4. (a) In the figure the current in resistance 6 is i6 = 1.45 A and the resistances are R1 = R2 = R3 = 1.90 N, R4 = 14.3 N, R5 = 8.79 N, and R6 = 4.55 N. What is the emf of the ideal battery? 5. R R2 R Rg R4 (b) In A certain parallel-plate capacitor is filled with a dielectric for whichK = 5.80. The area of each plate is 664 cm?, and the plates are separated by 1.97 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 247 kN/C. What is the maximum energy that can be stored in the capacitor?

4. (a) In the figure the current in resistance 6 is i6 = 1.45 A and the resistances are R1 = R2 = R3 = 1.90 N, R4 = 14.3 N, R5 = 8.79 N, and R6 = 4.55 N. What is the emf of the ideal battery? 5. R R2 R Rg R4 (b) In A certain parallel-plate capacitor is filled with a dielectric for whichK = 5.80. The area of each plate is 664 cm?, and the plates are separated by 1.97 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 247 kN/C. What is the maximum energy that can be stored in the capacitor?

College Physics

10th Edition

ISBN:9781285737027

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter18: Direct-Current Circuits

Section: Chapter Questions

Problem 18P: (a) Find the current in each resistor of Figure P18.18 by using the rules for resistors in series...

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(a) Find the current in each resistor of Figure P18.18 by using the rules for resistors in series and parallel. (b) Write three independent equations for the three currents using Kirchhoffs laws: one with the node rule; a second using the loop rule through the battery, the 6.0- resistor, and the 24.0- resistor; and the third using the loop rule through the 12.0- and 24.0- resistors. Solve to check the answers found in part (a). Figure Pl8.18
(a) Find the current in each resistor of Figure P18.18 by using the rules for resistors in series and parallel. (b) Write three independent equations for the three currents using Kirchhoffs laws: one with the node rule; a second using the loop rule through the battery, the 6.0- resistor, and the 24.0- resistor; and the third using the loop rule through the 12.0- and 24.0- resistors. Solve to check the answers found in part (a). Figure Pl8.18
When the switch is open in Figure 18.8, power Po is delivered to the resistor R1. When the switch is closed, which of the following is true about the power Pc delivered to R1? (Neglect the internal resistance of the battery.) (a) Pc Po (b) Pc = Po (c) Pc Po Figure 18.8 (Quick Quizzes 18.5 and 18.6)
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.
An electric eel generates electric currents through its highly specialized Hunters organ, in which thousands of disk-shaped cells called electrocytes are lined up in series, very much in the same way batteries are lined up inside a flashlight. When activated, each electrocyte can maintain a potential difference of about 150 mV at a current of 1.0 A for about 2.0 ms. Suppose a grown electric eel has 4.0 103 electrocytes and can deliver up to 3.00 102 shocks in rapid series over about 1.0 s. (a) What maximum electrical power can an electric eel generate? (b) Approximately how much energy does it release in one shock? (c) How high would a mass of 1.0 kg have to be lifted so that its gravitational potential energy equals the energy released in 3.00 102 such shocks?
Integrated Concepts (a) What energy is dissipated by a lightning bolt having a 20,000-A current, a voltage of 1.00102 MV, and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18.0°C to its boiling point and then evaporated by this energy, assuming sap has the same thermal characteristics as water?
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When an unknown resistance Rxis placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting to be 2500 . What is Rxif R2R1 = 0.625?
Power P0 = I0 V0 is delivered to a resistor of resistance R0. If the resistance is doubled (Rnew = 2R0) while the voltage is adjusted such that the current is constant, what are the ratios (a) Pnew/P0 and (b) Vnew/V0? If, instead, the resistance is held constant while Pnew = 2P0, what are the ratios (c) Vnew/V0, and (d) Inew/I0?
Power P0 = I0 V0 is delivered to a resistor of resistance R0. If the resistance is doubled (Rnew = 2R0) while the voltage is adjusted such that the current is constant, what are the ratios (a) Pnew/P0 and (b) Vnew/V0? If, instead, the resistance is held constant while Pnew = 2P0, what are the ratios (c) Vnew/V0, and (d) Inew/I0?
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