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circuit

The circuit is made of a variable resistor of 10kΩ value and two fixed resistors R1 and R2 .The output voltage V0 should be variable from 2V to 9V, i have to calculate the values of the two fixed resistors R1 and R2. I don't know how to start.

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A big hint (since this is a homework question).

The voltage varies from A to B, so the variation is (B - A) volts. That should give you a good idea as to what the current through the divider must be...

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Since this is a homework question, so the battery should be perfect with no internal resistor and the variable risistor should range from 0 to 10k. This come to these solution steps:

1, The V0 = 2V when the variable resistor is at zero, hence the equation: 2V = 12V*R2/(R1+R2+10k)

2, The V0 = 9V when the variable resistor is at its max, hence the second equation: 9V = 12V*(R2+10k)/(R1+R2+10k)

Hope this will help.

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  • \$\begingroup\$ Doesn't the denominator of your first equation need to include 10K, same as the second question? \$\endgroup\$ – tcrosley Jan 18 '16 at 1:51
  • \$\begingroup\$ Oh, that should be, since the other part still participate in the circuit. My error \$\endgroup\$ – Hung Nguyen Jan 18 '16 at 4:13
  • \$\begingroup\$ I don't think you can edit with a rep of just 1, so I corrected it. \$\endgroup\$ – tcrosley Jan 18 '16 at 4:15
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Start by assuming that the output current is approx zero, and think of the circuit as a pure voltage divider.

Since the extreme values are 2V and 9V, you can analyse each of those cases individually. Don't think of the pot as a pot, but just as a resistor where Vo is connected to one end or the other, and figure out the voltage-divider behaviour:

2 = 12*R2/(R1+R2+10k)

9 = 12*(R2+10k)/(R1+R2+10k)

That gives you a pair of simultaneous equations you can solve for R1 and R2.

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  • \$\begingroup\$ It's a homework question. \$\endgroup\$ – Chu Jan 18 '16 at 1:08

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