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Chapter 12: Problem 1774

What maximum power can be obtained from a battery of \(\mathrm{emf} \mathrm{E}\) and internal resistance \(\mathrm{r}\) connected with an external resistance \(\mathrm{R}\) ? (A) \(\left(E^{2} / 4 r\right)\) (B) \(\left(E^{2} / 3 r\right)\) (C) \(\left(E^{2} / 2 r\right)\) (D) \(\left(\mathrm{E}^{2} / \mathrm{r}\right)\)

Short Answer

Expert verified
The maximum power that can be obtained from a battery of emf E and internal resistance r connected with an external resistance R is \(P_{max} = \frac{E^{2}}{4r}\). The correct answer is (A).

Step by step solution

01

Write down the formula for power

Power (P) is given by the formula P = I²R, where I is the current through the external resistance R.
02

Write the expression for the current

Since we have a battery with emf E and internal resistance r, we can use Ohm's law to find the current (I) flowing through the circuit. When external resistance R is connected in the circuit, \( I = \frac{E}{r + R} \)
03

Write the expression for power using the current

Now substitute the expression for current in the power formula: \( P = \left(\frac{E}{r + R} \right)^{2}R \)
04

Apply the maximum power transfer theorem

According to the maximum power transfer theorem, for maximum power transfer, the external resistance should be equal to the internal resistance of the source. In other words, \( R = r \)
05

Find the maximum power

Substitute the value of R in the power formula: \( P_{max} = \left(\frac{E}{r + r} \right)^{2}r \) \( P_{max} = \left(\frac{E}{2r} \right)^{2}r \) \( P_{max} = \frac{E^{2}}{4r} \)
06

Find the correct option

Now, compare the result with the given options: (A) \(\left(E^{2} / 4 r\right)\): This option matches the result we obtained for maximum power. So the correct answer is (A).

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Most popular questions from this chapter

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) \(25 \%\)

A potentiometer wire has length \(10 \mathrm{~m}\) and resistance \(20 \Omega\). A \(2.5 \mathrm{~V}\) battery of negligible internal resistance is connected across the wire with an \(80 \Omega\) series resistance. The potential gradient on the wire will be: (A) \(2.5 \times 10^{-4} \mathrm{~V} / \mathrm{cm}\) (B) \(0.62 \times 10^{-4} \mathrm{~V} / \mathrm{mm}\) (C) \(1 \times 10^{-5} \mathrm{~V} / \mathrm{mm}\) (D) \(5 \times 10^{-5} \mathrm{~V} / \mathrm{mm}\)

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) \(1: 4\)

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

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at $500 \mathrm{~cm}$ on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The resistance of the voltmeter is (A) \(500 \Omega\) (B) \(290 \Omega\) (C) \(49 \Omega\) (D) \(20 \Omega\)
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