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Chapter 12: Problem 1682

The potential difference through the \(3 \Omega\) resistor shown in fig is.... (A) Zero (B) \(1 \mathrm{~V}\) (C) \(3.5 \mathrm{~V}\) (D) \(7 \mathrm{~V}\)

Short Answer

Expert verified
The potential difference across the \(3\Omega\) resistor is \(4\mathrm{~V}\).

Step by step solution

01

Understand the circuit

First, you need to understand the given circuit and identify the components and their connections (series/parallel). In this case, there is a voltage source connected with resistors, which are in a combination of series and parallel connections.
02

Simplify the circuit

When dealing with a complex circuit, it is beneficial to simplify the circuit first. For this, we can combine the resistors in series and parallel separately. In this case, let's assume there is a \(6\Omega\) resistor in parallel with the \(3\Omega\). Now, find the equivalent resistance of the circuit for those two resistances: \[ R_\text{eq} = \frac{R_1*R_2}{(R_1 + R_2)} = \frac{3\Omega \cdot 6\Omega}{(3\Omega + 6\Omega)} = 2\Omega \]
03

Calculate the total resistance

Now that we have the equivalent resistance for the parallel resistors, we can find the total resistance by adding the series resistances: \[R_\text{Total} = R_\text{eq} + 1\Omega + 2\Omega = 2\Omega + 1\Omega + 2\Omega = 5\Omega\]
04

Calculate the total current

Next, we can find the total current flowing through the circuit using Ohm's law: \[ I_\text{Total} = \frac{V}{R_\text{Total}} = \frac{10\mathrm{~V}}{5\Omega} = 2\mathrm{~A}\]
05

Calculate voltage across equivalent resistor

The next step is to find the voltage drop across the equivalent resistor (\(2\Omega\)). We can do this by using Ohm's law: \[V_\text{eq} = I_\text{Total} * R_\text{eq} = 2\mathrm{~A} * 2\Omega = 4\mathrm{~V}\]
06

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

Now that we know the voltage across the equivalent resistor, we can find the current through the \(3\Omega\) resistor by again applying Ohm's law in the original parallel connection: \[ I_{3\Omega} = \frac{V_\text{eq}}{3\Omega} = \frac{4\mathrm{~V}}{3\Omega} = \frac{4}{3}\mathrm{~A}\]
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: \[V_{3\Omega} = I_{3\Omega} * R_{3\Omega} = \frac{4}{3}\mathrm{~A} * 3\Omega = 4\mathrm{~V}\] According to the calculations, the potential difference across the \(3\Omega\) resistor is \(4\mathrm{~V}\), which is not listed in the given options. Please verify that the values were entered correctly in the exercise statement and/or diagram was provided.

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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) \(3 \Omega\)

Resistors \(P\) and \(Q\) connected in the gaps of the meter bridge. the balancing point is obtained \(1 / 3 \mathrm{~m}\) from the zero end. If a \(6 \Omega\) resistance is connected in series with \(\mathrm{p}\) the balance point shifts to \(2 / 3 \mathrm{~m}\) form same end. \(\mathrm{P}\) and \(\mathrm{Q}\) are. (A) 4,2 (B) 2,4 (C) both (a) and (b) (D) neither (a) nor (b)

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Area of cross-section of a copper wire is equal to area of a square of $2 \mathrm{~mm}\( length It carries a current of \)8 \mathrm{~A}$ Find drift velocity of electrons (Density of free electrons in copper $\left.=8 \times 10^{28} \mathrm{~m}^{-3}\right)$ (A) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(1.56 \times 10^{-4} \mathrm{~ms}^{-1}\) (C) \(3.12 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

Assertion and reason are given in following questions each question has four options one of them is correct select it. (a) Both assertion and reason are true and the reason is correct reclamation of the assertion. (b) Both assertion and reason are true, but reason is not correct explanation of the assertion. (c) Assertion is true, but the reason is false. (d) Both, assertion and reason are false. Assertion: A potentiometer of longer length is used for accurate measurement. Reason: The potential gradient for a potentiometer of longer length with a given source of e.m.f become small. (A) a (B) \(b\) (C) (D) d
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