query_24_060714

course PHY 202

?I??s???????????assignment #024

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?????€???????Physics II

07-15-2006

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13:09:07

Query problem set 3 #'2 1-6.

How do we determine the current in the circuit and the voltage across each resistor when we know the voltage across a series combination of two known resistances?

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RESPONSE -->

first you add the two resistance

then to find the current you take the voltage that is known and then divide it by the resistance this gives you the current. Then to find the voltage you take the current multiplied by the one of the resistance and then you do the same thing for the other resistor.

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13:09:20

** To get the current calculate I = V / R, where R is the sum of the two resistances.

To get the voltage across each resistor calculate V = I * R for each resistor. **

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RESPONSE -->

right

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13:13:18

How do we determine the current and voltage across each resistor when we know the voltage across a parallel combination of two known resistances?

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RESPONSE -->

I1 = V/R1 first current I2 = V/R2 second current

totale current I = I1 + I2 = V/R1 + V/ R2

power across resistors V * I1 = V * V/R1 = V^2/ R1 and V* I1 = V*V/ R2 = V^2 / R2

power across the circuit will be V*I = V*V*(1/R1 + 1/ R2)

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13:13:43

** The voltage across both resistors is the same and is equal to the voltage across the combination.

The current in each resistor is calculated by I = V / R.

The total current is the sum of the two currents. **

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RESPONSE -->

ok

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13:16:13

A series circuit contains a capacitor of known capacitance and a resistor of known resistance. The capacitor was originally uncharged before the source voltage was applied, and is in the process of being charged by the source. If we know the charge on the capacitor, how do we find the current through circuit?

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RESPONSE -->

capacitor hold charge Q then it coltage is C = Q/Vcap so Vcap = Q/C

Voltage across the resistor will be Vr=Vs-Vcap

current throught the resistor is I = (Vs - Vcap)/ R

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13:16:21

The voltage across the capacitor is equal to the charge divided by capacitance.

The voltage across the capacitor opposes the voltage of the source.

Since the voltage drop around the complete circuit must be zero, the voltage across the resistor is the difference between source voltage and the voltage across the capacitor.

Dividing the voltage across the resistor by the resistance we obtain the current thru the circuit.

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RESPONSE -->

right

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13:22:54

If we know the capacitance and initial charge on a capacitor in series with a resistor of known resistance then how to we find the approximate time required for the capacitor to discharge 1% of its charge through the circuit?

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RESPONSE -->

capacitor C = Q/ Vcap so Vcap = Q/C

Vr = Vs - Vcap

I = (Vs-Vcap)/ R = (Vs- Q/C)/ R

I = (Vs-Q/ C)/ R

rate I= dQ/dt

0.01Q 'dt = 0.1Q/ I= 0.1Q/[(Vs-Q/C)/R] = 0.01Q*R/(Vs-Q/C)

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13:23:09

** From capacitance and initial charge we find the voltage.

From the voltage and the resistance we find the current.

We take 1% of the initial charge and divide it by the current to get the approximate time required to discharge 1% of the charge.

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This result is a slight underestimate of the time required since as the capacitor discharges the current decreases. **

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RESPONSE -->

right

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