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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) \(2V\).

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Most popular questions from this chapter

Two electric bulbs marked \(25 \mathrm{~W}-220 \mathrm{~V}\) and $100 \mathrm{~W}-220 \mathrm{~V}\( are connected in series to a \)440 \mathrm{~V}$ supply which of the bulbs will fuse? (A) \(100 \mathrm{~W}\) (B) \(25 \mathrm{~W}\) (C) None of this (D) Both

Figure, shows a network of seven resistors number 1 to 7, each equal to $1 \Omega\( connection to a \)4 \mathrm{~V}$ battery of negligible internal resistance The current I in the circuit is.... (A) \(0.5 \mathrm{~A}\) (B) \(1.5 \mathrm{~A}\) (C) \(2.0 \mathrm{~A}\) (D) \(3.5 \mathrm{~A}\)

A bulb of \(300 \mathrm{~W}\) and \(220 \mathrm{~V}\) is connected with a source of \(110 \mathrm{~V}\). What is the \(\%\) decrease in power? (A) \(100 . \%\) (B) \(75 \%\) (C) \(70 \%\) (D) \(25 \%\)

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.

The reading of ammeter shown in figure is.... (A) \(2.18 \mathrm{~A}\) (B) \(3.28 \mathrm{~A}\) (C) \(6.56 \mathrm{~A}\) (D) \(1.09 \mathrm{~A}\)
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