P29_045

# P29_045 - . 62 mA when the total current in the resistor...

This preview shows page 1. Sign up to view the full content.

45. (a) The current in the galvanometer should be 1 . 62 mA when the potential diference across the resistor- galvanometer combination is 1 . 00 V. The potential diference across the galvanometer alone is iR g =(1 . 62 × 10 3 A)(75 . 3Ω) = 0 . 122 V, so the resistor must be in series with the galvanometer and the potential diference across it must be 1 . 00 V 0 . 122 V = 0 . 878V. The resistance should be R =(0 . 878 V) / (1 . 62 × 10 3 A) = 542 Ω. (b) The current in the galvanometer should be 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 62 mA when the total current in the resistor and galvanometer combination is 50 . 0 mA. The resistor should be in parallel with the galvanometer, and the current through it should be 50 . 0 mA 1 . 62 mA = 48 . 38 mA . The potential diference across the resistor is the same as that across the galvanometer, 0 . 122 V, so the resistance should be R = (0 . 122 V) / (48 . 38 10 3 A) = 2 . 52 ....
View Full Document

## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online