webct14

# Webct14 - 2 resistors 1 Two resistors R1 = 12 and R2 = 18 can be connected to an ideal battery with emf = 100 V in two different ways in series or

This preview shows pages 1–5. Sign up to view the full content.

2 resistors 1. Two resistors R 1 = 12 and R 2 = 18 can be connected to an ideal battery with emf ε = 100 V in two different ways: in series or in parallel. Find the total power dissipated in the system when the two resistors are connected in series. When the resistors are connected in series , the equivalent reistance of the circuit is: R = R1 + R2 R = 12 + 18 R = 30 ΩΩ The voltage across this equivalent resistor is ε = 100 V. Thus, the power dissipated is: 22 (100 V) 333 W 30 V P R == = 2. Find the total power dissipated in the system when the two resistors are connected in parallel. The equivalent resistane of the circuit when the resistors are in parallel is: 1 2 12 1 2 111 (12 )(18 ) 7.2 30 RR R R R R R =+ = + The voltage across this equivalent resistor (and across R 1 and R 2 too, since they are in parallel) is ε = 100 V. Thus, the power dissipated is: (100 V) 1390 W 7.2 V P R =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. Now imagine that the battery is not ideal. Its internal resistance is r = 5.0 . Use this non-ideal battery for the next two questions. What is in this case the power dissipated in the circuit (outside the battery) when the resistors are in series? For a non-ideal battery there will be an internal resistance which will be in series with battery, and also in series with the external circuit (which had an equivalent resistance of 30 Ohms, see question 1). This time the voltage across the 30 Ohm resistor is less than 100 V, because there is a potential drop inside the battery, due to its internal resistance. The current in the system is ext 100 V 2.86 A 35 V I Rr == = +Ω Thus, the power dissipated in the circuit outside the battery (so excluding the internal resistance of the battery) is ( )( ) 2 2.86 A 30 245 W PIR = 4. And what is in this case the power dissipated in the circuit (outside the battery) when the resistors are in parallel? In the case when the resistors in parallel, we have
This time the voltage across the 7.2 Ohm resistor is less than 100 V, because there is a potential drop inside the battery, due to its internal resistance. The current in the system is ext 100 V 8.2 A 12.2 V I Rr === +Ω Thus the power dissipated in the equivalent resistance outside the battery (so excluding the internal resistance of the battery) is, 2 (8.196 )(7.2 ) 484 0.48 pi pi pi pi PI R A P W P Pk W = = = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
circuit 1. Four resistors and an ideal battery have been assembled in the circuit below. The value of R 3 is unknown (but greater than zero and finite). R 1 = 10 , R 2 = 5 , R 4 = 5 . Compare the magnitude of the voltage across R 1 with the magnitude of the voltage across R 2 . a. | V 1 | < | V 2 | b. | V 1 | = | V 2 | *c. | V 1 | > | V 2 | R 1 > R 2 .
This is the end of the preview. Sign up to access the rest of the document.

## This homework help was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Spring '08 term at Iowa State.

### Page1 / 15

Webct14 - 2 resistors 1 Two resistors R1 = 12 and R2 = 18 can be connected to an ideal battery with emf = 100 V in two different ways in series or

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online