ssm_ch27 - Chapter 27 7 (a) Let i be the current in the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 27 7 (a) Let i be the current in the circuit and take it to be positive if it is to the left in R 1 . Use Kirchhoffs loop rule: E 1 iR 2 iR 1 E 2 = 0. Solve for i : i = E 1 E 2 R 1 + R 2 = 12 V 6 : 0 V 4 : + 8 : = 0 : 50 A : A positive value was obtained, so the current is counterclockwise around the circuit. (b) and (c) If i is the current in a resistor with resistance R , then the power dissipated by that resistor is given by P = i 2 R . For R 1 the power dissipated is P 1 = (0 : 50 A) 2 (4 : ) = 1 : 0 W and for R 2 the power dissipated is P 2 = (0 : 50 A) 2 (8 : ) = 2 : 0 W : (d) and (e) If i is the current in a battery with emf E , then the battery supplies energy at the rate P = i E provided the current and emf are in the same direction. The battery absorbs energy at the rate P = i E if the current and emf are in opposite directions. For battery 1 the power is P 1 = (0 : 50 A)(12 V) = 6 : 0 W and for battery 2 it is P 2 = (0 : 50 A)(6 : 0 V) = 3 : 0 W : (f) and (g) In battery 1, the current is in the same direction as the emf so this battery supplies energy to the circuit. The battery is discharging. The current in battery 2 is opposite the direction of the emf, so this battery absorbs energy from the circuit. It is charging. 13 (a) If i is the current and V is the potential difference, then the power absorbed is given by P = i V . Thus V = P i = 50 W 1 : 0 A = 50 V : Since energy is absorbed, point A is at a higher potential than point B; that is, V A V B = 50 V. (b) The end-to-end potential difference is given by V A V B = + iR + E , where E is the emf of element C and is taken to be positive if it is to the left in the diagram. Thus E = V A V B iR = 50 V (1 : 0 A)(2 : ) = 48 V. 164 Chapter 27 (c) A positive value was obtained for E , so it is toward the left. The negative terminal is at B. 21 (a) and (b) The circuit is shown in the diagram to the right. The current is taken to be positive if it is clockwise. The potential difference across battery 1 is given by V 1 = E ir 1 and for this to be zero, the current must be i = E =r 1 . Kirchhoffs loop rule gives 2 E ir 1 ir 2 iR = 0. Substitute i = E =r 1 and solve for R . You should get R = r 1 r 2 = 0 : 016 : 012 = 0 : 004 . Now assume that the potential difference across bat- tery 2 is zero and carry out the same analysis. You should find R = r 2 r 1 . Since r 1 > r 2 and R must E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . r 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....
View Full Document

Page1 / 8

ssm_ch27 - Chapter 27 7 (a) Let i be the current in the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online