A 2.00 ohm resistor (R_1), a 6.00 ohm resistor (R_2) and a 3.00 ohm resistor (R_

Question

A 2.00 ohm resistor (R_1), a 6.00 ohm resistor (R_2) and a 3.00 ohm resistor (R_3) are connected as shown with a 12.0 V battery (V). Assume the internal resistance and the resistance of the wires are all negligible. (a) What is the effective resistance of the circuit? (b) What is the current through the battery? (c) What is the current through the 2.00 ohm resistor? (d) What is the current through the 6.00 ohm resistor? (e) What is the current through the 3.00 ohm resistor? (f) How will the total power supplied by the battery compare to the total power dissipated by the resistors? For the circuit above, find the voltage dropped by each resistor.

Explanation / Answer

R2 and R3 are in parallel so effective resistance (R') = R2R3/(R2+R3) = 6*3/9 = 2 ohm

a.)R' and R1 are connected in series (R) = R'+R1 = 2 + 2 = 4 ohm

b.)current through circuit (I) = V/R = 12/4 = 3 A

c.) current through 2 ohm resistor = 3 A

d.) curent through 6 ohm resistor = 3A/3 = 1 A

e.) current through 3 ohm resistor = 3A/1.5 = 2A

f.) both will be same = I^2 * R = 3^2 * 4 = 36 W

voltage drop by 6 ohm resistor = I*R = 1*6 6 V

voltage drop by 3 ohm resistor = IR = 2*3 = 6 V

Volatge drop by 2 ohm resitor = IR = 3*2 = 6V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.