Physics Homework Problems Test #3

How many 12Ω resistors must be connected in series to give an equivalent resistance to five 600Ω resistors connected in parallel?

5 600Ω resistors in parallel: 1/R_eq = 1/600 + 1/600 + 1/600 + 1/600 + 1/600 R_eq = 120Ω For resistors in series: R_eq = sum of resistors 120Ω = x(12Ω) 10 = x

A 120-V hair dryer has two settings: 750W and 1250W. a) At which setting do you expect the resistance to be higher? b) Determine the resistance at the lower setting. c) Determine the resistance at the higher setting.

a) According to equation P = V^2/R, Power and Resistance are inversely proportional. So 750W. b) R = V^2/P c) Same as previous.

A hair dryer draws 9.7A when plugged into a 120V line. What is its resistance? How much charge passes through it in 17min?

a) R = V/I b) I = ∆Q/∆t ∆Q = I∆t

Three 2.30kΩ resistors can be connected together in four different ways, making combinations of series and/or parallel circuits. What is the net resistance of resistors connected in series? What is the net resistance of resistors connected in parallel? What is the net resistance of two resistors in series connected in parallel with the third? What is the net resistance of two resistors in parallel connected in series with the third?

A) Add them. B) 1/R_eq = 1/R_1 + 1/R_2 and so on. C) Get equivalence of the two in series then calculate in equivalence of the new resistor plus other in parallel. D) Opposite of (C).

Three equal resistors (R) are connected to a battery as shown in the figure. Qualitatively, what happens to the voltage drop across each of these resistors, when the switch S is opened, after having been closed for a long time? What happens to the current flow through each resistor, when the switch S is opened, after having been closed for a long time? What happens to the terminal voltage of the battery, when the switch S is opened, after having been closed for a long time?

A) Left resistor decreases. Central resistor increases. Right resistor goes to zero. B) Same as above. C) It increases.

A 660Ω and a 2200Ω resistor are connected in series with a 12V battery. What is the voltage across the 2200Ω resistor?

Find total resistance (add). Multiply (emf/Total R)(R of lone resistor).

A network of five equal resistors R is connected to a battery E as shown in the figure. Determine the current I that flows out of the battery. Use the value determined for I to find the single resistor Req that is equivalent to the five-resistor network.

Junction Rules: I = I_1 + I_2 (1) I_2 = I_3 + I_4 (2) I_1 + I_4 = I_5 (3) Loop Rules: Top loop: -I_2R - I_4R + I_1R = 0 (4) Bottom loop: -I_2R - I_3R + emf = 0 (5) Small loop: -I_3R + I_5R + I_4R = 0 (6) Substitute Junction Rule equations into Loop Rules to solve for the various currents. B) Voltage over current.

Determine the magnitudes and directions of the currents in each resistor shown in the figure. The batteries have emfs of E1 = 7.4V and E2 = 11.7V and the resistors have values of R1 = 23Ω , R2 = 36Ω , and R3 = 38Ω . First, ignore internal resistance. Then pretend they have 1.5Ω resistance.

Junction Rules: I_2 = I_1 + I_3 (1) I_1 = I_2 - I_3 (2) Loop Rules: E_1 - I_1R_1 - I_2R_2 = 0 (a) E_2 - I_3R_3 - I_2R_2 = 0 (b) Substitute (2) into (a) and solve for I_2. Plug result into (b). Then use (1) to calculate I_1. (Add internal resistance the next time).

Is the larger or smaller line the positive battery terminal?

Larger. (Moves from larger to smaller)

Suppose two batteries, with unequal emfs of 2.00 V and 3.00 V, are connected as shown in the figure. If each internal resistance is 0.13Ω , and R = 4.00Ω, what is the voltage across the resistor R?

Loop Rules (inner loop and outer loop). 2V - I_2r - (I_1 + I_2)R = 0 3V - I_3r - (I_1 + I_2)R = 0 Solve for currents. V = IR

An extension cord made of two wires of diameter 0.129 cm (no. 16 copper wire) and of length 2.7m (9ft ) is connected to an electric heater which draws 14.0A on a 120-V line. How much power is dissipated in the cord?

Needed Equations: P = I^2R R = ρl/A A = .25πd^2 P = I^2(4ρl/πd^2) Plug in and solve.

An ordinary flashlight uses two D-cell 1.5-V batteries connected in series as in the figures. The bulb draws 410mA when turned on. a) Calculate the resistance of the bulb. b) Calculate the power dissipated. c) By what factor would the power increase if four D-cells in series were used with the same bulb? (Neglect heating effects of the filament.)

