Electricity Demystified, 2nd edition |

Stan Gibilisco |

Explanations for Quiz Answers in Chapter 3 |

1. When we know the voltage and the resistance in a simple DC circuit, Ohm's Law tells
us that we can divide voltage by resistance to get current. We can use the formula
if we express the current
The correct answer is C. |

2. In a simple DC circuit such as this one, the current varies inversely in proportion to the resistance if we leave the voltage constant, and directly in proportion to the voltage if we keep the resistance constant. The voltage doesn't change in this situation, but the resistance doubles. Therefore, the current drops to half of its previous value. The correct choice is B. |

3. When we know the voltage and the current in a simple DC circuit, Ohm's Law allows
us to divide voltage by current to get resistance using the formula
as long as we express
The correct answer is A. |

4. In a simple DC circuit, the resistance varies inversely in proportion to the current if we keep the voltage constant, and directly in proportion to the voltage if we keep the current constant. We don't vary the voltage in this situation, but we do change the resistance to get three times as much current as we had before. In order to triple the current, we must reduce the potentiometer setting to 1/3 of its previous resistance. The correct choice is C. |

5. If we know the current and the resistance in a simple DC circuit, we can use Ohm's
law to calculate the voltage with the formula
when we express
The correct choice is D. |

6. If we double the voltage in a DC circuit while leaving all other factors constant, we double the current. If we cut the resistance in half while leaving all other factors constant, we also double the current. In this situation, we do both of the foregoing operations, so we "double the current twice over." In other words, we quadruple the current. The correct choice is C. |

7. When constant DC flows through a resistance from a voltage source, the dissipated
power equals the voltage times the current. That is,
if we express the power
The correct choice is A. |

8. Let's start by calculating the resistance in the scenario of Question 7. We had E
= 32.0 and I = 0.025 A, so the resistance was
We're told to quadruple the resistance
and
We can calculate the power dissipated by the potentiometer as
That's the same amount of power that we got in the solution to Question 7. Therefore, the correct choice is A; the dissipated power remains unchanged. |

9. In a simple DC circuit, the dissipated energy Q equals the product of the
voltage E, the current I, and the time t. That is,
when we express
The correct choice is B. |

10. We're told to double the battery voltage but leave the resistance unchanged. That
means we'll end up doubling the current through the resistance. (Remember, in a simple DC
circuit, the current through a constant resistance varies in direct proportion to the
voltage.) We now have E = 24.0 V and I = 200 mA = 0.200 A. If t
= 30 min = 0.50 h (the same time period as in the situation of Question 9), we can
calculate the new energy value as
The correct choice is D. |