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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 wire in a circular shape has \(10 \Omega\) resistance. The resistance per one meter is \(1 \Omega\) The resultant between \(A \& B\) is equal to \(2.4 \Omega\), then the length of the chord \(\mathrm{AB}\) will be equal to (A) \(2.4\) (B) 4 (C) \(4.8\) (D) 6

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 \(1 \Omega\)

In each of the following questions, match column \(\mathrm{I}\) and column II and select the correct match out of the four given choices Column I \(\quad\) Column II (a) The unit of electrical resistivity is (p) \(\mathrm{m}^{2} \mathrm{~S}^{-1} \mathrm{~V}^{-1}\) (b) The unit of current density is (q) \(\Omega^{-1} \mathrm{~m}^{-1}\) (c) The unit of electrical conductivity is (r) \(\mathrm{Am}^{-2}\) (d) The unit of electric mobility is (s) \(\Omega \mathrm{m}\) (A) \(a-p, b-q, c-r, d-s\) (B) \(a-s, b-r, c-q, d-p\) (C) \(a-r, b-q, c-p, d-s\) (D) \(a-q, b-r, c-s, d-p\)

A potentiometer wire of length \(1 \mathrm{~m}\) and resistance \(10 \Omega\) is connected in series with a cell of e.m.f \(2 \mathrm{~V}\) with internal resistance \(1 \Omega\) and a resistance box of a resistance \(R\) if potential difference between ends of the wire is \(1 \mathrm{~V}\) the value of \(R\) is. (A) \(4.5 \Omega\) (B) \(9 \Omega\) (C) \(15 \Omega\) (D) \(20 \Omega\)

How many dry cells, each of emf \(1.5 \mathrm{~V}\) and internal resistance $0.5 \Omega\(, much be joined in series with a resistor of \)20 \Omega$ to give a current of \(0.6 \mathrm{~A}\) in the circuit? (A) 2 (B) 8 (C) 10 (D) 12
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