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Chapter 12: Problem 1760

A wire is bent in the form of a circle of radius \(4 \mathrm{~m}\) Resistance per unit length of wire is \(1 / \pi \Omega / \mathrm{m}\) battery of $6 \mathrm{~V}\( is connected between \)\mathrm{A}\( and \)\mathrm{B} \angle \mathrm{AOB}=90^{\circ}$ Find the current through the battery (A) \(8 \mathrm{~A}\) (B) \(4 \mathrm{~A}\) (C) \(3 \mathrm{~A}\) (D) \(9 \mathrm{~A}\)

Short Answer

Expert verified
The current through the battery is \(1.5 A\). However, this result is not among the given options, indicating a possible error in the problem statement or calculations.

Step by step solution

01

1. Determine length of wire segments AO and OB

Since angle AOB is 90 degrees and OA and OB are radii of the circle, the problem can be visualized as a quarter circle. We can calculate the length of segments AO and OB separately by using the formula for the circumference of a circle, which is \(C = 2 \pi r\). Since we are only concerned with a quarter of the circle, we divide the circumference by 4: Length of wire segment AO = Length of wire segment OB = \(\frac{2 \pi r}{4} = \frac{1}{2}\pi r\) Plug in the radius, r = 4 meters: Length of AO = Length of OB = \(\frac{1}{2}\pi(4) = 2\pi m\)
02

2. Calculate resistance of wire segments AO and OB

We're given that the resistance per unit length of wire is \(1 / \pi \Omega / \mathrm{m}\). To calculate the resistance of wire segments AO and OB, we multiply the resistance per unit length (Rpl) by their respective lengths: Resistance of AO = Resistance of OB = Rpl × Length = \(\frac{1}{\pi} \Omega/m \times 2\pi m = 2 \Omega\)
03

3. Determine total resistance

Since the wire segments AO and OB are in series (connected end-to-end) with the battery, we can find the total resistance by adding the resistances of AO and OB: Total resistance = Resistance of AO + Resistance of OB = \(2 \Omega + 2 \Omega = 4 \Omega\)
04

4. Apply Ohm's Law to find the current

Finally, we can use Ohm's Law to calculate the current through the battery. Ohm's Law is given by the equation \(V = IR\), where V is the voltage, I is the current, and R is the resistance. We want to find the current, I: I = \(\frac{V}{R}\) Replace V with 6 V and R with 4 Ω: I = \(\frac{6V}{4 \Omega} = 1.5 A\) Since 1.5 A is not among the given options, there may have been a miscalculation or the problem may have been presented incorrectly. However, the step-by-step procedure for solving the problem is shown above.

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Most popular questions from this chapter

The drift velocity of free electrons through a conducting wire of radius \(\mathrm{r}\), carrying current \(\mathrm{I}\), is \(\mathrm{V}_{\mathrm{d}}\) if the same current is passed through a conductor of radius \(2 \mathrm{r}\) what will be the drift velocity? (A) \(\left(\mathrm{V}_{\mathrm{d}} / 4\right)\) (B) \(\mathrm{V}_{\mathrm{d}}\) (C) \(2 \mathrm{~V}_{\mathrm{d}}\) (D) \(24 \mathrm{~V}_{\mathrm{d}}\)

Three identical resistors connected in series with a battery, together dissipate \(10 \mathrm{~W}\) of power. What will be the power dissipated, if the same resistors are connected in paralle1 across the same battery? (A) \(60 \mathrm{~W}\) (B) \(30 \mathrm{~W}\) (C) \(90 \mathrm{~W}\) (D) \(120 \mathrm{~W}\)

Two conductors have the same resistance at \(0^{\circ} \mathrm{C}\) but their temperature coefficients of resistances are \(\alpha_{1}\) and \(\alpha_{2}\) The respective temperature coefficients of their series and parallel combinations are nearly.... (A) $\alpha_{1}+\alpha_{2},\left\\{\left(\alpha_{1} \alpha_{2}\right) /\left(\alpha_{1}+\alpha_{2}\right)\right\\}$ (B) $\left\\{\left(\alpha_{1}+\alpha_{2}\right) / 2\right\\},\left\\{\left(\alpha_{1}+\alpha_{2}\right) / 2\right\\}$ (C) $\left\\{\left(\alpha_{1}+\alpha_{2}\right) / 2\right\\}, \alpha_{1}+a_{2}$ (D) \(\alpha_{1}+\alpha_{2},\left\\{\left(\alpha_{1}+a_{2}\right) / 2\right\\}\)

The drift velocity of free electrons in a conductor is \(\mathrm{V}\), when a current. \(I\) is flowing in it If both the radius and current are doubled, then drift velocity will be. (A) \((\mathrm{V} / 4)\) (B) \((\mathrm{V} / 2)\) (C) \(4 \mathrm{~V}\) (D) \(2 \mathrm{~V}\)

There are n resistors having equal value of resistance \(\mathrm{r}\). First they are connected in such a way that the possible minimum value of resistance is obtained. Then they are connected in such a way that possible maximum value of resistance is obtained the ratio of minimum and maximum values of resistances obtained in these way is.... (A) \((1 / n)\) (B) \(\mathrm{n}\) (C) \(\mathrm{n}^{2}\) (D) \(\left(1 / \mathrm{n}^{2}\right)\)
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