Figure (attached) shows a potentiometer circuit. The uniform wire AB is 100 cm l
ID: 1393868 • Letter: F
Question
Figure (attached) shows a potentiometer circuit. The uniform wire AB is 100 cm long and has a resistance of 0.038ohm/cm. The standard cell has an emf of 1.35V. The galvonmeter registers zero when the sliding contact is at L = 48.0cm.
a) If the sliding contact were moved to L = 56.0cm, in what direction would electrons flow through the standard cell? Explain.
b) What is the resistance of 48.0cm of the uniform wire?
c) What is the current through the wire AB?
d) What is the voltage across 68.0cm of the uniform wire?
e) What is the terminal voltage of the battery that is connected across AB?
f) Another dry cell, whose emf is 1.50V, is substitued for the standard cell. At what value of L will the galvonmeter read zero?
Explanation / Answer
a)
Take all the three cases;
1. When the sliding contact is less than 48 cm.
electrons will flow from the negative terminal to the positive terminal of the standard cell, which is the case when a battery is connected to a load
2. When sliding contact is at 48 cm.
At this point, the two voltages will balance and no electrons will flow.
3. When the sliding contact is at more than 48 cm
At this point the voltage across the length 56 cm will exceed that of the standard cell and electrons, now will flow in the opposite direction, ie from positive terminal to the negative terminal of the standard cell.
b)
The resistance per unit length is 0.038 ohm/cm
Resistance at 48 cm is 0.038 ohm/cm x 48 cm = 1.824 ohm.
c)
When the sliding contact is at 48 cm,
Resistance R = 1.824 ohm, and the current flowing through is balanced by a voltage of 1.35 V.
I x R = 1.35
I = 1.35/1.824 = 0.74 A.
d)
Resistance of 68 cm of wire = 0.038 ohm/cm x 68 cm = 2.584 ohm.
Voltage = current x resistance
= 0.74 x 2.584 = 1.912 V
e)
Terminal voltage connected across the battery = Current x Total resistance
Total resistance = 0.038 ohm/cm x 100 cm = 3.8 ohm.
Terminal voltage = 0.74 x 3.8 = 2.8125 V.
f)
The current flowing through AB is determined by the battery, which is I = 0.74 A.
When the standard cell of 1.35 V balances the voltage at 48 cm,
I x 0.038 ohm/cm x 48 cm= 1.35 V ...(1)
Suppose the standard cell of 1.5 V balance it at L cm, then
I x 0.038 ohm/cm x L cm = 1.5 V ...(2)
(1)/(2) give
48/L = 1.35/1.5
L = 53.33 cm
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