(a) In Fig. 27-18a, with R 1 > R 2 , is the potential difference across R 2 more than, less than, or equal to the across R 1 ? (b) Is the current through resistor R 2 more than, less than, or equal to that through resistor R 1 ? Figure 27-18 Questions 1 and 2
(a) In Fig. 27-18a, with R 1 > R 2 , is the potential difference across R 2 more than, less than, or equal to the across R 1 ? (b) Is the current through resistor R 2 more than, less than, or equal to that through resistor R 1 ? Figure 27-18 Questions 1 and 2
(a) In Fig. 27-18a, with R1>R2, is the potential difference across R2 more than, less than, or equal to the across R1? (b) Is the current through resistor R2 more than, less than, or equal to that through resistor R1?
Figure 27-18 Questions 1 and 2
Expert Solution & Answer
To determine
To find:
a) The potential difference across R2 more than, less than or equal to that of across R1
b) The current across R2 more than, less than or equal to that of across R1
Answer to Problem 1Q
Solution:
a) The potential difference across R2 is same that of across R1.
b) The current across R2 is more than that of across R1.
Explanation of Solution
1) Concept:
If the resistances are connected in parallel, then the voltage across the resistances is same only the current gets divided. And the value of the current depends only on the resistance because the voltage is same in parallel arrangement.
V=IR where V is voltage, I is current and R is resistance.
As in parallel arrangement V is same so I=1R means greater the resistance, less is current and vise- versa.
2) Formula:
I=VR
3) Explanation:
a)
Here, in fig27-18a, the resistance R1&R2 are in parallel arrangement. So, the voltage difference across them is same.
b)
As the voltage is same in parallel arrangement, the current depends only on the resistance from Ohm’s law.
I=VR
V is constant for parallel arrangement. So,
I∝1R
in fig27-18a, the resistance R1>R2 hence,I2>I1 because the current is inversely proportional to the resistance.
Conclusion:
We can compare the current and the potential difference across the resistances using the concept of parallel arrangement of the resistances and the equation of Ohm’s law.
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Consider a image that is located 30 cm in front of a lens. It forms an upright image 7.5 cm from the lens. Theillumination is so bright that that a faint inverted image, due to reflection off the front of the lens, is observedat 6.0 cm on the incident side of the lens. The lens is then turned around. Then it is observed that the faint,inverted image is now 10 cm on the incident side of the lens.What is the index of refraction of the lens?
2. In class, we discussed several different flow scenarios for which we can make enough
assumptions to simplify the Navier-Stokes equations enough to solve them and obtain
an exact solution. Consulting the cylindrical form of the Navier-Stokes equations copied
below, please answer the following questions.
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a.) In class, we discussed how the Navier-Stokes equations are an embodiment of Newton's
2nd law, F = ma (where bolded terms are vectors). Name the 3 forces that we are considering in
our analysis of fluid flow for this class.
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do not simply state "it is the derivative of a with…
1. Consult the form of the x-direction Navier-Stokes equation below that we discussed in
class. (For this problem, only the x direction equation is shown for simplicity). Note that
the equation provided is for a Cartesian coordinate system. In the spaces below, indicate
which of the following assumptions would allow you to eliminate a term from the
equation. If one of the assumptions provided would not allow you to eliminate a
particular term, write "none" in the space provided.
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Flow is in the horizontal direction (e.g. patient lying
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We consider the flow to be between two flat,
infinitely wide plates
There is no pressure gradient
Flow is axisymmetric
Term(s) in equation
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