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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

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) \(1.92 \times 10^{-5} \mathrm{~J}\)

The network is made of uniform wire. The resistance of portion EL is $2 \Omega\(. Find the resistance of star between points \)F \& C .$ (A) \(0.985 \Omega\) (B) \(1.25 \Omega\) (C) \(1.946 \Omega\) (D) \(1.485 \Omega\)

The temperature co-efficient of resistance of a wire is $0.00125^{\circ} \mathrm{k}^{-1}\(. Its resistance is \)1 \Omega\( at \)300 \mathrm{~K}$. Its resistance will be \(2 \Omega\) at. (A) \(1400 \mathrm{~K}\) (B) \(1200 \mathrm{~K}\) (C) \(1000 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)

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) \(\left(1 / \mathrm{n}^{2}\right)\)

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) \(4.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)
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