can someone please help me with both of these questions Open-Response Homework P
ID: 1428890 • Letter: C
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can someone please help me with both of these questions
Open-Response Homework Problem 8.3 A 12V battery is connected to two parallel combinations of resistors. The two parallel combinations are connected in series. The first parallel combination. the one at the top of the page, is made up of a 200ft. 300ft. and a 500ft resistor. The second combination is made up of three 300ft resistors. The two parallel combinations are connected in series with the battery. Compute the equivalent resistance. R_eq, of the circuit. Compute the current drawn by the circuit. Compute the power dissipated by the 200ft resistor. Open-Response Homework Problem 8.4 Consider the system of resistors to the right. The battery voltage is Delta V_o = 12V. Even numbered resistors. R_2, R_4. and R_6 are 2ft and odd number resistors. R_1, R_3. and R_5 are 4Ohm. What is the equivalent resistance of circuit? What is the current through R_5? What is the current through R_3? What is A Vi, the potential difference across R_4? How much power does R_4 consume?Explanation / Answer
8.3)
a.)Resistance R1, R2 and R3 are in parallelso equivalent resisitance (R') = 1/(1/R1 + 1/R2 + 1/R3 ) = 96.774 ohm
eqivalent resistance(R") of R4,R5 and R6 = 1/(1/R4 + 1/R5 + 1/R6 ) = 100 ohm
Resistance R' and R" are in series so net resistance (R) = R" +R' = 96.774 + 100 = 197.774 ohm
b.) Current (I) = V/R = 12/ 197.774 = 0.0606 A
c.) current through 200 ohm resistor = I/ (5/4) = 0.04848 A
Power dissipiated = I^2 * R = 0.4700 watt
8.4)
a.) Equivalaent resistance(R') of R2 and R3 = R2R3/(R2+R3) = 1.33 ohm
now, R1 ,R' ,R4,R5 and R6 are conncetd in series
so net Resistance = R1+R'+R4+R5+R6 = 4 +1.33+2+4+2 = 13.33 ohm
b.) current in circuit == V/R = 12/13.33 = 0.900 A
current in R5 = 0.900 A connected in series so current will remain same
c.) current through R3 = I /(3/2) = 0.600 A
d.) potential difference acros R4 = IR4 = 0.900*2 = 1.8 V
e.) Power = I^2 * R = 0.900^2 * 2 = 1.62 watt
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