The battery voltage VB is 9 volts. In the absence of an external load, the outpu

Question

The battery voltage VB is 9 volts.
In the absence of an external load, the output
voltage Vo is 1.0 volts.
If the output terminals are short-circuited, the circuit is to take a current of not more than
9 mA from the battery.
The value of the resistance R1 must be as small as possible.

Work out the minimum value of R2, and find the corresponding value of R1.
If a load resistance RL with a value equal to R1 is connected to the output, workout the combined resistance of RL and R1, therefore allowing you to find the new value of the output voltage.

Explanation / Answer

Which one is R1 and which is R2? If I assume R1 is the ground side... R2 min = 9v/9ma = 1k when you short the output, all you have is the 9v battery and a resistor, so simple ohms law. current in divider is 8/1k = 8 mA when divider has no load, the output is 1 volt, so the ik has 8 volts across it. R1 is then 1v/8ma = 125 ohms two 125 ohms in parallel is 62.5 ohms New Vo is 9•62.5/1062.5 = 0.529 volts

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