answer question determine the value and direction of current in each branch of the network of fig.2.40,and the power dissipated by the 4 ohm load resistor.
Answers
Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then R1 in Figure 1(a) could be the resistance of the screwdriver’s shaft, R2 the resistance of its handle, R3 the person’s body resistance, and R4 the resistance of her shoes. Figure 2 shows resistors in series connected to
Explanation:
Suppose the voltage output of the battery in Figure 2 is 12.0 V, and the resistances are R1 = 1.00 Ω, R2 = 6.00 Ω, and R3 = 13.0 Ω. (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.
Strategy and Solution for (a)
The total resistance is simply the sum of the individual resistances, as given by this equation:
Rs=R1+R2+R3=1.00 Ω+6.00 Ω+13.0 Ω=20.0 ΩRs=R1+R2+R3=1.00 Ω+6.00 Ω+13.0 Ω=20.0 Ω.
Strategy and Solution for (b)
The current is found using Ohm’s law, V = IR. Entering the value of the applied voltage and the total resistance yields the current for the circuit:
I=VRs=12.0 V20.0 Ω=0.60 AI=VRs=12.0 V20.0 Ω=0.60 A.
Strategy and Solution for (c)
The voltage—or IR drop—in a resistor is given by Ohm’s law. Entering the current and the value of the first resistance yields
V1 = IR1 = (0.600A)(1.0 Ω) = 0.600 V.
Similarly,
V2 = IR2 = (0.600A)(6.0 Ω) = 3.60 V
and
V3 = IR3 = (0.600A)(13.0 Ω) = 7.80 V.
Discussion for (c)
The three IR drops add to 12.0 V, as predicted:
V1 + V2 + V3 = (0.600 + 3.60 + 7.80)V = 12.0 V.
Strategy and Solution for (d)
The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit is to use Joule’s law, P = IV, where P is electric power. In this case, each resistor has the same full current flowing through it. By substituting Ohm’s law V = IR into Joule’s law, we get the power dissipated by the first resistor as
P1 = I2R1 = (0.600 A)2(1.00 Ω) = 0.360 W.
Similarly,
P2 = I2R2 = (0.600 A)2(6.00 Ω) = 2.16 W.
and
P3 = I2R3 = (0.600 A)2(13.0 Ω) = 4.68 W.
Discussion for (d)
Power can also be calculated using either P = IV or