Suppose the resistors in Sample Problem No. 3 are connected in parallel. Draw the circuit diagram and find the (a) equivalent resistance of the combined resistors, (b) current flowing through each resistor, (c) voltage across each resistor, and (d) the power dissipated by each resistor. 3. Three resistors with values of 60.0 A, 30.0 A. 60.0 and 20.0 A, respectively, are connected in series to a 110.0 V battery of negligible internal resistance. Draw a circuit diagram and find the (a) equivalent resistance of the combined resistors, (b) current flowing through each resistor, (c) voltage drop across each resistor, and (d) the power dissipated by each resistor. 30.01 20.01 R, R R, 110.0 V Given: R, = 60.0 fN: R2 = 30.0 N: R, = 20.0 fL: V = 110.0 V

Physics for Scientists and Engineers: Foundations and Connections
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Chapter29: Direct Current (dc) Circuits
Section: Chapter Questions
Problem 64PQ: Ralph has three resistors, R1, R2, and R3, connected in series. When connected to an ideal emf...
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Suppose the resistors in Sample Problem No. 3 are connected in parallel. Draw the circuit diagram and find
the (a) equivalent resistance of the combined resistors, (b) current flowing through each resistor, (c) voltage
across each resistor, and (d) the power dissipated by each resistor.
3. Three resistors with values of 60.0 A, 30.0 A.
20.01
and 20.0 A, respectively, are connected in
series to a 110.0 V battery of negligible
internal resistance. Draw a circuit diagram
and find the (a) equivalent resistance of
the combined resistors, (b) current flowing
through each resistor, (c) voltage drop
across each resistor, and (d) the power
dissipated by each resistor.
60.02
30.0?
R,
R
R,
110.0V
Given:
R = 60.0 2: R2 = 30.0 N: R, = 20.0 £L' V = 110.0 V
Solution and Answer:
(a) Rtotat = R1 + R2 +R3 = 60.0 A + 30.0 A + 20.0 A = 110.0N
(b) I =
Because resistors are in series, I = 1, = 12 = l½ = 1.0 A
(c) V, = 1,R = (1.0 A)(60.0 N) = 60.0 V
V2 = l½R2 = (1.0 A)(30.0 n) = 30.0 V
V = hR3 = (1.0 A)(20.0 n) = 20.0 v
(d) P, = V, = (60.0 V)(1.0 A) = 60.0 W
P2 = Val2 = (30.0 V)(1.0 A) = 30.0 W
P = Vala = (20.0 V)(1.0 A) = 20.0 W
110.0 V
= 1.0 A
Reotal
110.0A
Transcribed Image Text:Suppose the resistors in Sample Problem No. 3 are connected in parallel. Draw the circuit diagram and find the (a) equivalent resistance of the combined resistors, (b) current flowing through each resistor, (c) voltage across each resistor, and (d) the power dissipated by each resistor. 3. Three resistors with values of 60.0 A, 30.0 A. 20.01 and 20.0 A, respectively, are connected in series to a 110.0 V battery of negligible internal resistance. Draw a circuit diagram and find the (a) equivalent resistance of the combined resistors, (b) current flowing through each resistor, (c) voltage drop across each resistor, and (d) the power dissipated by each resistor. 60.02 30.0? R, R R, 110.0V Given: R = 60.0 2: R2 = 30.0 N: R, = 20.0 £L' V = 110.0 V Solution and Answer: (a) Rtotat = R1 + R2 +R3 = 60.0 A + 30.0 A + 20.0 A = 110.0N (b) I = Because resistors are in series, I = 1, = 12 = l½ = 1.0 A (c) V, = 1,R = (1.0 A)(60.0 N) = 60.0 V V2 = l½R2 = (1.0 A)(30.0 n) = 30.0 V V = hR3 = (1.0 A)(20.0 n) = 20.0 v (d) P, = V, = (60.0 V)(1.0 A) = 60.0 W P2 = Val2 = (30.0 V)(1.0 A) = 30.0 W P = Vala = (20.0 V)(1.0 A) = 20.0 W 110.0 V = 1.0 A Reotal 110.0A
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