Question

When you connect an unknown resistor across the terminals of a 1.50 V D-cell battery having negligible internal resistance, you measure a current of 2.05 mA flowing through it. a. What is the resistance of this resistor? 731.7 Correct: Your answer is correct. Ω b. If you now place the resistor across the terminals of a 9.0V battery having no internal resistance, how much power dissipates across the resistor? .1107 Correct: Your answer is correct. W You now put the resistor across the terminals of a power supply with an unknown emf an internal resistance of 35 Ω and measure a current of 0.0036 A flowing through the resistor. c. What is the emf of the power supply? 2.76 Correct: Your answer is correct. V d. If the resistor is connected to the power supply for 18.0 minutes, how many kilowatt-hour is used by the resistor?

Answer 1

Answer:

Explanation:

Using ohm's law

a) V = IR where V is voltage in Volt, I is current in Ampere and R is resistance in ohms

R = V / I = 1.50 V/ ( 2.05 /1000) A = 731.71 ohms

b) Power = IV = $\frac{V}{R}$ × v = $\frac{V^{2} }{R}$ = $\frac{9^{2} }{731.71}$ = 0.1107 W

c) E = IR + Ir = ( 731.71 × 0.0036) + ( 35 × 0.0036) = 2.76 V

d) Power use by the resistor = I²R = 0.0036² × 731.71 = 0.00948 W = 0.00948 W = 0.000009483 kw × ( 18 / 60 ) H = 2.84 × 10⁻⁶ KW-H