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Homework 6 solution

Lama K Banna
Lama K Banna

This document contains the solutions to homework problems in a General Physics for Medical Sciences course. It includes the solutions to 5 questions: 1) calculating equivalent resistances in a circuit, 2) finding currents in resistors, 3) analyzing a circuit diagram with resistors in series and parallel, 4) calculating charge, current, and time constants in a discharging capacitor circuit, and 5) identifying voltage, current, and time constant expressions and values for an RC circuit. The instructor is Dr. Hassan Ashour and the homework is due on January 10, 2010.

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General Physics for Medical Sciences Homework 6 Instructor: Dr. Hassan Ashour Due Date: Jan 10, 2010 Solve the following questions Q1: find the equivalent resistance Solution: The equivalent resistance on your right hand is, 5 + 3 = 8立, 8//12 = 4.8立 4.8 + 5.2 = 10立 The equivalent resistance on your left hand is 45 + 15 = 60立 60 // 40 = 24立 24 + 6 = 30立 Now, unfold the resistance 1.5立and 1立, the resistance would appear as follow,
Thus, = 30//10 + 1.5 + 1 = 10立 Q2: find the currents in 5立,10立, 20立 resistances. Solution: To find any current, we need to find the equivalent resistance, = . + + + + + = . = , = . . = . = . = . = . = . = .
Q3: question 5, on page 162, Consider the circuit shown in Figure below. Find (a) the current in the 20.0立 resistor and (b) the potential difference between points a and b. Solution: Before attaching this question! I want you to take a closer look how resistors are connected: I mean which of these connected in parallel and which are connected in series, hum. With me so far, I hope so. Look at the connecting points and , what about them, hum. The resistors 20立 5立 are in series, I think you saw that clearly. Redraw the circuit in you mind by replaces these resistors with their equivalent25立. Now what do you think, yah. The resistor25立, 50立 and 10立 are in parallel. SO there equivalent resistance is = + + = + + = . The 25.0 Volt source will distributed on 2.94立 and the10立, via the voltage divider law we have . = . . + = . So, the potential difference between a and b is 5.68 Volt, and the current in the 20立 can be found by finding the potential difference across it
= . + = . = . = . Or, we can find the branch current, that is = . = . This current passes through the 20立 5立 resistances. Q4: question 10, A 2.00-nF capacitor with an initial charge of 5.10 is discharged through a 1.30 立 resistor. (a) Calculate the current through the resistor 9.00 after the resistor is connected across the terminals of the capacitor. (b) What charge remains on the capacitor after 8.00 ? (c) What is the maximum current in the resistor? Solution: a. We know the discharging current is given by = = Thus, = = 5.10 2 1.3 立 = 1.962 (9 ) = 1.962 9 10 2 10 1.3 10 (9 ) = 62 b. The charge after 8 microseconds
( ) = exp (8 ) = 5.10 8 10 2 10 1.3 10 (8 ) = 0.235 c. The maximum current is = 1.962 Q5: question 12, 1. In circuit the voltage and current expressions are = 100exp(1000 ) , 0; = 5 exp(1000 ) , 0 Find , , and the time constant Solution: In this circuit = 5 = = = = 20 立 The time constant of this circuit is = = 10 = 10 20 立 = 50 Also we can find the initial charge on the capacitor's plates = = 50 100 = 5

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Homework 6 solution

  • 1. General Physics for Medical Sciences Homework 6 Instructor: Dr. Hassan Ashour Due Date: Jan 10, 2010 Solve the following questions Q1: find the equivalent resistance Solution: The equivalent resistance on your right hand is, 5 + 3 = 8立, 8//12 = 4.8立 4.8 + 5.2 = 10立 The equivalent resistance on your left hand is 45 + 15 = 60立 60 // 40 = 24立 24 + 6 = 30立 Now, unfold the resistance 1.5立and 1立, the resistance would appear as follow,
  • 2. Thus, = 30//10 + 1.5 + 1 = 10立 Q2: find the currents in 5立,10立, 20立 resistances. Solution: To find any current, we need to find the equivalent resistance, = . + + + + + = . = , = . . = . = . = . = . = . = .
  • 3. Q3: question 5, on page 162, Consider the circuit shown in Figure below. Find (a) the current in the 20.0立 resistor and (b) the potential difference between points a and b. Solution: Before attaching this question! I want you to take a closer look how resistors are connected: I mean which of these connected in parallel and which are connected in series, hum. With me so far, I hope so. Look at the connecting points and , what about them, hum. The resistors 20立 5立 are in series, I think you saw that clearly. Redraw the circuit in you mind by replaces these resistors with their equivalent25立. Now what do you think, yah. The resistor25立, 50立 and 10立 are in parallel. SO there equivalent resistance is = + + = + + = . The 25.0 Volt source will distributed on 2.94立 and the10立, via the voltage divider law we have . = . . + = . So, the potential difference between a and b is 5.68 Volt, and the current in the 20立 can be found by finding the potential difference across it
  • 4. = . + = . = . = . Or, we can find the branch current, that is = . = . This current passes through the 20立 5立 resistances. Q4: question 10, A 2.00-nF capacitor with an initial charge of 5.10 is discharged through a 1.30 立 resistor. (a) Calculate the current through the resistor 9.00 after the resistor is connected across the terminals of the capacitor. (b) What charge remains on the capacitor after 8.00 ? (c) What is the maximum current in the resistor? Solution: a. We know the discharging current is given by = = Thus, = = 5.10 2 1.3 立 = 1.962 (9 ) = 1.962 9 10 2 10 1.3 10 (9 ) = 62 b. The charge after 8 microseconds
  • 5. ( ) = exp (8 ) = 5.10 8 10 2 10 1.3 10 (8 ) = 0.235 c. The maximum current is = 1.962 Q5: question 12, 1. In circuit the voltage and current expressions are = 100exp(1000 ) , 0; = 5 exp(1000 ) , 0 Find , , and the time constant Solution: In this circuit = 5 = = = = 20 立 The time constant of this circuit is = = 10 = 10 20 立 = 50 Also we can find the initial charge on the capacitor's plates = = 50 100 = 5