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Chapter 12: Problem 1816

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) \(2V\).

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Most popular questions from this chapter

The resistivity of a potentiometer wire is $40 \times 10^{-8} \mathrm{ohm} \mathrm{m}\( and its area of cross-section is \)8 \times 10^{-6} \mathrm{~m}^{2}\(. If \)0.2$ amp current is flowing through the wire, the potential gradient will be. (A) \(10^{-2}\) Volt \(/ \mathrm{m}\) (B) \(10^{-1}\) Volt \(/ \mathrm{m}\) (C) \(3.2 \times 10^{-2}\) Volt \(/ \mathrm{m}\) (D) 1 Volt \(/ \mathrm{m}\)

The circuit shown in fig consists of the following \(E_{1}=6, E_{2}=2, E_{3}=3\) Volt \(\mathrm{R}_{1}=6, \mathrm{R}_{4}=3 \mathrm{Ohm}\) \(\mathrm{R}_{3}=4, \mathrm{R}_{2}=2 \mathrm{Ohm}\) $\mathrm{C}=5 \mu \mathrm{F}$$E_{1}=6 \mathrm{~V} \quad E_{2}=2 \mathrm{~V} \quad E_{3}=3 \mathrm{~V} \quad \mathrm{R}_{1}=6 \Omega$ \(R_{2}=2 \Omega \quad R_{3}=4 \Omega \quad R_{4}=3 \Omega\) The energy stored in the capacitor is. (A) \(4.8 \times 10^{-6} \mathrm{~J}\) (B) \(9.6 \times 10^{-6} \mathrm{~J}\) (C) \(1.44 \times 10^{-5} \mathrm{~J}\) (D) \(1.92 \times 10^{-5} \mathrm{~J}\)

In the circuit shown, the current sources are of negligible internal resistances. What is the potential difference between the points \(\mathrm{B}\) and \(\mathrm{A}\) ? (A) \(-4.0 \mathrm{~V}\) (B) \(4.0 \mathrm{~V}\) (C) \(-8.0 \mathrm{~V}\) (D) \(8.0 \mathrm{~V}\)

Formula for current flowing through a wire is \(I=6 t^{2}+4 t-2\) here \(\mathrm{t}\) is in second and I is an ampere. In this wire, what is the quantity of electric charge passing in time interval \(1 \mathrm{sec}\) to 2 sec? (A) \(8 \mathrm{C}\) (B) \(18 \mathrm{C}\) (C) \(20 \mathrm{C}\) (D) \(24 \mathrm{C}\)

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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