside it. What is the voltage drop across the resistance applied?

Ans. Ohm’s Law states that V=IR so, V = 750 μA * 11 KΩ = 8.25 V.

Ques 4. While measuring current in a particular circuit a voltage drop was noticed. Initially the current was  40 mA operated at 18 V. Later the current was dropped to 20 mA. Calculate the voltage dropped.

Ans. Applying Ohm’s Law V=IR;

R= \frac{V}{I} = \frac{18 V}{40 mA} = 450 \Omega

Now using this Resistance Value to calculate the Voltage when current was 20 mA.

V= IR = \frac{20 mA}{450 \Omega} = 9 V

Voltage Dropped = V1 – V2 = 18 V – 9 V = 9 Volts.

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