Fall 2011 Homework 9 Solutions

Fall 2011 Homework 9 Solutions - Physics 21 Fall, 2011...

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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-9 26-27 In the circuit shown in the figure the batteries have negligible internal resistance and the meters are both ideal- ized. With the switch S open, the voltmeter reads 15 V. (a) Find the emf E of the battery. (b) What will the ammeter read when the switch is closed? (a) For this part we can ignore the battery with the switch, since the switch is open and there will be no current through it. So let’s draw the circuit that we are concerned with, showing also the currents in each branch and the loops we will use to write Kirchhoff’s equations: We can find I 3 since we were given the measured voltage across the 50 Ω resistor. Using Ohm’s Law, V = IR = ⇒ I 3 = V R = 15 V 50 Ω = 0 . 30 A . Now, let’s write out the loop and node equations: Node : I 1 = I 2 + I 3 Left loop : E - (20 Ω) I 1- (75 Ω) I 2 = 0 Right loop : E - (20 Ω) I 1- (30 Ω) I 3- (50 Ω) I 3 = 0 Since we already know I 3 , there are three unknowns in this system, E , I 1 , and I 2 . Here are the equations after substi- tuting for I 3 and combining some terms: Node : I 1 = I 2 + 0 . 3 Left loop : E - 20 I 1- 75 I 2 = 0 Right loop : E - 20 I 1- 24 = 0 Now we have three equations for three unknowns. Next we simplify the equations and solve for E . The solution is E = 36 . 4 V , I 1 = 0 . 62 A , I 2 = 0 . 32 A . (b) The circuit is different for this part and is shown below. Conveniently, to find the new current in the ammeter, we only need to consider the one loop shown....
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This note was uploaded on 12/18/2011 for the course PHYS 021 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Fall 2011 Homework 9 Solutions - Physics 21 Fall, 2011...

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