
In the circuit shown in the figure, find the current in \[45\Omega \].

A. \[4\;{\rm{A}}\]
B. \[2.5\;{\rm{A}}\]
C. \[2\;{\rm{A}}\]
D. None of these
Answer
124.8k+ views
Hint: In this case, we have to calculate the voltage drops across the resistance elements using Ohm's Law. The values of these currents are then entered into the Current law expression to obtain a suitable value for current. We'll keep simplifying the circuit by employing equivalent resistances instead of a resistor combination. On the last simplest circuit, we will discover the current, divide it into branches, and then find the current via each branch.
Formula used:
The formula for finding the current in \[45\Omega \] are:
\[v = i \times R\]
Equivalent resistance in series is given by
\[R = {R_1} + {R_2} + ....... + {R_9}\]
Equivalent resistance in parallel is given by
\[\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ....... + \dfrac{1}{{{R_n}}}\]
Complete step by step solution:
We have been given the circuit diagram in the question. Now, let us split the circuit into branches for easy understanding, we obtain

In the given circuit, we have been understood that voltage is
\[180\;{\rm{V}}\]
The sign “I” is commonly used to represent current. Ohm's law states that the current flowing through a conductor is proportional to the voltage V and resistance R.
That is, \[{\rm{V = IR}}\]
Ohm's law can also be stated as,
\[{\rm{i = }}\dfrac{{\rm{V}}}{{\rm{R}}}\]
Now, we have to substitute the values given in the circuit in the above formula, we get
\[i = \dfrac{{180}}{{90}}\]
Now, let’s simplify the above expression, we get
\[{\rm{i = 2A}}\]
Therefore, the current in \[45\Omega \] is \[{\rm{2}}\] Ampere.
Hence, the option C is correct.
Note: Keep in mind that current does not always divide evenly between the two branches of the circuit. It is determined by the amount of resistance between them. The current was divided equally in this case because the resistance on the two branches of the circuit was equal.
Formula used:
The formula for finding the current in \[45\Omega \] are:
\[v = i \times R\]
Equivalent resistance in series is given by
\[R = {R_1} + {R_2} + ....... + {R_9}\]
Equivalent resistance in parallel is given by
\[\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ....... + \dfrac{1}{{{R_n}}}\]
Complete step by step solution:
We have been given the circuit diagram in the question. Now, let us split the circuit into branches for easy understanding, we obtain

In the given circuit, we have been understood that voltage is
\[180\;{\rm{V}}\]
The sign “I” is commonly used to represent current. Ohm's law states that the current flowing through a conductor is proportional to the voltage V and resistance R.
That is, \[{\rm{V = IR}}\]
Ohm's law can also be stated as,
\[{\rm{i = }}\dfrac{{\rm{V}}}{{\rm{R}}}\]
Now, we have to substitute the values given in the circuit in the above formula, we get
\[i = \dfrac{{180}}{{90}}\]
Now, let’s simplify the above expression, we get
\[{\rm{i = 2A}}\]
Therefore, the current in \[45\Omega \] is \[{\rm{2}}\] Ampere.
Hence, the option C is correct.
Note: Keep in mind that current does not always divide evenly between the two branches of the circuit. It is determined by the amount of resistance between them. The current was divided equally in this case because the resistance on the two branches of the circuit was equal.
Recently Updated Pages
JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

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 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

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

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
