Need Help figuring two things out??? ASAP - Electronic Circuits Projects Diagrams Free

BuzzKill · 25th November 2003, 06:33 AM

ok I have attached three schematics and need the 3 answers for each of the schematics I have attached? TIA

1. What is the correct total Capacitance of the schematic

BuzzKill · 25th November 2003, 06:34 AM

2. Total resistance of the circuit below?

BuzzKill · 25th November 2003, 06:36 AM

3. a) In the circuit below is the Diode Biased correctly?
b) If battery voltage is increased to 16V, what would the voltage come across the Diode?

Ravi · 25th November 2003, 07:21 AM

Here is an easy formula. In the first case, caps are connected in parallel. The effective capacitance of the network is simply equal to the sum of the individual capacitances.Hence, C=0.1+0.2+0.3

In the second case, resistors are in parallel. Hence the reciprocal of the effective resistance of the combination is equal to the sum of the reciprocals of the individual resistances. Therefore, 1/R= 1/100+1/100+1/100+1/100

BuzzKill · 25th November 2003, 07:25 AM

thanks man

BuzzKill · 25th November 2003, 07:27 AM

so the second is 400 ohms or is it 25 ohms?

Sebi · 25th November 2003, 07:31 AM

25ohms

If the battery symbol is standard, the zener is forward biased.If the battery positive upper side, the zener reverse biased and You can measure 10V. Increase the battery voltage not change the voltage, only increase the current.

Ravi · 25th November 2003, 07:43 AM

It is 25 Ohms. However,if there are only two resistors connected in parallel the formula could be re-arranged as:

R=R1XR2/R1+R2 that is, product of individual resistance divided by sum of individual resistance.

In general whenever two resistors of identical value are wired in parallel, the resistance of the combination will be equal to half that of each individual resistor, whilst the power rating of the combinations will be double that of each individual component.

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25th November 2003, 07:21 AM	(permalink)
Ravi	Here is an easy formula. In the first case, caps are connected in parallel. The effective capacitance of the network is simply equal to the sum of the individual capacitances.Hence, C=0.1+0.2+0.3 In the second case, resistors are in parallel. Hence the reciprocal of the effective resistance of the combination is equal to the sum of the reciprocals of the individual resistances. Therefore, 1/R= 1/100+1/100+1/100+1/100 __________________ \"Waste your money and you\'re only out of money,but waste your time and you\'re lost part of your life\"

25th November 2003, 07:25 AM	(permalink)
BuzzKill	thanks man

25th November 2003, 07:27 AM	(permalink)
BuzzKill	so the second is 400 ohms or is it 25 ohms?

25th November 2003, 07:43 AM	(permalink)
Ravi	It is 25 Ohms. However,if there are only two resistors connected in parallel the formula could be re-arranged as: R=R1XR2/R1+R2 that is, product of individual resistance divided by sum of individual resistance. In general whenever two resistors of identical value are wired in parallel, the resistance of the combination will be equal to half that of each individual resistor, whilst the power rating of the combinations will be double that of each individual component. __________________ \"Waste your money and you\'re only out of money,but waste your time and you\'re lost part of your life\"