Resistors in Series and Parallel and Potential Dividers Resistors in Series and Parallel Potential...
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Transcript of Resistors in Series and Parallel and Potential Dividers Resistors in Series and Parallel Potential...
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Resistors in Series and Parallel and Potential Dividers
• Resistors in Series and Parallel• Potential Dividers
![Page 2: Resistors in Series and Parallel and Potential Dividers Resistors in Series and Parallel Potential Dividers.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0131a28abf838ccc9e4/html5/thumbnails/2.jpg)
Rule for resistors in series
R
V
Resistors in Series
• If you put Extra Resistance in Series with the same PD, the current will reduce
R2R1
I
V1 V2
• R = V / I The Resistance will get higher
• The Total resistance is the sum of the individual resistances in series
R = R1 + R2 + …
III
![Page 3: Resistors in Series and Parallel and Potential Dividers Resistors in Series and Parallel Potential Dividers.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0131a28abf838ccc9e4/html5/thumbnails/3.jpg)
Rule for resistors in parallel
R
V
Resistors in Parallel
• If you add more resistors in parallel, for the same PD the current increases
I
• R = V / I • The Resistance will get lower • The Total Resistance is always less than the
resistance of the least resistive path1/R= 1/R1 + 1/R2 + …
Note: the resistance in each case is the total resistance of each path.
R2
R1I1
I2
I
![Page 4: Resistors in Series and Parallel and Potential Dividers Resistors in Series and Parallel Potential Dividers.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0131a28abf838ccc9e4/html5/thumbnails/4.jpg)
…then work out the resistance of these two effective resistors in series
…then work out the effective resistance of the two branches
Work out the resistance of this branch…
How you would solve these…
4
2 3
Questions from APfY Page 216, Q4, 6. Tough ext. Q2 P274
Work out the effective resistance of the two branches…
4
2
3
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Current in parallel branches…
Questions from APfY Page 217, Q7, 8 & 3
4
2
1I
• All of the current flows through the 2 resistor• 4x as much current will flow through the 1 resistor than the 4
resistor• 4/5 of the total will flow through the 1 resistor and 1/5 of the
total will flow through the 4 resistor
4
2 3I = 9mA
• In groups, state and explain which resistor will have the highest current
• Now work out how much current flows through the each resistor.
Ratio of 4:1
4+1=5
4/5 of total current
1/5 of total current
Ratio of 4:5
4+5=9
4/9 of total current = 4mA
5/9 of total current = 4mA
![Page 6: Resistors in Series and Parallel and Potential Dividers Resistors in Series and Parallel Potential Dividers.](https://reader036.fdocuments.in/reader036/viewer/2022083009/5697c0131a28abf838ccc9e4/html5/thumbnails/6.jpg)
Potential Dividers• From last session we know that
– The current is the same in all parts of the circuit
– The 6V PD is shared between the resistors
• The 3 resistor is 3/5 of the total resistance
• Because the current is the same in all parts of the circuit and V = I R then…
• …the PD across the 3 resistor will be 3/5 of the total PD
23
6V
V
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Experiment• Start off by getting a battery, a voltmeter
and 4 leads• Return to your desk and measure the PD
of the battery• Predict the PD across all resistors on
some of the circuits• Go and measure them• Note: 2K2 is the way that a 2.2k resistor
is represented.
• Now have a look at Q16-18 on Page 217