Chapter 26 Problem 36

Problem

A circuit

(a) Determine the currents $I_{1}$,$I_{2}$ and $I_{3}$. Assume the internal resistance of each battery is $r=1\Omega$

(b) What is the terminal voltage of the 6.0-V battery?

solution

(a)

$I_{1}=I_{2}+I_{3}$

$12.0{\textrm {V}}-I_{2}(12\Omega )+12.0{\textrm {V}}-I_{1}(35\Omega )=0$ $\rightarrow 24=35I_{1}+12I_{2}$

$12.0{\textrm {V}}-I_{2}(12\Omega )-6.0{\textrm {V}}-I_{3}(34\Omega )=0$ $\rightarrow 6=12I_{2}-34I_{3}$

$24=35I_{1}+12I_{2}=35(I_{2}+I_{3})+12I_{2}=47I_{2}+35I_{3}$

Solve for $I_{3}$:

$6=12I_{2}-34I_{3}\rightarrow I_{2}={\frac {6+34I_{3}}{12}}$

$24=47{\frac {6+34I_{3}}{12}}+35I_{3}$

$I_{3}=2.97{\textrm {mA}}$

then the rest is straightforward

$I_{2}={\frac {6+34I_{3}}{12}}=0.508{\textrm {A}}$

$I_{1}=I_{2}+I_{3}=0.511{\textrm {A}}$

(b)

The terminal voltage at that battery is

$6.0{\textrm {V}}-I_{3}r=6.0{\textrm {V}}-(2.97{\textrm {mA}})(1.0\Omega )=5.997{\textrm {V}}$