(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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103 In Fig. 27-83, E, = 6.00 V, E, =
12.0 V, R, = 200 N, and R, = 100 N.
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R.
What are the (a) size and (b) direction
(up or down) of the current through
resistance 1, the (c) size and (d) direc-
tion of the current through resistance
2, and the (e) size and (f) direction of
the current through battery 2?
Figure 27-83 Problem 103.
In Fig. 27-26, the ideal batterieshave emfs E1=150 V and E2=50 Vand the resistances are R1 = 3.0 0 andR2 = 2.0 0. If the potential at P is 100 V,what is it at Q?
6 Res-monster maze. In Fig. 27-21, all the resistors have a resis-
tance of 4.0 0 and all the (ideal) batteries have an emf of 4.0 V.
What is the current through resistor R? (If you can find the proper
loop through this maze, you can answer the question with a few
seconds of mental calculation.)
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