In the figure below, find the following. (Let &₁ = 20.0 V, E₂ = 10.0 V, and R = 8.0 M.) &₁ = 13: P2 II P3 || E₂ (a) the current in each resistor I1 A 12 A A = + || + www R 28.0 Ω (b) the power delivered to each resistor P₁ W W W 12.0 Ω www 1₁ 1₂ 13
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- Consider the circuit shown in the figure, where 1 = 23.7 V,2 = 13.3 V and R = 12.0 Ω. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)(a)What is the current (in A) in each resistor? I1= A, I2= A, I3= A. (b)What is the power (in W) delivered to each resistor? P1= W, P2= W, P3= W.(c)What is the total power (in W) supplied by both batteries? (d)What is the relationship between the total power supplied by the batteries and the power delivered to each resistor? The total power supplied by the batteries is equal to the sum of the power delivered to each resistor .The total power supplied by the batteries is greater than the sum of the power delivered to each resistor. The total power supplied by the batteries is less than the sum of the power delivered to each resistor.For the circuit shown below left, find the magnitude and the direction of the current through each resistor and the power supplied by each battery, using the following values: R1 = 4.00 Ω, R2 = 6.00 Ω, R3 = 8.00 Ω, R4 = 6.00 Ω, R5 = 5.00 Ω, R6 = 10.0 Ω, R7 = 3.00 Ω, emf1 = 6.00 V, and emf2 = 12.0 V.A wire of resistance 5.0 ohm is connected to a battery whose emf is 2.0 V and whose internal resistance is 1.0. In 2.0 min, how much energy is (a) transferred from chemical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?
- A 10-gauge copper wire (diameter = 2.588 mm, ρ = 1.7 × 10–8 Ω · m) has a total resistance of 0.32 Ω. How long is the wire? A wire 10 m long with a diameter of 1.0 mm has a resistance of 5.0 Ω. What is the resistance of a second wire made of the same material, but 3.0 m long with a diameter of 4.0 mm? Please provide formula used and steps in orderA resistor with resistance R is connected to a battery that has emf 13.0 Vand internal resistance r= 0.390 Ω. Part A For what two values of R will the power dissipated in the resistor be 78.0 W? Enter your answers separated by a comma.For the circuit shown in the figure shown below, find the magnitude and the direction of the current through each resistor and the power supplied by each battery, using the following values: R1 = 4 ?, R2 = 6 ?, R3 = 8 ?, R4 = 6 ?, R5 = 5 ?, R6 = 10 ?, R7 = 3 ?, Vemf,1 = 6 V, and V emf,2 = 12 V.
- In the circuit shown, V1= 1.50 V, V2= 2.50 V, R1= 4.00 ohms and R2= 5.00 ohms. What is the current flowing through resistor R1?A gold wire has a length of 23.0 m and a resistance of 4.70 Ω at 20.0°C. Assuming a circular cross-section, what is the wire's diameter (in mm)? The resistivity of gold at 20.0°C is 2.44 ✕ 10−8 Ω · mFor the circuit shown below, find the magnitude and the direction of the current through each resistor and the power supplied by each battery, using the following values: R1 = 4.00 Ω, R2 = 6.00 Ω, R3 = 8.00 Ω, R4 = 6.00 Ω, R5 = 5.00 Ω, R6 = 10.0 Ω, R7 = 3.00 Ω, emf1 = 6.00 V, and emf2 = 12.0 V.
- A gold wire carries 22 A of current with a drift velocity of 4.3 × 10-3 m/s. Assume there is one free electron per gold atom, and the density of gold is 1.93 × 104 kg/m3. The molecular weight of gold is 196.97 g/mol. Part (b) For the problem above, calculate the resistance of the wire under those conditions. Assume the wire is 3.7 m in length, and the resistivity of gold is 2.44 × 10-8 Ω⋅m. R = ΩUse the equation: R= (ρ*L)/A to calculate the resistance of a thin tube filled with saline solution that is 4 cm in length and has a cross-sectional area of 1 x 10^-3 m^2 ? Assume the resistivity of saline solution to be 0.1 Ω m. a.) 4Ω b.) 40Ω c.) 0.25Ω d.) 0.4ΩConsider an aluminum wire of diameter 0.600 mm and length 15.0 m. The resistivity of aluminum at 20.0°C is 2.82 × 10−8 Ω · m. (a) Find the resistance of this wire at 20.0°C. (b) If a 9.00 V battery is connected across the ends of the wire, find the current in the wire.