Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 1815

In the circuit shown in fig the potential difference across \(3 \Omega\) is. (A) \(2 \mathrm{~V}\) (B) \(4 \mathrm{~V}\) (C) \(8 \mathrm{~V}\) (D) \(16 \mathrm{~V}\)

Short Answer

Expert verified
The potential difference across the \(3\Omega\) resistor is \(2.4V\), which is closest to option (A) \(2V\). Therefore, the answer is (A) \(2V\).

Step by step solution

01

Identify the series and parallel components of the circuit

Observe the circuit and identify which resistors are in series and which are in parallel. In this case, we can see that the \(3\Omega\) and \(6\Omega\) resistors are in parallel, and the combination is in series with the \(12\Omega\) resistor.
02

Calculate the equivalent resistance of the parallel components

For resistors connected in parallel, the equivalent resistance is given by: \[\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}.\] In our case, \(R_1 = 3\Omega\) and \(R_2 = 6\Omega\). Substitute the values and calculate the equivalent resistance of the parallel combination: \[\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2},\] so the equivalent resistance is: \[R_{eq} = 2\Omega.\]
03

Calculate the total resistance of the circuit

Now that we have found the equivalent resistance of the parallel combination, the total resistance in the circuit is the sum of the resistance of this combination and the resistance of the \(12\Omega\) resistor: \[R_{total} = 2\Omega + 12\Omega = 14\Omega.\]
04

Calculate the current flowing through the circuit

Using Ohm's Law (\(V = IR\)), we can find the current flowing through the circuit. We are given that the total voltage in the circuit is \(28V\): \[I = \frac{V}{R} = \frac{28V}{14\Omega} = 2A.\] The same current flows through the \(12\Omega\) resistor and the parallel combination of the \(3\Omega\) and \(6\Omega\) resistors.
05

Calculate the voltage across the parallel combination

The voltage across the parallel combination is the same as the voltage across the \(12\Omega\) resistor, as they are in series. Using Ohm's Law, we can calculate the voltage across the \(12\Omega\) resistor: \[V_{12\Omega} = I \times R_{12\Omega} = 2A \times 12\Omega = 24V.\]
06

Calculate the current through the \(3\Omega\) resistor

Using the current divider rule for parallel circuits, the current through the \(3\Omega\) resistor is given by: \[I_{3\Omega} = I_{total} \times \frac{R_{eq}}{R_{3\Omega} + R_{eq}} = 2A \times \frac{2\Omega}{3\Omega + 2\Omega} = 2A \times \frac{2}{5} = \frac{4}{5}A.\]
07

Calculate the potential difference across the \(3\Omega\) resistor

Finally, we can use Ohm's Law to calculate the potential difference across the \(3\Omega\) resistor: \[V_{3\Omega} = I_{3\Omega} \times R_{3\Omega} = \frac{4}{5}A \times 3\Omega = \frac{12}{5}A = 2.4V.\] The potential difference across the \(3\Omega\) resistor is \(2.4V\), which is closest to option (A) \(2V\). Therefore, the answer is (A) \(2V\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A circuit with an infinite no of resistance is shown in fig. the resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\), when $\mathrm{R}_{1}=1 \Omega\( and \)\mathrm{R}_{2}=2 \Omega$ will be (A) \(4 \Omega\) (B) \(1 \Omega\) (C) \(2 \Omega\) (D) \(3 \Omega\)

Two resistances \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have effective resistance \(\mathrm{R}_{\mathrm{s}}\) when connected in sires combination and \(R_{p}\) when connected in parallel combination if $\mathrm{R}_{8} \mathrm{R}_{\mathrm{p}}=16\( and \)\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)=4\( the values of \)\mathrm{R}_{1}\( and \)\mathrm{R}_{2}$ are (A) \(2 \Omega\) and \(0.5 \Omega\) (B) \(1 \Omega\) and \(0.25 \Omega\) (C) \(8 \Omega\) and \(2 \Omega\) (D) \(4 \Omega\) and \(1 \Omega\)

If \(\sigma_{1}, \sigma_{2}\), and \(\sigma_{3}\) are the conductance's of three conductor then equivalent conductance when they are joined in series, will be. (A) \(\sigma_{1}+\sigma_{2}+\sigma_{3}\) (B) $\left(1 / \sigma_{1}\right)+\left(1 / \sigma_{2}\right)+\left(1 / \sigma_{3}\right)$ (C) $\left\\{\left(\sigma_{1} \sigma_{2} \sigma_{3}\right) /\left(\sigma_{1}+\sigma_{2}+\sigma_{3}\right)\right\\}$ (D) None of these.

A wire of length \(L\) is drawn such that its diameter is reduced to half of its original diameter. If the resistance of the wire were \(10 \Omega\), its new resistance would be. (A) \(40 \Omega\) (B) \(60 \Omega\) (C) \(120 \Omega\) (D) \(160 \Omega\)

4 cell each of emf \(2 \mathrm{v}\) and internal resistance of \(1 \Omega\) are connected in parallel to a load resistor of \(2 \Omega\) Then the current through the load resistor is.... (A) \(2 \mathrm{~A}\) (B) \(1.5 \mathrm{~A}\) (C) \(1 \mathrm{~A}\) (D) \(0.888 \mathrm{~A}\)
See all solutions

Recommended explanations on English Textbooks

Text Comparison

Read Explanation

Pragmatics

Read Explanation

Listening and Speaking

Read Explanation

English Grammar Summary

Read Explanation

Argumentative Essay

Read Explanation

Semiotics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free