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 formulaI = E / R if we express the current I in amperes, the voltage E in volts, and the resistance R in ohms. In this situation, we know that E = 9.00 V and R = 90.0 ohms, so I = 9.00 / 90.0 = 0.100 A = 100 mA 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 formulaR = E / I as long as we express R in ohms, E in volts, and I in amperes. We know that E = 13.5 V and I = 250 mA. When we convert the current to amperes, we get I = 0.250 A. Now we can use the formula to get R = 13.5 / 0.250 = 54.0 ohms 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 formulaE = I R when we express E in volts, I in amperes, and R in ohms. We know that I = 700 µA (700 microamperes), which translates to I = 0.000700 A. We also know that R = 1.25 k = 1250 ohms. Therefore E = 0.000700 x 1250 = 0.875 V = 875 mV 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,P = E I if we express the power P in watts, the voltage E in volts, and the current I in amperes. In this case, we have E = 32.0 V and I = 25.0 mA = 0.025 A, so P = 32.0 x 0.025 = 0.800 W = 800 mW 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 wasR = E / I = 32.0 / 0.025 = 1280 ohms We're told to quadruple the resistance R and, at the same time, double the voltage E. That means we'll end up with R = 1280 x 4 = 5120 ohms and E = 32.0 x 2 = 64.0 V We can calculate the power dissipated by the potentiometer as P = E 2 / R = 64.02 / 5120 = 4096 / 5120 = 0.800 W = 800 mW 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,Q = E I t when we express Q in watt-hours, E in volts, I in amperes, and t in hours. In this case, we're told that E = 12 V, I = 100 mA = 0.100 A, and t = 30 min = 0.50 h. Therefore Q = 12 x 0.100 x 0.50 = 0.60 Wh 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 asQ = 24.0 x 0.200 x 0.50 = 2.4 Wh The correct choice is D.