Question

In Figure , the ideal batteries have emfs $E_1=20.0V$, $E_2=10.0V$, $E_3=5.00V$, and $E_4=5.00V$, and the resistances are each $2.00\Omega$. What are the (a) size and (b) direction (left or right) of current $i_1$ and the (c) size and (d) direction of current $i_2$? (This can be answered with only mental calculation.) (e) At what rate is energy being transferred in battery 4, and (f) is the energy being supplied or absorbed by the battery?

Answer 1

Reducing the bottom two series resistors to a single $R^{\prime }=4.00\Omega$ (with current $i_1$ through it), we see we can make a path (for use with the loop rule) that passes through $R$,

the ${\epsilon }_4=5.00V$ battery, the ${\epsilon }_1=20.0V$ battery, and the ${\epsilon }_3=5.00V.$ This leads to

$i_1=\frac{{\epsilon }_1+{\epsilon }_3+{\epsilon }_4}{R^{\prime }}=\frac{20.0V+5.00V+5.00V}{40.0\Omega }=\frac{30.0V}{4.0\Omega }=7.50A$

(b) The direction of $i_1$ is leftward.

(c) The voltage across the bottom series pair is $i_1{R}^{\prime }=30.0V$. This must be the same as the voltage across the two resistors directly above them, one of which has current $i_2$ through it and the other (by symmetry) has current $\frac{1}{2}{i}_2$ through it. Therefore,

$30.0V=i_2\left(2.00\Omega \right)+\frac{1}{2}{i}_2\left(2.00\Omega \right)$

which leads to $i_2=\left(30.0V\right)/\left(3.00\Omega \right)=10.0A.$

(d) The direction of $i_2$ is also leftward.

(e) We use $P_4=(i1+i2){\epsilon }_4=(7.50A+10.0A)\left(5.00V\right)=87.5W.$

(f) The energy is being supplied to the circuit since the current is in the "forward" direction through the battery.