The ideal battery in Fig. 27-39 a has emf ℰ = 6.0 V. Plot 1 in Fig. 27-39 b gives the electric potential difference V that can appear across resistor 1 versus the current i in that resistor when the resistor is individually tested by putting a variable potential across it. The scale of the V axis is set by V s = 18.0 V. and the scale of the i axis is set by i s = 3.00 mA. Plots 2 and 3 are similar plots for resistors 2 and 3, respectively, when they are individually tested by putting a variable potential across them. What is the current in resistor 2 in the circuit of Fig. 27-39 a ? Figure 27-39 Problem 28.
The ideal battery in Fig. 27-39 a has emf ℰ = 6.0 V. Plot 1 in Fig. 27-39 b gives the electric potential difference V that can appear across resistor 1 versus the current i in that resistor when the resistor is individually tested by putting a variable potential across it. The scale of the V axis is set by V s = 18.0 V. and the scale of the i axis is set by i s = 3.00 mA. Plots 2 and 3 are similar plots for resistors 2 and 3, respectively, when they are individually tested by putting a variable potential across them. What is the current in resistor 2 in the circuit of Fig. 27-39 a ? Figure 27-39 Problem 28.
The ideal battery in Fig. 27-39a has emf ℰ = 6.0 V. Plot 1 in Fig. 27-39b gives the electric potential difference V that can appear across resistor 1 versus the current i in that resistor when the resistor is individually tested by putting a variable potential across it. The scale of the V axis is set by Vs = 18.0 V. and the scale of the i axis is set by is = 3.00 mA. Plots 2 and 3 are similar plots for resistors 2 and 3, respectively, when they are individually tested by putting a variable potential across them. What is the current in resistor 2 in the circuit of Fig. 27-39a?
2. A thin Nichrome wire is used in an experiment to test Ohm's
law using a power supply ranging from 0 to 12 V in steps of 2 V.
Why isn't the graph of I vs V linear?
1.
Nichrome wire does obey Ohm's law. Explain how that can that be true given the
results above
1. The average KE and temperature in Kelvin of the molecules of a gas are related by the
equation KE = 3/2 KT where k is the Boltzmann constant 1.38 x 10 m² kg s².
The diagram shows the energy levels for a Hydrogen atom.
Energy/eV
0.00
-1.51
3.39
13.58
Use this information to show that Hydrogen at room temperature will not emit light.
2. When hydrogen burns in oxygen 241.8 kJ of energy are released per mole. Show that this
reaction can produce light.
3. By using the fact that around any closed loop the sum of the EMFS = the sum of the PDs. Write
equations for the two loops shown in the cct below.
40
ΔΩ
I₂
4V
(loop1
20 (loop2) 2v
I+12
Use these equations to show that the current flowing through the 20 resistor is 0.75A
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DC Series circuits explained - The basics working principle; Author: The Engineering Mindset;https://www.youtube.com/watch?v=VV6tZ3Aqfuc;License: Standard YouTube License, CC-BY