Answered: Determine the current through the 50…

Determine the current through the 50 load resistor in the circuit shown below. 1.0811 A 1.0769 A 1.0865 A 1.0799 A Battery 1 r1 1 Ohm V1 olo 믐 12 V Battery 2 r2 0.5 Ohm V2 -=-|0|0 5 V R3 5 Ohms

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter18: Direct-Current Circuits

Section: Chapter Questions

Problem 5CQ: Electric current I enters a node with three resistors connected in parallel (Fig. CQ18.5). Which one...

See similar textbooks

Similar questions

Consider the circuit shown in Figure P28.21 on page 860. (a) Find the voltage across the 3.00-0 resistor, (b) Find the current in the 3.00-12 resistor.
The power dissipated in a resistor is given by P = V2/R, which means power decreases if resistance increases. Yet this power is also given by P = I2R, which means power increases if resistance increases. Explain why there is no contradiction here.
A child's electronic toy is supplied by three 1.58-V alkaline cells having internal resistances of 0.0200 inseries with a 1.53-V carbon-zinc dry cell having a 0.100- internal resistance. The load resistance is 10.0 . (a) Draw a circuit diagram of the toy and itsbatteries, (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load?
A child’s electronic toy is supplied by three 1.58-V alkaline cells having internal resistances of 0.0200 in series with a 1.53-V carbon-zinc dry cell having a 0.100- internal resistance. The load resistance is 10.0 . (a) Draw a circuit diagram of the toy and its batteries, (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load?
Currents of approximately 0.06 A can be potentially fatal. Currents in that range can make the heart fibrillate (beat in an uncontrolled manner). The resistance of a dry human body can be approximately 100 k . (a) What voltage can cause 0.06 A through a dry human body? (b) When a human body is wet, the resistance can fall to 100 . What voltage can cause harm to a wet body?
A 240-kV power transmission line carrying 5.00x102. A is hung from grounded metal towers by ceramic insulators, each having a 1.00x 109- resistance. Figure 21.51. (a) What is the resistance to ground of 100 of these insulators? (b) Calculate the power dissipated by 100 of them, (c) What fraction of the power carried by the line is this? Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors.
A person with body resistance between his hands of 1.00 k accidentally grasps the terminals of a 20.0-kV power supply. (Do NOT do this!) (a) Draw a circuit diagram to represent the situation. (b) If the internal resistance of the power supply is 2000 , what is the current through his body? (c) What is the power dissipated in his body? (d) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in this situation to be 1.00 mA or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-re si stance devices? Explain your reasoning,
Integrated Concepts A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery’s internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in °C/min ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of 0.300 kcal/kg. °C, assuming no heat escapes?
When a straight wire is heated, its resistance changes according to the equation R = R0 [1 + (T T0)] (Eq. 17.7), where is the temperature coefficient of resistivity. (a) Show that a more precise result, which includes the length and area of a wire change when it is heated, is R=R0[1+(TT0)][1+(TT0)][1+2(TT0)] where is the coefficient of linear expansion. (See Topic 10.) (b) Compare the two results for a 2.00-m-long copper wire of radius 0.100 mm, starting at 20.0C and heated to 100.0C.
The label on a battery-powered radio recommends the use of a rechargeable nickel-cadmium cell (nicads), although it has a 1.25-V emf, whereas an alkaline cell has a 1.58-V emf. The radio has a 3.20 resistance.(a) Draw a circuit diagram of the radio and its battery. Now, calculate the power delivered to the radio (b) when using a nicad cells, each having an internal resistance of 0.0400 , and (c) when using an alkaline cell, having an internal resistance of 0.200 . (d) Does this differenceseem significant, considering that the radio's effective resistance is lowered when its volume is aimed up?
An automobile starter motor has an equivalent resistance of 0.0500 and is supplied by a 12.0-V battery with a 0.0100- internal resistance, (a) What is thecurrent to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add 0.0900 to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)
Three 60.0-W, 120-V lightbulbs are connected across a 120-V power source, as shown in Figure P18.50. Find (a) the total power delivered to the three bulbs and (b) the potential difference across each. Assume the resistance of each bulb is constant (even though, in reality, the resistance increases markedly with current). Figure P18.50