English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

A 55.0- resistor is connected in parallel with a 115.0- resistor. This parallel group is connected in series with a 24.0- resistor. The total combination is connected across a 15.0-V battery. Find:
(a) the current in the 115.0- resistor
(b) the power dissipated in the 115.0- resistor.

2007-05-01 19:28:48 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

find the equivelant resistor Rt
= 55 // 115 - 24

=(55 * 115 ) / ( 55 + 115 ) + 24
= 37.205 + 24 = 61.205ohm
total current in circuit It =
= 15 / 61.205= 0.245 A

use current divider rule to find the current in 115

I = 0.245 * 55 / ( 55 + 115 ) = 0.0792 A

power dissipated in dc circuit P is ( I^2 * R )

power dissipated = 0.0792^2 * 115 = 0.721 watt

2007-05-02 08:11:10 · answer #1 · answered by ? 3 · 0 0

Start by reducing the parallel resistors to equivalent
R=115*55/(115+55)
=37.2 ohm
The current through the 24 resistor is I
using sum of voltages
15=I*(R+24)
I=15/(R+24)
=.245 A
The voltage across the 115 resistor is
15-I*24
or
9.12 V

The current then is
9.11/115
0.0792 Amp
and power is
V*I
=.722 W

j

2007-05-02 14:53:33 · answer #2 · answered by odu83 7 · 0 1

fedest.com, questions and answers