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Chapter 18: Problem 96

Chelsea inadvertently bumps into a set of batteries with an emf of $100.0 \mathrm{V}\( that can supply a maximum power of \)5.0 \mathrm{W} .$ If the resistance between the points where she contacts the batteries is $1.0 \mathrm{k} \Omega,$ how much current passes through her?

Short Answer

Expert verified
Question: Calculate the current passing through a person when she comes into contact with a set of batteries with an emf of 100 V and a maximum power transfer of 5 W, given that the resistance at the contact points is 1.0 k惟. Answer: The current passing through the person is 0.0667 A or 66.7 mA.

Step by step solution

01

Recall the maximum power transfer theorem formula

The Maximum power transfer theorem states that for a load to get the maximum power from a source, the internal resistance of the source should be equal to the resistance of the load. The maximum power transfer formula is given by: \(P_{max} = \frac{V^2}{4R_i}\), where \(P_{max}\) is the maximum power, \(V\) is the emf, and \(R_i\) is the internal resistance of the battery.
02

Calculate the internal resistance of the battery

Using the maximum power transfer formula, we can find the internal resistance of the battery as follows: \(R_i = \frac{V^2}{4P_{max}}\), Plug in the values: \(V = 100V\), \(P_{max} = 5W\), \(R_i = \frac{(100.0)^2}{4(5.0)}\), \(R_i = \frac{10000}{20} = 500\ \Omega\), So, the internal resistance of the battery is \(500\ \Omega\).
03

Calculate the total resistance

The total resistance in the circuit is the sum of the internal resistance of the battery and the resistance between the contact points where Chelsea touches the batteries. Let's denote the resistance at the contact points as \(R_C\), which is given to be \(1.0\ k\Omega = 1000\ \Omega\). Thus, the total resistance is: \(R_{total} = R_i + R_C = 500\ \Omega + 1000\ \Omega = 1500\ \Omega\).
04

Calculate the current passing through Chelsea

Now, we can use Ohm's Law to determine the current passing through Chelsea: \(I = \frac{V}{R_{total}}\), Plug in the values: \(V = 100V\), \(R_{total} = 1500\ \Omega\), \(I = \frac{100V}{1500\ \Omega} = 0.0667A\). Therefore, the current passing through Chelsea is \(0.0667\ A\) or \(66.7\ mA\).

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Most popular questions from this chapter

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