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Chapter 12: Problem 1726

How many dry cells, each of emf \(1.5 \mathrm{~V}\) and internal resistance $0.5 \Omega\(, much be joined in series with a resistor of \)20 \Omega$ to give a current of \(0.6 \mathrm{~A}\) in the circuit? (A) 2 (B) 8 (C) 10 (D) 12

Short Answer

Expert verified
\(10\) cells are required to produce a current of \(0.6A\) in the circuit (C).

Step by step solution

01

Understanding the problem

The problem presents a circuit with multiple cells in series, along with an external resistor. Every cell in the circuit has an emf of 1.5V and an internal resistance of 0.5 Ohms. The circuit has an additional resistance of 20 Ohms. The aim is to find out the number of cells needed to generate a current of 0.6A.
02

Formulate the equation

The total emf of the cells (\(E_{\text{Total}}\)) and the total resistance of the circuit (\(R_{\text{Total}}\)) can be used to get an equation for the current (I) from Ohm's law: \(I = E_{\text{Total}} / R_{\text{Total}}\). Since the cells are in series, the total emf and the total resistance add up: \(E_{\text{Total}} = nE\) (where \(E\) is the emf of a single cell and \(n\) is the number of cells) and \(R_{\text{Total}} = nR + R_e\) (where \(R\) is the internal resistance of a single cell and \(R_e\) is the external resistance).
03

Substitute known values and solve

We know that \(E = 1.5V\), \(R = 0.5\Omega\), \(R_e = 20\Omega\), and \(I = 0.6A\). We substitute these values into the equation from step 2 and solve for \(n\): \[ \begin{align*} 0.6 &= \frac{n(1.5)}{n(0.5)+20} \\ 0.6n(0.5) + 20(0.6) &= n(1.5) \\ 0.3n + 12 &= 1.5n \\ 12 &= 1.2n \\ n &= 12 / 1.2 \\ n &= 10. \end{align*} \] Hence, the answer is (C) - 10 cells are required to produce a current of 0.6A in the circuit.

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Most popular questions from this chapter

Two heater wires of equal length are first connected in series and then in parallel The ratio of heat produced in the two cases is (A) \(2: 1\) (B) \(1: 2\) (C) \(4: 1\) (D) \(1: 4\)

The tungsten filament of bulb has resistance equal to \(18 \Omega\) at \(27^{\circ} \mathrm{C}\) temperature \(0.25 \mathrm{~A}\) of current flows, when \(45 \mathrm{~V}\) is connected to it If $\alpha=4.5 \times 10^{-3} \mathrm{~K}^{-1}$ for a tungsten then find the temperature of the filament. (A) \(2160 \mathrm{~K}\) (B) \(1800 \mathrm{~K}\) (C) \(2070 \mathrm{~K}\) (D) \(2300 \mathrm{~K}\)

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.

Which of the following set up can be used to verify the ohm's law?

An electric kettle has two coils when one of these is switched on. the water in the kettle boils in 6 minutes. When the other coil is switched on, boils in 3 minutes If the two coils are connected in parallel, the time taken to boil water in the kettle is. (A) 3 minutes (B) 6 minutes (C) 2 minutes (D) 9 minutes
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