Needed Equations: V = IR P = IV a) R = V/I b) P = IV c) Adding two more d-cells, the voltage will increase to 6V total. Use this number in P = IV to receive power. Then set up a ratio (P_4/P_2) to receive the factor.

An ac voltage, whose peak value is 180 V, is across a 360−Ω resistor. What are the rms and peak currents in the resistor?

Needed Equations: V = IR ---> I = V/R I_rms = I/sqrt(2)

Determine the terminal voltage of each battery in the figure.

Needed Equations: V_term = E - Ir Top Battery: Use Kirchoff's Loop Rule to determine the current. Plug into V_term. Bottom Battery) Use equation V_term = E + Ir, changing the - to a + because their is a voltage gain across the internal resistance of the battery.

A hollow cylindrical resistor with inner radius r1 and outer radius r2, and length ℓ, is made of a material whose resistivity is ρ. Find the resistance for current that flows radially outward. [Hint: Divide the resistor into concentric cylindrical shells and integrate.] Evaluate the resistance R for such a resistor made of carbon whose inner and outer radii are 1.2mm and 1.7mm and whose length is 4.7cm . (Choose ρ=15×10−5Ω⋅m.)

Needed equations: A = 2πrl (area of thin cylindrical shell) R = ρl/A l is the direction of flow, so replace it with dr. R = ρdr/2πrl Integrate through the range of radii to receive a formula (shown in pic). a) Formula = R = (ρ/2π)ln(r_2/r_1) b) Plug numbers into (a).

A hollow cylindrical resistor with inner radius r1 and outer radius r2, and length ℓ, is made of a material whose resistivity is ρ. What is the resistance in part B for current flowing parallel to the axis? [Part B - Evaluate the resistance R for such a resistor made of carbon whose inner and outer radii are 1.2mm and 1.7mm and whose length is 4.7cm . (Choose ρ=15×10−5Ω⋅m.)]

Needed equations: A = π(r_2^2 - r_1^2) (Area of a hollow faced cylindrical shell). R = ρl/A R = ρl/π(r_2^2 - r_1^2) Plug in.

The heating element of an electric oven is designed to produce 3.8kW of heat when connected to a 240-V source. What must be the resistance of the element?

Needed equations: P = V^2/R (Relates voltage and power) Plug in and solve for R.

Compute the voltage drop along a 33m length of household no. 14 copper wire (used in 15A circuits). The wire has diameter 1.628 mm and carries a 14A current.

Needed equations: V = IR R = (ρl/A) Cross-sectional area of wire (A) = πr^2 d/2 = r Plug in for r. V = I(4ρl/πd^2). Plug in numbers and solve for V.

What is the maximum instantaneous power dissipated by a 5.0−hp pump connected to a 240−V_rms AC power source? What is the maximum current passing through the pump?

Needed equations: P_0 = V_0I_0 P_rms = V_rmsI_rms I = sqrt(2)I_rms Must convert horsepower to watts (746W = 1hp). a) In picture. b) P_rms = V_rmsI_rms Solve for I_rms. Plug that into I = sqrt(2)I_rms I = maximum current.

Three equal resistors (R) are connected to a battery as shown in the figure. If the emf of the battery is 8.0V , what is its terminal voltage when the switch is closed if the internal resistance r is 0.52Ω and R = 5.28Ω? What is the terminal voltage when the switch is open?

Needed equations: V_terminal = E - Ir A) I = E/R R for this circuit = (R/2) + R + r = (3/2)R + r Plug in. B) R for this circuit = 2R + r Same as above.

A bird stands on a dc electric transmission line carrying 2800A . The line has 2.6×10−5Ω resistance per meter, and the bird's feet are 3.7cm apart. What is the potential difference between the bird's feet?

R = (Resistance per meter)(Distance in meters) V = IR

A service station charges a battery using a current of 6.7A for 3.0h. How much charge passes through the battery?

Use definition of current. I = ∆Q/∆t. Rearrange to ∆Q = I∆t. Solve.

Determine the currents I1, I2, and I3 in the figure. Assume the internal resistance of each battery is 1.0Ω . What is the terminal voltage of the 6.0-V battery?

Use the junction and loops rules to solve. Substitute. b) V_term = E - Ir Answer: I1, I2, I3 = 0.51, 0.51, 3.0×10−3A b) V_t = 6.0V

To what value would you have to raise the temperature of a copper wire (originally at 20 ˚C) to increase its resistance by 20%?

ρ = ρ_0(1 + α∆t) Multiply by 1/A to change from resistivity to resistance. R = R_0(1 + α∆t) R = 1.2R_0 (Resistance of new R is equal to 20% more than original R) 1.2R_0 = R_0(1 + α∆t)