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Chapter 12: Problem 1683

when a cell is connected to a resistance \(R_{1}\) the rate at which heat is generated in it is the same as when the cell is connected to a resistance \(\mathrm{R}_{2}\left(<\mathrm{R}_{1}\right)\) the internal resistance of the cell is.... (A) \(\left(R_{1}-R_{2}\right)\) (B) \((1 / 2)\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)\) (C) $\left\\{\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) /\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)\right.$ (D) \(\sqrt{R}_{1} R_{2}\)

Short Answer

Expert verified
The internal resistance of the cell is: \(r = \cfrac{R_1 R_2}{R_1 + R_2}\)

Step by step solution

01

Establish relationships for power in both cases

In case 1, the cell is connected to resistance R₁, and in case 2, it is connected to resistance R₂. Let's denote the internal resistance of the cell as r, the total resistance of the circuit in case 1 as R₁' and in case 2 as R₂'. Then: \(R_1' = R_1 + r\) \(R_2' = R_2 + r\) The power in each situation, using the formula P = I²R is: \(P_1 = I_1^2 R_1\) \(P_2 = I_2^2 R_2\) Since the problem states that the rate of heat generation in both cases is the same, we can set the power equal to each other. \(P_1 = P_2\) Now we have two equations representing the power in each case. We need to find a relationship between the currents and resistances to determine the internal resistance r.
02

Find current in terms of resistances in both cases

We can use Ohm's law, V = IR, to find the current in each case. For case 1, we have: \(I_1 = \cfrac{E}{R_1'}\), where E is the EMF of the cell. Similarly, for case 2, we have: \(I_2 = \cfrac{E}{R_2'}\) Now we can write the power equations in terms of resistances: \(P_1 = \left(\cfrac{E}{R_1'}\right)^2 R_1\) \(P_2 = \left(\cfrac{E}{R_2'}\right)^2 R_2\) Since P₁ = P₂: \(\left(\cfrac{E}{R_1'}\right)^2 R_1 = \left(\cfrac{E}{R_2'}\right)^2 R_2\)
03

Solve for internal resistance r

Since E is nonzero, we can divide both sides by E²: \(\cfrac{R_1}{R_1'^2} = \cfrac{R_2}{R_2'^2}\) Now substitute the expressions for total resistances found in Step 1: \(\cfrac{R_1}{(R_1 + r)^2} = \cfrac{R_2}{(R_2 + r)^2}\) To solve for r, cross-multiply and simplify: \(R_1 (R_2 + r)^2 = R_2 (R_1 + r)^2\) Expanding the squares and simplifying further, we have: \(R_1 R_2^2 + 2 R_1 R_2 r + R_1 r^2 = R_2 R_1^2 + 2 R_2 R_1 r + R_2 r^2\) Canceling equal terms and solving for r, we get: \(r = \cfrac{R_1 R_2}{R_1 + R_2}\) The correct answer is: (C) \(\left\\{\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) /\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)\right.)$

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Most popular questions from this chapter

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) \(8.0 \mathrm{~V}\)

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) \(16 \mathrm{~V}\)

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) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

The resistance of the wire made of silver at \(27^{\circ} \mathrm{C}\) temperature is equal to \(2.1 \Omega\) while at \(100^{\circ} \mathrm{C}\) it is \(2.7 \Omega\) calculate the temperature coefficient of the resistivity of silver. Take the reference temperature equal to \(20^{\circ} \mathrm{C}\) (A) \(4.02 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (B) \(0.402 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (C) \(40.2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\) (D) \(4.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)

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) \(120 \mathrm{~W}\)
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