(a)
The diagram of the circuit.
(a)
Answer to Problem 83P
The diagram is shown in figure (1).
Explanation of Solution
Calculation:
The diagram of the electric circuit is shown below,
Figure (1)
Conclusion:
Therefore, the diagram is shown in figure (1).
(b)
The current in each branch of the circuit.
(b)
Answer to Problem 83P
The current
Explanation of Solution
Formula used:
The expression for Kirchhoff’s law in first loop is given by,
The expression for Kirchhoff’s law in second loop is given by,
The expression for current in loop 1 is given by,
Calculation:
The expression for Kirchhoff’s law in first loop is calculated as,
The expression for Kirchhoff’s law in second loop is calculated as,
From equation (1) and (2),
Substitute
The
Conclusion:
Therefore, the current
(c)
The power supplied by second battery.
(c)
Answer to Problem 83P
The power supplied by second battery is
Explanation of Solution
Formula used:
The expression for power received by first battery is given by,
The expression for power received by second battery is given by,
The expression for power dissipated in load resistance is given by,
Calculation:
The expression for power received by first battery is calculated as,
The expression for power received by second battery is calculated as,
The power dissipated in load resistance is calculated as,
Conclusion:
Therefore, the power supplied by second battery is
Want to see more full solutions like this?
Chapter 25 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- A circuit loop containing an external resistor of value R = 25.0 (ohms) is connected to a real battery with an internal emf = 20.0 V that has an internal resistance of R = 1.50 (ohms). How much power is generated by the power supply in Watts?arrow_forwardA charged capacitor of C=45.0 μF is connected to a resistor of R=2.8 MΩ as shown in the figure. The switch S is closed at time t=0. Find the time (in seconds) it takes the current to fall to 0.25 of its initial value.arrow_forwardIn the figure R1 = 9.81 kΩ, R2 = 15.4 kΩ, C = 0.433 μF, and the ideal battery has emf ε = 18.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 3.60 ms?arrow_forward
- Consider the circuit shown in the figure below. Here, R1 = 2.00 Ω, R2 = 10.00 Ω, R3 = 5.00 Ω, R4 = 4.00 Ω, R5 = 3.00 Ω, and emf = 8.00 V. (a) Calculate the equivalent resistance of the R2 = 10.00 Ω and R3 = 5.00 Ω, resistors connected in parallel. (b) Using the result of part (a), calculate the combined resistance of the R2 = 10.00 Ω , R3 = 5.00 Ω , and R4 = 4.00 Ω resistors. (c) Calculate the equivalent resistance of the combined resistance found in part (b) and the parallel R5 = 3.00 Ω resistor.arrow_forwardIn the figure, suppose the switch has been closed for a length of time sufficiently long for the capacitor to become fully charged. For this circuit, R1 = 12.0 kΩ, R2 = 15.0 kΩ, R3 = 3.000 kΩ, C = 10.0 μF , and emf = 9.00 V. Find (d) the potential differance across R2. (e) the charge on the capacitor.arrow_forwardGiven a double-loop circuit above, with emfs B1 = 49.09V, B2 = 31.31V, and resistors R1 = 49.41Ω, R2 = 18.36Ω and R3 = 53.45Ω, The values of currents I1, I2 and I3 are ?arrow_forward
- A battery applies an EMF of 30 V and has an internal resistance of 1.6 Ohms. The battery is connected to three 9.7 Ohm resistors in parallel with eachother. What is the voltage drop across one of the resistors? Answer in volts. Question 7 options: A) 6.21E+00 B) 9.48E+00 C) 2.58E+01 D) 3.00E+01 E) 2.01E+01 F) 9.77E-01 G) 9.28E+00 H) None of these answers. Two straight wires of length 2.75 m, carrying identical 21 A currents lie parallel to eachother on a table top and carry antiparallel currents. The wires are separated by 25 cm. What is the magnitude of the force between the wires? Answer in Newtons. Question 8 options: A) 9.70E-04 B) 4.62E-05 C) 9.70E-06 D) 0.00E+00 E) 3.53E-04 F) 3.53E+03 G) 6.10E-03 H) None of these answers.arrow_forwardA heart defibrillator being used on a patient has an RC time constant of 11 ms due to the resistance of the patient's body and the capacitance of the defibrillator. t = 11 ms C = 8.5 uF V = 14 kV (b) If the initial voltage is 14 kV, how long does it take to decline to 6.00 x 10^2 V in ms?arrow_forwardTwo resistors, R1 = 2.00 kΩ and R2 = 3.00 kΩ, are connected in parallel and their combination is connected in series to a fully charged, 150-µF capacitor. When the switch is opened, the capacitor begins to discharged. What is the time constant for the discharge? Choices: A) 16 s B) 0.17 s C) 0.18 s D) 0.19 sarrow_forward
- An electronic flashgun has a capacitor with capacitance CC that is charged to a voltage VV. Answer the questions below: Given that C=C= 1441 μFμF and V=V= 317 VV. When the photographer takes a picture, the flash fires for t=t= 0.0032 ss. Find the power delivered by the flash tube. After the picture is taken, the capacitor has to be recharged by a power supply that delivers a maximum current of I=I= 27 mAmA. How long will it take to charge the capacitor?arrow_forwardCalculate the equivalent resistance of four resistors R1=100 ohms, R2=250 ohms, R3=350 ohms and R4=200 ohms such that R1 and R2 in parallel will be connected in series with R3 and R4 in parallel.arrow_forwardYou are to connect resistors R1 and R2, with R1 > R2, to a battery, first individually, then in series, and then in parallel. Rank those arrangements according to the amount of current through the battery, greatest first.arrow_forward